457454 physical chemistry quantum chemistry
TRANSCRIPT
Chemistry 351 and 352
Physical Chemistry I and II
Darin J. Ulness
Fall 2006 – 2007
Contents
I Basic Quantum Mechanics 15
1 Quantum Theory 161.1 The “Fall” of Classical Physics . . . . . . . . . . . . . . . . . . . . 16
1.2 Bohr’s Atomic Theory . . . . . . . . . . . . . . . . . . . . . . . . 17
1.2.1 First Attempts at the Structure of the Atom . . . . . . . . 17
2 The Postulates of Quantum Mechanics 222.1 Postulate I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 How to normalize a wavefunction . . . . . . . . . . . . . . . . . . 23
2.3 Postulates II and II . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 The Setup of a Quantum Mechanical Problem 273.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2 The Quantum Mechanical Problem . . . . . . . . . . . . . . . . . 27
3.3 The Average Value Theorem . . . . . . . . . . . . . . . . . . . . . 29
3.4 The Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . 30
4 Particle in a Box 314.1 The 1D Particle in a Box Problem . . . . . . . . . . . . . . . . . . 31
4.2 Implications of the Particle in a Box problem . . . . . . . . . . . 34
5 The Harmonic Oscillator 385.1 Interesting Aspects of the Quantum Harmonic Oscillator . . . . . 40
i
5.2 Spectroscopy (An Introduction) . . . . . . . . . . . . . . . . . . . 42
II Quantum Mechanics of Atoms and Molecules 45
6 Hydrogenic Systems 466.1 Hydrogenic systems . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.2 Discussion of the Wavefunctions . . . . . . . . . . . . . . . . . . . 49
6.3 Spin of the electron . . . . . . . . . . . . . . . . . . . . . . . . . . 51
6.4 Summary: the Complete Hydrogenic Wavefunction . . . . . . . . 52
7 Multi-electron atoms 557.1 Two Electron Atoms: Helium . . . . . . . . . . . . . . . . . . . . 55
7.2 The Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . 56
7.3 Many Electron Atoms . . . . . . . . . . . . . . . . . . . . . . . . 58
7.3.1 The Total Hamiltonian . . . . . . . . . . . . . . . . . . . . 59
8 Diatomic Molecules and the Born Oppenheimer Approximation 608.1 Molecular Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
8.1.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . 61
8.1.2 The Born—Oppenheimer Approximation . . . . . . . . . . 62
8.2 Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 63
8.2.1 The Morse Oscillator . . . . . . . . . . . . . . . . . . . . . 64
8.2.2 Vibrational Spectroscopy . . . . . . . . . . . . . . . . . . . 66
9 Molecular Orbital Theory and Symmetry 679.1 Molecular Orbital Theory . . . . . . . . . . . . . . . . . . . . . . 67
9.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
10 Molecular Orbital Diagrams 7210.1 LCAO–Linear Combinations of Atomic Orbitals . . . . . . . . . 72
10.1.1 Classification of Molecular Orbitals . . . . . . . . . . . . . 73
10.2 The Hydrogen Molecule . . . . . . . . . . . . . . . . . . . . . . . 74
10.3 Molecular Orbital Diagrams . . . . . . . . . . . . . . . . . . . . . 76
10.4 The Complete Molecular Hamiltonian and Wavefunction . . . . . 78
11 An Aside: Light Scattering–Why the Sky is Blue 7911.1 The Classical Electrodynamics Treatment of Light Scattering . . . 79
11.2 The Blue Sky . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
11.2.1 Sunsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
11.2.2 White Clouds . . . . . . . . . . . . . . . . . . . . . . . . . 83
III Statistical Mechanics and The Laws of Thermody-namics 88
12 Rudiments of Statistical Mechanics 8912.1 Statistics and Entropy . . . . . . . . . . . . . . . . . . . . . . . . 89
12.1.1 Combinations and Permutations . . . . . . . . . . . . . . . 90
12.2 Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
13 The Boltzmann Distribution 9413.1 Partition Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 96
13.1.1 Relation between the Q and W . . . . . . . . . . . . . . . 97
13.2 The Molecular Partition Function . . . . . . . . . . . . . . . . . . 99
14 Statistical Thermodynamics 103
15 Work 10715.1 Properties of Partial Derivatives . . . . . . . . . . . . . . . . . . . 107
15.1.1 Summary of Relations . . . . . . . . . . . . . . . . . . . . 107
15.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
15.2.1 Types of Systems . . . . . . . . . . . . . . . . . . . . . . . 108
15.2.2 System Parameters . . . . . . . . . . . . . . . . . . . . . . 109
15.3 Work and Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
15.3.1 Generalized Forces and Displacements . . . . . . . . . . . 110
15.3.2 PV work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
16 Maximum Work and Reversible changes 11316.1 Maximal Work: Reversible versus Irreversible changes . . . . . . . 113
16.2 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
16.3 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . 116
16.3.1 Example 1: The Ideal Gas Law . . . . . . . . . . . . . . . 116
16.3.2 Example 2: The van der Waals Equation of State . . . . . 117
16.3.3 Other Equations of State . . . . . . . . . . . . . . . . . . . 118
17 The Zeroth and First Laws of Thermodynamics 11917.1 Temperature and the Zeroth Law of Thermodynamics . . . . . . . 119
17.2 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . 121
17.2.1 The internal energy state function . . . . . . . . . . . . . . 121
18 The Second and Third Laws of Thermodynamics 12418.1 Entropy and the Second Law of Thermodynamics . . . . . . . . . 124
18.1.1 Statements of the Second Law . . . . . . . . . . . . . . . . 127
18.2 The Third Law of Thermodynamics . . . . . . . . . . . . . . . . . 127
18.2.1 The Third Law . . . . . . . . . . . . . . . . . . . . . . . . 128
18.2.2 Debye’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 129
18.3 Times Arrow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
IV Basics of Thermodynamics 134
19 Auxillary Functions and Maxwell Relations 13519.1 The Other Important State Functions of Thermodynamics . . . . 135
19.2 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
19.2.1 Heuristic definition: . . . . . . . . . . . . . . . . . . . . . . 137
19.3 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . 137
19.3.1 Heuristic definition: . . . . . . . . . . . . . . . . . . . . . . 138
19.4 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 138
19.4.1 Heuristic definition: . . . . . . . . . . . . . . . . . . . . . . 139
19.5 Heat Capacity of Gases . . . . . . . . . . . . . . . . . . . . . . . . 139
19.5.1 The Relationship Between CP and CV . . . . . . . . . . . 139
19.6 The Maxwell Relations . . . . . . . . . . . . . . . . . . . . . . . . 140
20 Chemical Potential 14220.1 Spontaneity of processes . . . . . . . . . . . . . . . . . . . . . . . 142
20.2 Chemical potential . . . . . . . . . . . . . . . . . . . . . . . . . . 144
20.3 Activity and the Activity coefficient . . . . . . . . . . . . . . . . . 146
20.3.1 Reference States . . . . . . . . . . . . . . . . . . . . . . . 147
20.3.2 Activity and the Chemical Potential . . . . . . . . . . . . 148
21 Equilibrium 15121.0.3 Equilibrium constants in terms of KC . . . . . . . . . . . . 153
21.0.4 The Partition Coefficient . . . . . . . . . . . . . . . . . . . 153
22 Chemical Reactions 15622.1 Heats of Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 156
22.1.1 Heats of Formation . . . . . . . . . . . . . . . . . . . . . . 157
22.1.2 Temperature dependence of the heat of reaction . . . . . . 157
22.2 Reversible reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 158
22.3 Temperature Dependence of Ka . . . . . . . . . . . . . . . . . . . 159
22.4 Extent of Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . 160
23 Ionics 16123.1 Ionic Activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
23.1.1 Ionic activity coefficients . . . . . . . . . . . . . . . . . . . 162
23.2 Theory of Electrolytic Solutions . . . . . . . . . . . . . . . . . . . 163
23.3 Ion Mobility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
23.3.1 Ion mobility . . . . . . . . . . . . . . . . . . . . . . . . . . 165
24 Thermodynamics of Solvation 16924.1 The Born Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
24.1.1 Free Energy of Solvation for the Born Model . . . . . . . . 173
24.1.2 Ion Transfer Between Phases . . . . . . . . . . . . . . . . . 174
24.1.3 Enthalpy and Entropy of Solvation . . . . . . . . . . . . . 174
24.2 Corrections to the Born Model . . . . . . . . . . . . . . . . . . . . 175
25 Key Equations for Exam 4 177
V Quantum Mechanics and Dynamics 180
26 Particle in a 3D Box 18126.1 Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
26.2 The 3D Particle in a Box Problem . . . . . . . . . . . . . . . . . . 183
27 Operators 18727.1 Operator Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
27.2 Orthogonality, Completeness, and the Superposition Principle . . 191
28 Angular Momentum 19228.1 Classical Theory of Angular Momentum . . . . . . . . . . . . . . 192
28.2 Quantum theory of Angular Momentum . . . . . . . . . . . . . . 193
28.3 Particle on a Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
28.4 General Theory of Angular Momentum . . . . . . . . . . . . . . . 195
28.5 Quantum Properties of Angular Momentum . . . . . . . . . . . . 199
28.5.1 The rigid rotor . . . . . . . . . . . . . . . . . . . . . . . . 200
29 Addition of Angular Momentum 20129.1 Spin Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . 201
29.2 Addition of Angular Momentum . . . . . . . . . . . . . . . . . . . 202
29.2.1 The Addition of Angular Momentum: General Theory . . 202
29.2.2 An Example: Two Electrons . . . . . . . . . . . . . . . . . 203
29.2.3 Term Symbols . . . . . . . . . . . . . . . . . . . . . . . . . 204
29.2.4 Spin Orbit Coupling . . . . . . . . . . . . . . . . . . . . . 205
30 Approximation Techniques 20730.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . 207
30.2 Variational method . . . . . . . . . . . . . . . . . . . . . . . . . . 209
31 The Two Level System and Quantum Dynamics 21131.1 The Two Level System . . . . . . . . . . . . . . . . . . . . . . . . 211
31.2 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 214
VI Symmetry and Spectroscopy 220
32 Symmetry and Group Theory 22132.1 Symmetry Operators . . . . . . . . . . . . . . . . . . . . . . . . . 222
32.2 Mathematical Groups . . . . . . . . . . . . . . . . . . . . . . . . . 222
32.2.1 Example: The C2v Group . . . . . . . . . . . . . . . . . . 223
32.3 Symmetry of Functions . . . . . . . . . . . . . . . . . . . . . . . . 223
32.3.1 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . 225
32.4 Symmetry Breaking and Crystal Field Splitting . . . . . . . . . . 225
33 Molecules and Symmetry 22833.1 Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 228
33.1.1 Normal Modes . . . . . . . . . . . . . . . . . . . . . . . . 229
33.1.2 Normal Modes and Group Theory . . . . . . . . . . . . . . 229
34 Vibrational Spectroscopy and Group Theory 23134.1 IR Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
34.2 Raman Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . 233
35 Molecular Rotations 23535.1 Relaxing the rigid rotor . . . . . . . . . . . . . . . . . . . . . . . . 236
35.2 Rotational Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . 236
35.3 Rotation of Polyatomic Molecules . . . . . . . . . . . . . . . . . . 237
36 Electronic Spectroscopy of Molecules 24036.1 The Structure of the Electronic State . . . . . . . . . . . . . . . . 240
36.1.1 Absorption Spectra . . . . . . . . . . . . . . . . . . . . . . 241
36.1.2 Emission Spectra . . . . . . . . . . . . . . . . . . . . . . . 241
36.1.3 Fluorescence Spectra . . . . . . . . . . . . . . . . . . . . . 242
36.2 Franck—Condon activity . . . . . . . . . . . . . . . . . . . . . . . 243
36.2.1 The Franck—Condon principle . . . . . . . . . . . . . . . . 243
37 Fourier Transforms 24537.1 The Fourier transformation . . . . . . . . . . . . . . . . . . . . . 245
VII Kinetics and Gases 249
38 Physical Kinetics 25038.1 kinetic theory of gases . . . . . . . . . . . . . . . . . . . . . . . . 250
38.2 Molecular Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . 252
39 The Rate Laws of Chemical Kinetics 25439.1 Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
39.2 Determination of Rate Laws . . . . . . . . . . . . . . . . . . . . . 258
39.2.1 Differential methods based on the rate law . . . . . . . . . 259
39.2.2 Integrated rate laws . . . . . . . . . . . . . . . . . . . . . . 259
40 Temperature and Chemical Kinetics 26140.1 Temperature Effects on Rate Constants . . . . . . . . . . . . . . . 261
40.1.1 Temperature corrections to the Arrhenious parameters . . 262
40.2 Theory of Reaction Rates . . . . . . . . . . . . . . . . . . . . . . 262
40.3 Multistep Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 265
40.4 Chain Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
41 Gases and the Virial Series 26941.1 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . 269
41.2 The Virial Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
41.2.1 Relation to the van der Waals Equation of State . . . . . . 271
41.2.2 The Boyle Temperature . . . . . . . . . . . . . . . . . . . 272
41.2.3 The Virial Series in Pressure . . . . . . . . . . . . . . . . . 272
41.2.4 Estimation of Virial Coefficients . . . . . . . . . . . . . . . 273
42 Behavior of Gases 27442.1 P, V and T behavior . . . . . . . . . . . . . . . . . . . . . . . . . 274
42.1.1 α and κT for an ideal gas . . . . . . . . . . . . . . . . . . . 275
42.1.2 α and κT for liquids and solids . . . . . . . . . . . . . . . . 275
42.2 Heat Capacity of Gases Revisited . . . . . . . . . . . . . . . . . . 276
42.2.1 The Relationship Between CP and CV . . . . . . . . . . . 276
42.3 Expansion of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . 279
42.3.1 Isothermal and Adiabatic expansions . . . . . . . . . . . . 279
42.3.2 Heat capacity CV for adiabatic expansions . . . . . . . . . 280
42.3.3 When P is the more convenient variable . . . . . . . . . . 281
42.3.4 Joule expansion . . . . . . . . . . . . . . . . . . . . . . . . 282
42.3.5 Joule-Thomson expansion . . . . . . . . . . . . . . . . . . 283
43 Entropy of Gases 28643.1 Calculation of Entropy . . . . . . . . . . . . . . . . . . . . . . . . 286
43.1.1 Entropy of Real Gases . . . . . . . . . . . . . . . . . . . . 288
VIII More Thermodyanmics 292
44 Critical Phenomena 29344.1 Critical Behavior of fluids . . . . . . . . . . . . . . . . . . . . . . 293
44.1.1 Gas Laws in the Critical Region . . . . . . . . . . . . . . . 294
44.1.2 Gas Constants from Critical Data . . . . . . . . . . . . . . 295
44.2 The Law of Corresponding States . . . . . . . . . . . . . . . . . . 296
44.3 Phase Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 296
44.3.1 The chemical potential and T and P . . . . . . . . . . . . 297
44.3.2 The Clapeyron Equation . . . . . . . . . . . . . . . . . . . 298
44.3.3 Vapor Equilibrium and the Clausius-Clapeyron Equation . 298
44.4 Equilibria of condensed phases . . . . . . . . . . . . . . . . . . . . 299
44.5 Triple Point and Phase Diagrams . . . . . . . . . . . . . . . . . . 300
45 Transport Properties of Fluids 30145.1 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
45.2 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
45.3 Thermal conductivity . . . . . . . . . . . . . . . . . . . . . . . . . 305
45.3.1 Thermal Conductivity of Gases and Liquids . . . . . . . . 306
45.3.2 Thermal Conductivity of Solids . . . . . . . . . . . . . . . 307
46 Solutions 30846.1 Measures of Composition . . . . . . . . . . . . . . . . . . . . . . . 308
46.2 Partial Molar Quantities . . . . . . . . . . . . . . . . . . . . . . . 308
46.2.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
46.2.2 Partial Molar Volumes . . . . . . . . . . . . . . . . . . . . 310
46.3 Reference states for liquids . . . . . . . . . . . . . . . . . . . . . . 311
46.3.1 Activity (a brief review) . . . . . . . . . . . . . . . . . . . 311
46.3.2 Raoult’s Law . . . . . . . . . . . . . . . . . . . . . . . . . 312
46.3.3 Ideal Solutions (RL) . . . . . . . . . . . . . . . . . . . . . 314
46.3.4 Henry’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 316
46.4 Colligative Properties . . . . . . . . . . . . . . . . . . . . . . . . . 318
46.4.1 Freezing Point Depression . . . . . . . . . . . . . . . . . . 318
46.4.2 Osmotic Pressure . . . . . . . . . . . . . . . . . . . . . . . 319
47 Entropy Production and Irreverisble Thermodynamics 32247.1 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
47.2 The Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
47.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
47.3.1 Entropy Production due to Heat Flow . . . . . . . . . . . 326
47.3.2 Entropy Production due to Chemical Reactions . . . . . . 328
47.4 Thermodynamic Coupling . . . . . . . . . . . . . . . . . . . . . . 330
47.5 Echo Phenonmena . . . . . . . . . . . . . . . . . . . . . . . . . . 331
Chemistry 351: PhysicalChemistry I
1
1
Solved Problems
I make-up most of the problems on the problems sets, so it might be helpful to
you to see some of these problems worked out.
Even though there aren’t many “book” problems assigned during the year, you can
still learn a lot be working these and looking that their solutions in the solution
manual.
Keep in mind this chapter provides some examples of how to solve problems for
both physical chemistry I and physical chemistry II. Consequently early in the
course some of the examples might seem very itimidating. Simply skip those
examples as you scan through this chapter.
Tips for solving problems
Working problem sets is the heart and sole of learning physical chemistry. The
only way that you can be sure that you understand a concept at to be able to
solve the problems associated with it.
This takes time and hard work.
But there are some things that you can do to help yourself with these problems.
Tips
2
2
1. Remember nobody cares if you solve any particular problem on the problem
set. They have all been solved before, so if you solve them you will not
become famous nor will you save the world. The only reason you work them
is to learn.
2. Budget your time so that you don’t have to work on an overwhelming number
of problems at a time. Try to whip-off a few on the same day that you get
the problem set. Then work on them consistently during the week. This
will make the problem sets much more efficient at helping you learn.
3. You can do the problem. I don’t assign problems that you cannot do. If you
think you can’t do the problem then maybe you need try a different way of
thinking about it.
4. Part of the trouble is simply understanding what the problem is asking you
to do. There is a tendency to try to start solving the problem before fully
understanding the question.
• Read the question carefully
• Try to think about what topic(s) in lecture and in the notes the problemis dealing with.
• Do not worry about not knowing how to solve it yet.
• Just identify the general ideas that you think you might need.
• Determine wether you need to approach the problem mathematically
or conceptually or both.
• If the question is long, try to identify subsections of it.
5. For problems that require a mathematical approach...
• Do not be afraid. Try to figure out what mathematical techniques youneed to express the solution to the problem.
3
• Do the math; either you will be able to do this or you won’t. It mighttake some review on your part.
• Always check to see if the math makes sense when you are done.
6. For problems that require a conceptual approach...
• Make sure that the physical idea that you are using in your argument iscorrect. If you are not sure, start with a related concept that is better
known by you.
• Look for self-consistency. Does you final answer jive with what youknow.
Problems Dealing With Quantum Mechanics
Problem: What is the periodicity of the following functions
• f(x) = sin2 x
• f(x) = cosx
• f(x) = e−2ix
Solution: For the first function it is easiest to see the periodicity by writing thefunction as f(x) = (sinx)(sinx). We know that this function will repeat zeros
when ever sinx = 0. This occurs at x = nπ, n = 0,±1,±2 . . ., so the periodicityis π. The second function we should remember from trig as having a period of 2π.
Finally for the last function it is best to used Euler’s identity and write
e−2ix = cos 2x+ i sin 2x (1)
The real part of this function, cos 2x, has a period of π as does the imaginary
part, sin 2x. Therefore the entire function has a period of π.
4
Problem: Which of the following functions are eigenfunction of the momentumoperator, px = −i~ d
dx.
• ψ(x) = eikx
• ψ(x) = e−αx2
• ψ(x) = cos kx
Solution: We need to determine if pxψ(x) = λψ(x) where λ is a constant. If
this equation is true then the function is an eigenfunction with eigenvalue λ. For
the case of momentum all we need to do is take the derivative of each function,
multiply by −i~ and check to see if the eigenvalue equation holds.For the first function
pxψ(x) = −i~dψ(x)
dx= −i~de
ikx
dx= ~keikx = ~kψ(x), (2)
so, yes, this function is an eigenfunction of the momentum operator.
For the second function
pxψ(x) = −i~dψ(x)
dx= −i~de
−αx2
dx= 2i~αxe−αx2 = 2i~α↓xψ(x), (3)
so, no, this function is not and eigenfunction of the momentum operator.
For the last function
pxψ(x) = −i~dψ(x)
dx= −i~d cos kx
dx= −i~k
6=cos kxz | sin kx, (4)
so, no, this function is not an eigenfunction of the momentum operator.
Problem: A quantum object is described by the wavefunction ψ(x) = e−αx2.
What is the probability of finding the object further than α away from the origin
(x = 0)?
5
Solution: First of all we do not know if this wavefunction is normalized, so weshould assume that it isn’t. We could normalize this wavefunction, but we won’t.
We are interested in finding the probability that the object is outside of the region
−α < x < α. To do this using an unnormalized wavefunction we must evaluate
P (|x| > α) =
R −α−∞ |ψ(x)|
2 dx+R∞α|ψ(x)|2 dxR∞
−∞ |ψ(x)|2 dx
. (5)
The first integral in the numerator gives the probability that the object is at a
position x < −α and the second integral in the numerator gives the probabilityfor x > α. The denominator accounts for the fact that the wavefunction is un-
normalized. The limits of the integral in the denominator represent all space for
the object. If you were working with a normalized wavefunction the denominator
would be equal to 1 and hence not needed. Plugging in the wavefunctions we have
P (|x| > α) =
R −α−∞ e−2αx
2dx+
R∞α
e−2αx2dxR∞
−∞ e−2αx2dx. (6)
Mathematica can assist with these integrals to give the final answer of
P (|x| > α) = erfc[√2α
32 ]. (7)
Problem: A quantum object is described by the wavefunction ψ(x) = e−γx over
the range 0 ≤ x <∞. Normalize this wavefunction.
Solution: Following our general procedure from the notes if we have some unnor-malized wavefunction, ψunnorm we know that this function must simply be some
constant N multiplied by the normalized version of this function:
ψunnorm = Nψnorm (8)
We have shown generally that N is given by
N =
sZspace
|ψunnorm(x)|2 dx. (9)
6
Which for this case is
N =
sZ ∞
0
|e−γx|2 dx =sZ ∞
0
e−2γxdx =
r1
2γ(10)
So finally we get the normalized wavefunction by rearanging ψunnorm = Nψnorm:
ψnorm(x) =p2γe−γx. (11)
Problem: A quantum object is described by the wavefunction ψ(x) = e−γx over
the range 0 ≤ x <∞. What is the average position of the object?
Solution: We need to work with the normalized wavefunction that we found inthe previous problem, ψ(x) =
√2γe−γx. Generally and average is calculated as
hoi =Zspace
ψ∗(x)oψ(x), (12)
which in this case is
hxi =Z ∞
0
p2γe−γxx
p2γe−γxdx = 2γ
Z ∞
0
xe−2γxdx =1
2γ. (13)
So on average you will find the object at x = 12γ.
Problem: What is the probability of finding an electron in the 1s state of hydrogenfurther than one Bohr radius away from the nucleus?
Solution: We need to evaluate
P (r > a0) =
Z 2π
0
Z π
0
Z ∞
a0
|ψ1s|2 r2 sin θdrdθdφ. (14)
Remember the extra r2 sin θ is needed when integrating in spherical polar coordi-
nates. The normalized 1s wavefunction is
ψ1s =1pπa30
er/a0 . (15)
7
We can do this integral by hand or have Mathematica help us to give
P (r > a0) =5
e2= 0.677. (16)
So, about 68% of the time the electron would be found at some distance greater
then one Bohr radius from the proton.
Problem: A free particle in three dimensions is described by the Hamiltonian,
H = −~22m∇2. Express the wavefunction (in Cartesian coordinates) as a product
state.
Solution: This problem appears hard at first since we are not studying three
dimensional systems, but all it is asking is to express the wavefunction, which is
a function of the three spatial dimensions, Ψ(x, y, z) as a product state. We know
that if the wavefunction is to be a product state then the Hamiltonian must be
made up of a sum of independent terms. To see this we write out the Laplacian
to get
H =−~22m
µ∂2
∂x2+
∂2
∂y2+
∂2
∂z2
¶. (17)
We see that indeed the Hamiltonian is a sum of term that depends only on x,
a term depending only on y and a term that depends only on z. Therefore the
appropriate product state is
Ψ(x, y, z) = ψ(x)ψ(y)ψ(z). (18)
Problem: Expand the Morse potential in a Taylor’s series about Req. Verify that
the coefficient for the linear term is zero. What is the force constant associated
with the Morse potential?
Solution: The Morse potential is
V (x) = De
£1− e−β(R−Req)
¤. (19)
8
The Taylor series about Req for this function is
V (x) = V (x)|Req| z = 0
+dV (x)
dx
¯Req| z
= 0
(R−Req) +1
2!
d2V (x)
dx2
¯Req| z
= β2De
(R−Req)2 + · · · . (20)
So, yes the coefficient of the linear term (the term involving (R − Req) to the
first power) is zero. This will always be true when you perform a Taylor series
expansion about a minimum (or maximum). The force constant is given by the
coefficient of the quadratic term so in this case k = β2De.
Problem: Without performing any calculations, compare hRi as a function ofthe vibrational quantum number for a diatomic modelled as a harmonic oscillator
versus a Morse oscillator.
Solution: This problem requires the we think qualitatively about the wavefunc-
tions and the potentials for the harmonic oscillator and the Morse oscillator. The
potential for the harmonic oscillator is described by a parabola centered about the
equilibrium bond length. Hence no mater what the vibrational quantum number is
there is just as much of the wavefunction on either side equilibrium thus hRi = Req
for any quantum number. The Morse potential does not have this symmetry. It
is steeper on the “short” side of equilibrium and softer on the “long” side of equi-
librium and this “softness” increases with increasing quantum number. Therefore
without performing any calculations we can at least say that hRi increases as thequantum number increases.
Problems Dealing With Statistical Mechanics and Thermo-
dynamics
Problem: A vial containing 1020 benzene molecules is at 300K. How many mole-cules are in the first excited state of the ‘ring breathing’ mode (992 cm−1)? How
9
many are in the first excited state of the symmetric C—H vibrational mode (3063
cm−1)?
Solution: This is a problem that deals with the Boltzmann distribution. So,
N rbv=1 =
µ2 sinh
992
2× 208
¶×³e−
3×9922×208
´× 1020 = 8.41× 1017 (21)
and
NC—Hv=1 =
µ2 sinh
3063
2× 208
¶×³e−
3×30632×208
´× 1020 = 4.02× 1013. (22)
We see that about 8.41×1017
1020× 100% = 0.841% of the benzene molecules are in the
first vibrational excited state for the ring breathing mode and 4.02×10131020
× 100% =0.0000402% of the benzene molecules are in the first excited state for the C—H
stretching mode.
Problem: Consider a linear chain of N atoms. Each of the atoms can be in one
of three states A, B or C, except that an atom in state A can not be adjacent to
an atom in state C. Find the entropy per atom for this system as N → ∞. To
solve this problem it is useful to define the set of three dimensional column vectors
V (j) such that the three elements are the total number of allowed configurations of
a j-atom chain having the jth atom in state A, B or C. For example,
V (1) =
⎡⎢⎣ 111
⎤⎥⎦ , V (2) =
⎡⎢⎣ 232
⎤⎥⎦ , V (3) =
⎡⎢⎣ 575
⎤⎥⎦ , · · · . (23)
The V (j+1) can be found from the V (j) vector using the matrix equation,
V (j+1) =MV (j), (24)
where for this example
M =
⎡⎢⎣ 1 1 0
1 1 1
0 1 1
⎤⎥⎦ . (25)
10
The matrix M is the so-called transfer matrix for this system. It can be shownthat the number of configurations W = Tr[MN ]. Now for large N, Tr[MN ] ≈ λNmax,
where λmax is the largest eigenvalue of M. So
W = limN→∞
λNmax. (26)
1. 1. Use M to find V (4)
2. Verify V (3) explicitly by drawing all the allowed 3-atom configurations.
3. Verify W = Tr[MN ] for N = 1 and N = 2.
4. Use Boltzmann’s equation to find the entropy per atom for this chain
as N goes to infinity.
Solution: For part (a) we simply use the transfer matrix as directed in the
problem (we are given V (3)):
V (4) =
⎡⎢⎣ 1 1 0
1 1 1
0 1 1
⎤⎥⎦⎡⎢⎣ 575
⎤⎥⎦ =⎡⎢⎣ 121712
⎤⎥⎦ .For part (b) we need to list all states for the case of N = 3 and verify the we get
the same result as calculated using the transfer matrix. Remembering that V (3)
gives us the number of sequences that end in a given state we should organize our
list in the same manner
States ending in A States ending in B States ending in C
AAA AAB ABC
ABA ABB BBC
BAA BAB BCC
BBA BBB CBC
CBA BCB
CBB
CCB
5 states√
7 states√
5 states√
.
11
States like AAC are not allowed because A and C are neighbors.
For part (c) we evaluate W = Tr[MN ] for N = 1 and 2. For N = 1, W =
Tr[M ] = 3 This corresponds to the three distinguishable microstates A, B, and
C. For N = 2,
W = Tr[M2] = Tr
⎡⎢⎣⎡⎢⎣ 1 1 0
1 1 1
0 1 1
⎤⎥⎦⎡⎢⎣ 1 1 0
1 1 1
0 1 1
⎤⎥⎦⎤⎥⎦ = Tr
⎡⎢⎣⎡⎢⎣ 2 2 1
2 3 2
1 2 2
⎤⎥⎦⎤⎥⎦ = 7 (27)
This corresponds to the seven distinguishable microstates AA, AB, BA, BB, BC,
CB and CC (Remember C and A cannot be neighbors).
For part (d) we use
S
N=
k
NlnW = lim
N→∞
k
NlnλNmax = lim
N→∞
k
NN lnλmax = k lnλmax. (28)
So, we simply need to find the maximum eigenvalue of the Transfer matrix. Using
Mathematica we find λmax = 1 +√2. Therefore the limiting entropy per atom
isS
N= k ln
³1 +√2´. (29)
Problem: Using the classical theory of light scattering, calculate the positions ofthe Rayleigh, Stokes and anti-Stokes spectral lines for benzene. Assume benzene
has only two active modes (992cm−1 and 3063cm−1) and assume the Laser light
used to do the scattering is at 20000cm−1 (this is 500nm–green light).
Solution: Since there are two vibrational modes we expect two Stokes lines tothe red of 20000cm−1, one at 20000cm−1 − 992cm−1 = 19008cm−1 and one at
20000cm−1 − 3063cm−1 = 16937cm−1. Likewise we expect two anti-Stokes lines,one at 20000cm−1 + 992cm−1 = 20992cm−1 and one at 20000cm−1 + 3063cm−1 =
23063cm−1. There is only one Rayleigh line and it is at the same frequency at the
input laser beam which, in this case, is 20000cm−1.
12
Problem: A simple model for a crystal is a “gas” of harmonic oscillators. De-termine A, S, and U from the partition function for this model.
Solution: For this model the crystal is modelled as a collection of harmonicoscillators so we need the partition function for the harmonic oscillator.
Qcrystal = qNHO =
Ã1
2 sinh β~ω2
!(30)
From our formulas for statistical thermodynamics
A = −kT lnQcrystal = +NKT ln
µ2 sinh
β~ω2
¶, (31)
where we used properties of logs to pull the N out front and move the sinh term
from to the numerator,
S = −kβ∂Qcrystal
∂β+ k lnQcrystal (32)
=Nkβ~ω2
cothβ~ω2− k ln
µ2 sinh
β~ω2
¶and
U = −∂Qcrystal
∂β=
N~ω2
cothβ~ω2
. (33)
Problem: Express the equation of state for internal energy for a Berthelot gas.
Solution: The equation representing a Berthelot gas is
P =nRT
V − nb− n2a
TV 2. (34)
We are interesting in an equation of state for U(T, V ). Writing out the total
derivative of U(T, V ) we get
dU =
µ∂U
∂T
¶V
dT +
µ∂U
∂V
¶T
dV. (35)
13
Now¡∂U∂T
¢Vis just heat capacity, CV , but
¡∂U∂V
¢Tis nothing convenient so we must
proceed. We employ the “useful relation”µ∂U
∂V
¶T
= T
µ∂P
∂T
¶V
− P (36)
to eliminate U in favor of P so that we can use the equation of state for a Berthelot
gas. One obtains
T
µ∂P
∂T
¶V
− P = T
µnR
V − nb+
n2a
T 2V 2
¶− nRT
V − nb+
n2a
TV 2=2n2a
TV 2. (37)
Hence the equation of state for internal energy of a Berthelot gas is
dU = CV dT +2n2a
TV 2dV (38)
Problem: Use the identities for partial derivatives to eliminate the¡∂P∂T
¢Vfactor
in
Cp = Cv + T
µ∂V
∂T
¶P
µ∂P
∂T
¶V
(39)
so that all derivatives are at constant pressure or temperature.
Solution: Here we either remember an identity or turn to our handout of partialderivative identities to employ the cyclic rule to
¡∂P∂T
¢V:µ
∂P
∂T
¶V
= −µ∂P
∂V
¶T
µ∂V
∂T
¶P
. (40)
This eliminates the constant V term and so,
Cp = Cv − T
µ∂V
∂T
¶2P
µ∂P
∂V
¶T
. (41)
14
Part I
Basic Quantum Mechanics
15
15
1. Quantum Theory
The goal of science is unification.
• Many phenomena described by minimal and general concepts.
1.1. The “Fall” of Classical Physics
A good theory:
• explain known experimental results
• self consistent
• predictive
• minimal number of postulates
Around the turn of the century, experiments were being performed in which the re-
sults defied explanation by means of the current understanding of physics. Among
these experiments were
1. The photoelectric effect
2. Low temperature heat capacity
3. Atomic spectral lines
4. Black body radiation and the ultraviolet catastrophe
16
16
5. The two slit experiment
6. The Stern-Gerlach experiment
∗ ∗ See Handouts ∗ ∗
1.2. Bohr’s Atomic Theory
1.2.1. First Attempts at the Structure of the Atom
The “solar system” model.
• The electron orbits the nucleus with the attractive coulomb force balancedby the repulsive centrifugal force.
Flaws of the solar system model
• Newton: OK √
• Maxwell: problem √
17
— As the electron orbits the nucleus, the atom acts as an oscillating dipole
• — The classical theory of electromagnetism states that oscillating dipolesemit radiation and thereby lose energy.
— The system is not stable and the electron spirals into the nucleus. Theatom collapses!
Bohr’s model: Niels Bohr (1885—1962)
18
• Atoms don’t collapse =⇒ what are the consequences
Experimental clues
• Atomic gases have discrete spectral lines.
• If the orbital radius was continuous the gas would have a continuous spec-trum.
• Therefore atomic orbitals must be quantized.
r =4π 0N
2~2
Zmee2(1.1)
where Z is the atomic number, me and e are the mass and charge of the
electron respectively and 0 is the permittivity of free space. N is a positive
real integer called the quantum number. ~ = h/2π is Planck’s constant
divided by 2π.
The constant quantity 4π 0~2mee2
appears often and is given the special symbol a0 ≡4π 0~2mee2
= 0.52918 Å and is called the Bohr radius.
The total energy of the Bohr atom is related to its quantum number
EN = −Z2µ
e2
2a0
¶1
N2. (1.2)
Tests of the Bohr atom
• Ionization energy of Hydrogen atoms
— The Ionization energy for Hydrogen atoms (Z = 1) is the minium
energy required to completely remove an electron form it ground state,
i.e., N = 1→ N =∞
Eionize = E∞ −E1 =−Z2e22a0
µ1
∞2− 1
12
¶=
e2
2a0(1.3)
19
— Eionize =e2
2a0= 13.606 eV= 109,667 cm−1 =R. R is called the Rydberg
constant.
— Eionize experimentally observed from spectroscopy is 13.605 eV (very
good agreement)
• Spectroscopic lines fromHydrogen represent the difference in energy betweenthe quantum states
— Bohr theory: Difference energies
Ej −Ek =e2
2a0
µ1
N2j
− 1
N2k
¶= R
µ1
N2j
− 1
N2k
¶(1.4)
Initial state Nk Final States Nj Series Name
1 2,3,4,· · · Lyman
2 3,4,5,· · · Balmer
3 4,5,6,· · · Pachen
4 5,6,7,· · · Brackett
5 6,7,8,· · · Pfund
• — Since the orbitals are quantized, the atom may only change its orbital
radius by discrete amounts.
— Doing this results in the emission or absorption of a photon with energy
v =4E
hc(1.5)
Failure of the Bohr model
• No fine structure predicted (electron-electron coupling)
• No hyperfine structure predicted (electron-nucleus coupling)
• No Zeeman effect predicted (response of spectrum to magnetic field)
20
• Spin is not included in theory
The Bohr quantization idea points to a wavelike behavior for the electron.
The wave must satisfy periodic boundary conditions much like a vibrating ring
∗ ∗ ∗ See Fig. 11.9 Laidler&Meiser ∗ ∗∗
The must be continuous and single valued
Particles have wave-like characteristics
The Bohr atom was an important step towards the formulation of quantum theory
• Erwin Schrödinger (1887—1961): Wave mechanics
• Werner Heisenberg (1902—1976): Matrix mechanics
• Paul Dirac (1902—1984): Abstract vector space approach
21
2. The Postulates of Quantum
Mechanics
2.1. Postulate I
Postulate I: The state of a system is defined by a wavefunction, ψ, which con-
tains all the information that can be known about the system.
We will normally take ψ to be a complex valued function of time and coordi-
nates: ψ(t, x, y, z) and, in fact, we will most often deal with time independent
“stationary” states ψ(x, y, z)
Note: In general the wavefunction need not be expressed as a function of coordi-
nate. It may, for example, be a function of momentum.
The wavefunction ψ represents a probability amplitude and is not directly observ-
able.
However the mod-square of the wavefunction, ψ∗ψ = |ψ|2 , represents a probabilitydistribution which is directly observable.
That is, the probability of finding a particle which is described by ψ(x, y, z) at the
position between x and x+dx, y and y+dy and z and z+dz is |ψ(x, y, z)|2 dxdydz(or |ψ(r, θ, φ)|2 r2 sin θdrdθdφ in spherical coordinates).
22
22
Properties of the wavefunction
• Single valueness
• continuous and finite
• continuous and finite first derivative
•Rspace |ψ(x, y, z)|
2 dxdydz <∞
Normalization of the wavefunction
In order for |ψ(x, y, z)|2 to be exactly interpreted as a probability dis-tribution, ψ(x, y, z) must be normalizable.
That is, ψunnorm = Nψnorm, whereN =qR
space |ψunnorm(x, y, z)|2 dxdydz
This assures thatRspace |ψnorm|
2 dxdydz = 1 as expected for a proba-
bility distribution
From now on we will always normalize our wavefunctions.
2.2. How to normalize a wavefunction
If we have some unnormalized wavefunction, ψunnorm we know that this function
must simply be some constant N multiplied by the normalized version of this
function:
ψunnorm = Nψnorm. (2.1)
Now, we take the mod-square of both sides and then integrate both sides of this
equation over all spaceZspace
|ψunnorm|2 dxdydz =Zspace
|Nψnorm|2 dxdydz. (2.2)
23
The N is just a constant so it can be pulled out of both the mod-square and the
integral Zspace
|ψunnorm|2 dxdydz = N2
Zspace
|ψnorm|2 dxdydz, (2.3)
but Zspace
|ψnorm|2 dxdydz = 1 (2.4)
because that is the very definition of a normalized wavefunction. Thus wherever
we seeRspace |ψnorm|
2 dxdydz we can replace it with 1. So,Zspace
|ψunnorm|2 dxdydz = N2 × 1 = N2. (2.5)
This gives us an expression for N. Taking the square root of both sides gives.
N =
sZspace
|ψunnorm(x, y, z)|2 dxdydz. (2.6)
So finally we get the normalized wavefunction by reagranging ψunnorm = Nψnorm:
ψnorm =1
Nψunnorm. (2.7)
Notice that no where did we ever specify what ψunnorm or ψnorm actually were,
therefore this is a general procedure that will work for any wavefunction.
To find the probability for the particle to be in a finite region of space we simple
evaluate (here a 1D case)
P (x1 < x < x2) =
R x2x1|ψ(x)|2 dxR∞
−∞ |ψ(x)|2 dx
if ψ(x)=⇒
normalized
Z x2
x1
|ψ(x)|2 dx (2.8)
2.3. Postulates II and II
Postulate II: Every physical observable is represented by a linear (Hermitian)operator.
24
An operator takes a function and turns it into another function
Of(x) = g(x) (2.9)
This is just like how a function takes a number and turns it into another number.
So in quantum mechanics operators act on the wavefunction to produce a new
wavefunction
The two most important operators as far as we are concerned are
• x = x
• px = −i~ ∂∂x
and of course the analogous operators for the other coordinates (y, z) and coordi-
nate systems (spherical, cylindrical, etc.).
Nearly all operators we will need are algebraic combinations of the above.
Postulate III: The measurement of a physical observable will give a result thatis one of the eigenvalues of the corresponding operator.
There is a special operator equation called the eigenvalue equation which is
Of(x) = λf(x) (2.10)
where λ is just a number.
For a given operator only a special set of function satisfy this equation. These
functions are called eigenfunctions.
25
The number that goes with each function is called the eigenvalue.
So solution of the eigenvalue equation gives a set of eigenfunctions and a set of
eigenvalues.
Example
Let O in the eignevalue equation be the operator that takes the derivative: O =
d = ddx.
So we want a solution to
df(x) = λf(x) (2.11)df(x)
dx= λf(x)
So, we ask ourselves what function is proportional to its own derivative? ⇒f(x) = eλx.
So the eigenfunctions are the set of functions f(x) = eλx and the eigenvalues are
the numbers λ
26
3. The Setup of a Quantum
Mechanical Problem
3.1. The Hamiltonian
The most important physical observable is that of the total energy E.
The operator associated with the total energy is called the Hamiltonian operator
(or simply the Hamiltonian) and is given the symbol H.
The eigenvalue equation for the Hamiltonian is
Hψ = Eψ. (3.1)
This equation is the (time independent) Schrödinger equation.
This equation is the most important equation of the course and we will use it many
times throughout our discussion of quantum mechanics and statistical mechanics.
3.2. The Quantum Mechanical Problem
Nearly every problem one is faced with in elementary quantum mechanics is han-
dled by the same procedure as given in the following steps.
1. Define the classical Hamiltonian for the system.
27
27
• The total energy for a classical system is
Ecl = T + V, (3.2)
where T is the kinetic energy and V is the potential energy.
• The kinetic energy is always of the form
T =1
2m
¡p2x + p2y + p2z
¢(3.3)
• The potential energy is almost always a function of coordinates only
V = V (x, y, z) (3.4)
• Note: Some quantum systems don’t have classical analogs so the Hamil-tonian operator must be hypothesized.
2. Use Postulate II to replace the classical variables, x, px etc., with their
appropriate operators. Thus,
T =−~22m∇2 = −~
2
2m∇2, (3.5)
where ∇2 ≡ ∂2
∂x2+ ∂2
∂y2+ ∂2
∂z2, and
V = V (x, y, z) = V (x, y, z). (3.6)
So,
H = T + V =−~22m∇2 + V (x, y, z) (3.7)
3. Solve the Schrödinger equation, Hψ = Eψ, which is now a second order
differential equation of the form∙−~22m∇2 + V (x, y, z)
¸ψ = Eψ
⇒ −~22m∇2ψ + (V (x, y, z)−E)ψ = 0 (3.8)
28
• Note: It is solely the form of V (x, y, z) which determines whether this
is easy or hard to do.
• For one-dimensional problems
−~22m
d2
dx2ψ + (V (x)−E)ψ = 0 (3.9)
3.3. The Average Value Theorem
Postulate III implies that if ψ is an eigenfunction of a particular operator rep-
resenting a physical observable, then all measurements of that physical property
will yield the associated eigenvalue.
However, If ψ is not an eigenfunction of a particular operator, then all measure-
ments of that physical property will still yield an eigenvalue, but we cannot predict
for certain which one.
We can, however, give an expectation, or average, value for the measurement.
This is given by
hαi =Zspace
ψ∗αψdxdydz (3.10)
For example,
hxi =Zspace
ψ∗xψdxdydz =
Zspace
x |ψ|2 dxdydz (3.11)
and
hpxi =Zspace
ψ∗pxψdxdydz = −i~Zspace
ψ∗∂ψ
∂xdxdydz (3.12)
29
3.4. The Heisenberg Uncertainty Principle
In quantum mechanics certain pairs of variables can not, even in principle, be
simultaneously known to arbitrary precision. Such variables are called compli-
mentary.
This idea is the Heisenberg uncertainty principle and is of profound im-portance.
The general statement of the Heisenberg uncertainty principle is
δαδβ ≥ 12
¯Dhα, β
iE¯, (3.13)
where the notationhα, β
imeans the commutator of α and β. The commutator is
defined as hα, β
i≡ αβ − βα. (3.14)
The most important example of complimentary variables is position and momen-
tum. We see
δpxδx ≥1
2|h[px, x]i| =
1
2|hpxx− xpxi| (3.15)
=1
2
¯Zψ∗~i
µ∂
∂xx− x
∂
∂x
¶ψdx
¯=
¯~2i
¯=~2.
So, at the very best we can only hope to simultaneously know position and momen-
tum such that the product of the uncertainty in each is ~2. (n.b., δpxδy = 0, we can
know, for example, the y position and the x momentum to arbitrary precision.)
Suppose we know the position of a particle perfectly, what can we say about its
momentum?
30
4. Particle in a Box
We now will apply the general program for solving a quantum mechanical problem
to our first system: the particle in a box.
This system is very simple which is one reason for beginning with it. It also can
be used as a “zeroth order” model for certain physical systems.
We shall soon see that the particle in a box is a physically unrealistic system and,
as a consequence, we must violate one of our criteria for a good wavefunction.
Nevertheless it is of great pedagogical and practical value.
4.1. The 1D Particle in a Box Problem
Consider the potential, V (x), shown in the figure and given by
V (x) =
⎧⎪⎨⎪⎩∞ x ≤ 00 0 < x < a
∞ x ≥ a
. (4.1)
Because of the infinities at x = 0 and x = a, we need to partition the x-axis into
the three regions shown in the figure.
31
31
Now, in region I and III, where the potential is infinite, the particle can never
exist so, ψ must equal zero in these regions.
The particle must be found only in region II.
The Schrödinger equation in region II is (V (x) = 0)
Hψ = Eψ =⇒ −~2
2m
d2ψ(x)
dx2= Eψ, (4.2)
which can be rearranged into the form
d2ψ(x)
dx2+2mE
~2ψ(x) = 0. (4.3)
The general solution of this differential equation is
ψ(x) = A sin kx+B cos kx, (4.4)
where k =q
2mE~2 .
Now ψ must be continuous for all x. Therefore it must satisfy the boundary
conditions (b.c.): ψ(0) = 0 and ψ(a) = 0.
32
From the ψ(0) = 0 b.c. we see that the constant B must be zero because
cos kx|x=0 = 1.
So we are left with ψ(x) = A sin kx for our wavefunction.
As can be inferred from the following figure, the second b.c., ψ(a) = 0, places
certain restrictions on k.
In particular,
kn =nπ
a, n = 1, 2, 3, · · · . (4.5)
The values of k are quantized. So, now we have
ψn(x) = A sinnπx
a. (4.6)
The constant A is the normalization constant. We obtain A fromZ ∞
−∞ψ∗n(x)ψn(x) = 1 =
Z a
0
A2 sinnπx
asin
nπx
adx. (4.7)
Letting u = πxa, du = π
adx, this becomes
1 = A2a
π
Z π
0
sin2 nudu = A2a
π/π/2=
A2a
2. (4.8)
33
Solving for A gives
A =
r2
a. (4.9)
Thus our normalized wavefunctions for a particle in a box are
ψn(x) =
⎧⎪⎪⎨⎪⎪⎩0 Iq
2asin nπx
aII
0 III
. (4.10)
Is this wavefunction OK?
We can get the energy levels from kn =q
2mEn~2 and kn =
nπa:
En =n2π2~2
2ma2~= h
2π=n2h2
8ma2. (4.11)
4.2. Implications of the Particle in a Box problem
Zero Point Energy
34
The smallest value for n is 1 which corresponds to an energy of
E1 =h2
8ma26= 0. (4.12)
That is, the lowest energy state, or ground state, has nonzero energy. This residual
energy is called the zero point energy and is a consequence of the uncertainty
principle.
If the energy was zero then we would conclude that momentum was exactly zero,
δp = 0. But we also know that the particle is located within a finite region of
space, so δx 6=∞.
Hence, δxδp = 0 which violates the uncertainty principle.
Features of the Particle in a Box Energy Levels
• The energy level spacing is
4E = En+1 − En =(n+ 1)2h2
8ma2− n2h2
8ma2= (n2/ + 2n+ 1− n2/ ) h2
8ma2
4E = (2n+ 1)h2
8ma2(4.13)
• This spacing increases linearly with quantum level n
• This spacing decreases with increasing mass
• This spacing decreases with increasing a
• It is this level spacing that is what is measured experimentally
The Curvature of the Wavefunction
35
The operator for kinetic energy is T = −~22m
d2
dx2. The important part of this is d2
dx2.
From freshman calculus we know that the second derivative of a function describes
its curvature so, a wavefunction with more curvature will have a larger second
derivative and hence it will posses more kinetic energy.
This is an important concept for the qualitative understanding of wavefunctions
for any quantum system.
Applying this idea to the particle in a box we an anticipate both zero point energy
and the behavior of the energy levels with increasing a.
• We know the wavefunction is zero in regions I and III. We also know thatthe wave function is not zero everywhere. Therefore it must do something
between x = 0 and x = a. It must have some curvature and hence some zero
point energy.
• As a is increased, the wavefunction is less confined and so the curvature doesnot need to be as great to satisfy the boundary conditions. Therefore the
energy levels decrease in energy as does their difference.
The particle in a box problem illustrates some of the many strange features of
quantum mechanics.
We have already seen such nonclassical behavior as quantized energy and zero
point energy.
As another example consider the expectation value of position for a particle in
the second quantum level:
hxi =Z ∞
−∞ψ∗2(x)xψ2(x)dx =
2
a
Z a
0
x sin2[2π
ax]dx =
a
2(4.14)
36
yet the probability of finding the particle at x = a2is zero: ψ2(
a2) = 0. There is
a node at x = a2. So even though the particle may be found anywhere else in the
box and it may get from the left side of the node to the right side, it can never
be found at the node.
37
5. The Harmonic Oscillator
The harmonic oscillator model which is simply a mass undergoing simple harmonic
motion. The classical example is a ball on a spring
The harmonic oscillator is arguably the single most important model in all of
physics.
We shall begin by reviewing the classical harmonic oscillator and than we will
turn our attention to the quantum oscillator.
The force exerted by the spring in the above figure is F = −k(R−Req), where k
is the spring constant and Req is the equilibrium position of the ball.
Setting x = R − Req we can measure the displacement about the equilibrium
position.
38
38
From Newton’s law of motion F = ma = md2xdt2
, we get
md2x
dt2= −kx⇒ d2x
dt2+
k
mx = 0 (5.1)
This is second order differential equation which we already know the solutions to:
x = A sinωt+B cosωt, (5.2)
where ω =q
kmand A and B are constants which are determined by the initial
conditions.
For quantum mechanics it is much more convenient to talk about energy rather
than forces, so in going to the quantum oscillator, we need to express the force of
the spring in terms of potential energy V . We know
V = −Z
Fdx =1
2kx2 + C. (5.3)
Since energy is on an arbitrary scale we can set C = 0. Thus V = 12kx2.
By postulate III the Schrödinger equation becomes
Hψ = Eψ ⇒
⎛⎜⎝−~22m
d2
dx2| z K.E.
+1
2kx2| z P.E.
⎞⎟⎠ψ = Eψ. (5.4)
This can be rearrange into the form
−~22m
d2ψ
dx2+
µ1
2kx2 − E
¶ψ = 0 (5.5)
This differential equation is not easy to solve (you can wait to solve it in graduate
school).
39
The equation is very close to the form of a know differential equation called Her-
mite’s differential equation the solutions of which are called the Hermite polynom-
inals.
As it turns out, the solutions (the eigenfunctions) to the Schrödinger equation for
the harmonic oscillator are
ψn(y) = AnHn(y)e− y2
2 , y =
µkm
~2
¶ 14
x, An =1p
2nn!√π, (5.6)
where An is the normalization constant for the nth eigenfunction and Hn(y) are
the Hermite polynomials.
The eigenvalues (the energy levels) are
En = (n+1
2)~ω, (5.7)
where again ω =q
km.
Note the energy levels are often written as
En = (n+1
2)hν0, (5.8)
where ν0 = 12π
qkmand is called the vibrational constant.
∗ ∗ ∗ See Fig. 11.12 Laidler&Meiser ∗ ∗∗
5.1. Interesting Aspects of the Quantum Harmonic Oscilla-
tor
It is interesting to investigate some of the unintuitive properties of the oscillator
as we have gone quantum mechanical
40
1. Consider the ground state (the lowest energy level)
• There is residual energy in the ground state because
E0 = (0 +1
2)~ω.
• Just like for the particle in a box, this energy is called the zero pointenergy.
• It is a consequence of uncertainty principle
— If the ground state energy was really zero, then we would concludethat the momentum of the oscillator was zero.
— On the other hand, we would conclude the particle was located atthe bottom of the potential well (at x = 0)
— Thus we would have δp = 0, δx = 0, so δpδx = 0 Not allowed!
— The uncertainty principle forces there to be some residual zeropoint energy.
2. Consider the wavefunctions.
• The wavefunctions penetrate into the region where the classical particleis forbidden to go
— The wavefunction is nonzero past the classical turning point.
• The probability distribution |ψ|2 becomes more and more like what isexpected for the classical oscillator when v →∞.
— This is a manifestation of the correspondence principle whichstates that for large quantum numbers, the quantum system must
behave like a classical system. In other words the quantum me-
chanics must contain classical mechanics as a limit.
3. Interpretation of the wavefunctions and energy levels
41
• Remember the wavefunctions are time independent and the energy lev-els are stationary
• If a molecule is in a particular vibrational state it is NOT vibrating.
5.2. Spectroscopy (An Introduction)
The primary method of measuring the energy levels of a material is through the
use of electromagnetic radiation.
Experiments involving electromagnetic radiation—matter interaction are called
spectroscopies.
Atoms and molecules absorb or emit light only at specific (quantized) energies.
These specific values correspond to the energy level difference between the initial
and final states.
42
Key Equations for Exam 1
Listed here are some of the key equations for Exam 1. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• The short cut for getting the normalization constant (1D, see above for 3D).
N =
sZspace
|ψunnorm(x)|2 dx. (5.9)
• The normalized wavefunction:
ψnorm =1
Nψunnorm. (5.10)
• The Schrödinger equation (which should be posted on your refrigerator),
Hψ = Eψ. (5.11)
43
43
• The Schrödinger equation for 1D problems as a differential equation,
−~22m
d2
dx2ψ + (V (x)−E)ψ = 0. (5.12)
• How to get the average value for some property (1D version),
hαi =Zspace
ψ∗αψdx. (5.13)
• The momentum operator
px = −i~∂
∂x. (5.14)
• Normalized wavefunctions for the 1D particle in a box,
ψn(x) =
r2
asin
nπx
a. (5.15)
• The energy levels for the 1D particle in a box,
En =n2π2~2
2ma2~= h
2π=n2h2
8ma2. (5.16)
• The energy level spacing for the 1D particle in a box,
4E = (2n+ 1)h2
8ma2(5.17)
• The wavefunctions for the harmonic oscillator are
ψn(y) = AnHn(y)e−y2
2 , y =
µkm
~2
¶ 14
x, An =1p
2nn!√π, (5.18)
where An is the normalization constant for the nth eigenfunction and Hn(y)
are the Hermite polynomials.
• The energy levels are
En = (n+1
2)~ω, ω =
rk
m(5.19)
44
Part II
Quantum Mechanics of Atomsand Molecules
45
45
6. Hydrogenic Systems
Now that we have developed the formalism of quantum theory and have discussed
several important systems, we move onto the quantum mechanical treatment of
atoms.
Hydrogen is the only atom for which we can exactly solve the Schrödinger equation
for. So this will be the first atomic system we discuss.
The Schrödinger equation for all the other atoms on the periodic table must be
solved by approximate methods.
6.1. Hydrogenic systems
Hydrogenic systems are those atomic systems which consist of a nucleus and one
electron. The Hydrogen atom (one proton and one electron) is the obvious exam-
ple
Ions such as He+ and Li2+ are also hydrogenic systems.
These system are centrosymmetric. That is they are completely symmetric about
the nucleus.
The obvious choice for the coordinate system is to use spherical polar coordinates
46
46
with the origin located on the nucleus.
The classical potential energy for these hydrogenic systems is
V (r) =−Ze2(4π 0)r
. (6.1)
So the Hamiltonian is
H =−~22me∇2 + −Ze2
(4π 0)r. (6.2)
Schrödinger’s equation (in spherical polar coordinates) becomes
Eψ = Hψ (6.3)
Eψ =
µ−~22me∇2 + −Ze2
(4π 0)r
¶ψ
Eψ =
µ−~22me
∙1
r2∂
∂rr2
∂
∂r+1
r2
µ1
sin θ
∂
∂θsin θ
∂
∂θ+
1
sin2 θ
∂2
∂φ2
¶¸+−Ze2(4π 0)r
¶ψ
The Hamiltonian is (almost) the sum of a radial part (only a function of r) and
an angular part (only a function of θ and φ):
H = Hrad +1
r2Hang, (6.4)
Hrad =−~22me
∙1
r2∂
∂rr2
∂
∂r− Ze2
(4π 0)r
¸(6.5)
and
Hang =−~22me
µ1
sin θ
∂
∂θsin θ
∂
∂θ+
1
sin2 θ
∂2
∂φ2
¶(6.6)
Since the Hamiltonian is the sum of two terms, ψ must be a product state.
ψ(r, θ, φ) = ψrad(r)ψang(θ, φ) (6.7)
It turns out that solving the Schrödinger equation,
Hangψang(θ, φ) = Eψang(θ, φ), (6.8)
47
yields
ψang(θ, φ) = Ylm(θ, φ), (6.9)
where the Ylm(θ, φ)’s are the spherical harmonic functions characterized by quan-
tum numbers l andm. The spherical harmonics are known functions. (Mathematica
knows them and you can use them just like any other built-in function like sine
or cosine.)
We shall use the spherical harmonics more next semester when we develop the
quantum theory of angular momentum.
It also turns out that the energy associated with Hang is found to be
E = El =l(l + 1)~2
2me. (6.10)
So,
Hangψang(θ, φ) =l(l + 1)~2
2meψang(θ, φ) (6.11)
Now let’s denote the radial part of the wavefunction as ψrad(r) = R(r).
The full Schrödinger equation becomes
Hψ(r, θ, φ) = Eψ(r, θ, φ) (6.12)
HR(r)Ylm(θ, φ) = ER(r)Ylm(θ, φ)µHrad +
1
r2Hang
¶R(r)Ylm(θ, φ) = ER(r)Ylm(θ, φ),
Operating with Hang we getµHrad +
l(l + 1)~2
2mer2
¶R(r)Ylm(θ, φ) = ER(r)Ylm(θ, φ) (6.13)
48
The Ylm(θ, φ) can now be cancelled to leave a one dimensional differential equation:
−~22me
µ1
r2∂
∂rr2
∂
∂r− Ze2
4π 0r− l(l + 1)
r2
¶R(r) = ER(r). (6.14)
This differential equation is very similar to a known equation called Laguerre’s
differential equation which has as solutions the Laguerre polynomials Lln(x).
In fact, the solutions to our differential equation are closely related to the Laguerre
polynomials.
Rnl(σ) = Anl
µ2σ
n
¶l
e−σ/nL2l+1n+1
µ2σ
n
¶, (6.15)
where the normalization constant, Anl, depends on the n and l quantum numbers
as
Anl = −
sµ2Z
na0
¶3(n− l − 1)!2n[(n+ l)!]3
(6.16)
The energy eigenvalues, i.e., the energy levels are given by
En = −Z2Rn2
(6.17)
Note: The energy levels are determined by n alone–l drops out.
Also Note: the energy levels are the same as for the Bohr model.
So, the total wavefunction that describes a hydrogenic system (ignoring the spin
of the electron, which will be briefly discussed later) is
ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ) (6.18)
6.2. Discussion of the Wavefunctions
We are now very close to having the atomic orbitals familiar from freshman chem-
istry.
49
We have explicitly derived the “physicists” picture of the atomic orbitals
orbital n l m wavefunctions (σ = r/a0)
1s 1 0 0 ψ1s = ψ100 = e−σ
2s 2 0 0 ψ2s = ψ200 =¡1− σ
2
¢e−σ/2
2p 2 1 0 ψ2p0 = ψ210 = σe−σ/2 cos θ
2 1 ±1 ψ2p±1 = ψ21±1 = σe−σ/2 sin θe±iφ
3d 3 2 0 ψ3d0 = ψ320 = σ2e−σ/3 (3 cos2 θ − 1)3 2 ±1 ψ3d±1 = ψ32±1 = R32(r) cos θ sin θe
±iφ
3 2 ±2 ψ3d±2 = ψ32±2 = R32(r) sin2 θe±i2φ
The wavefunctions in the “physicists” picture are complex (they have real and
imaginary components). The wavefunctions that chemists like are pure real. So
one needs to form linear combinations of these orbitals such that these combina-
tions are pure real.
The atomic orbital you are used to from freshman chemistry are the “chemists”
picture of atomic orbitals
In the above table ψ1s, ψ2s, ψ2p0, ψ3d0 are pure real and so these are the same in
the “chemists” picture as in the “physicists” picture.
The table below lists the atomic orbitals in the “chemists” picture as linear com-
binations of the “physicists” picture wave functions.
50
orbital n l m wavefunctions (σ = r/a0)
1s 1 0 0 ψ1s = ψ1s
2s 2 0 0 ψ2s = ψ2s
2p 2 1 0 ψpz = ψ2p02 1 ±1 ψ2px =
1√2
£ψ2p1 + ψ2p−1
¤2 1 ±1 ψ2py =
1i√2
£ψ2p1 − ψ2p−1
¤3d 3 2 0 ψ3dz2 = ψ3d0
3 2 ±1 ψ3dxz =1√2
£ψ3d1 + ψ3d−1
¤3 2 ±1 ψ3dyz =
1i√2
£ψ3d1 − ψ3d−1
¤3 2 ±2 ψ3dxy =
1√2
£ψ3d2 + ψ3d−2
¤ψ3dx2−y2 =
1i√2
£ψ3d2 − ψ3d−2
¤6.3. Spin of the electron
As we know from freshman chemistry, electrons also posses an intrinsic quantity
called spin.
Spin is actually rather peculiar so we will put off a more detailed discussion until
next semester.
For now we must be satisfied with the following:
• There are two quantum numbers associated with spin: s and ms
• s is the spin quantum number and for an electron s = 1/2 (always).
• ms is the spin orientation quantum number and ms = ±1/2 for electrons.
The spin wavefunction is a function in spin space not the usual coordinate space,
so we can not write down an explicit function of the coordinate space variables.
51
We simply denote the spin wavefunction generally as χs,msand “tack it on” as
another factor of the complete wavefunction.
When a particular spin state is needed a further notation is commonly used:
α ≡ χ 12, 12(the “spin-up” state) and β ≡ χ 1
2,−1
2(the “spin-down” state)
6.4. Summary: the Complete Hydrogenic Wavefunction
We are now in position to fully describe all properties of hydrogenic systems
(except for relativistic effects)
The full wave function is
Ψn,l,m,s,ms = ψn,l,mχs,ms(6.19)
= Rnl(r)Yl,m(θ, φ)χ
The energy is given by
En = −Z2Rn2
, (6.20)
where recall. Again note that for a free hydrogenic system the total energy depends
only on the principle quantum number n.
The quantum numbers of the hydrogenic system
• The principle quantum number, n: determines the total energy of the sys-
tems and the atomic shells.
— The principle quantum number, n, can take on values of 1,2,3. . .
• The angular momentum quantum numbers, l: determines the total angular
momentum of the system. It also determines the atomic sub-shells
52
— The angular momentum quantum number, l, can take on values of 0,
1, . . . (n− 1)
— For historical reasons l = 0 is called s, l = 1 is called p, l = 2 is called
d, l = 3 is called f etc.
• The orientation quantum number, m: determine the projection of the an-
gular momentum onto the z-axis. It also determines the orientation of the
atomic sub-shells
— The magnetic quantum number, m, can take on values of 0, ±1, . . .± l.
• The spin quantum number, s: determines the total spin angular momentum.
— For electrons s = 1/2.
• The spin orientation quantum number, ms: determines the projection of the
spin angular momentum onto the z-axis (i.e., spin-up or spin-down).
— For electrons ms = ±1/2
We have accomplished quite a bit. We have determined all that we can about the
hydrogen atom within Schrödinger’s theory of quantum mechanics.
This is not the full story however. The Schrödinger theory is a non-relativistic
one; that is, it can not account for relativistic effects which show up in spectral
data. We also had to add spin in an ad hoc manner to account for what we know
experimentally–spin did not fall out of the theory naturally.
Dirac, in the late 1920’s, developed a relativistic quantum theory in which the
well established phenomenon of spin arose naturally. His theory also made the
53
bold prediction of the existence anti-matter that has now been verified time and
again.
The Dirac theory was still not fully complete, because there still existed exper-
imental phenomena that was not properly described. In 1948 Richard Feynman
developed the beginnings of quantum electrodynamics (QED). QED is the best
theory ever developed in terms of matching with experimental data.
Both the relativistic Dirac theory and QED are beyond our reach, so we limit
ourselves to the non-relativistic Schrödinger theory.
54
7. Multi-electron atoms
7.1. Two Electron Atoms: Helium
We now consider a system consisting of two electrons and a nucleus; for example,
helium.
Although the extension from hydrogen to helium seems simple it is actually ex-
tremely complicated. In fact, it is so complicated that it can’t be solved exactly.
The helium atom is an example of the “three-body-problem”–difficult to handle
even in classical mechanics–one can not get a closed form solution.
The Hamiltonian for helium is
H = − ~2
2me∇21| z
K.E of electron 1
− ~2
2me∇22| z
K.E of electron 2
− Ze2
4π 0r1| z P.E of electron 1
− Ze2
4π 0r2| z P.E of eletcron 2
+e2
4π 0r12| z elec.—elec. repulsion
, (7.1)
where r12 = |r1 − r2| is the distance between the electrons.
The electron—electron repulsion term is responsible for the difficulty of the prob-
lem. It makes a closed form solution impossible.
The problem must be solved by one of the following methods
• Numerical solutions (we will not discuss this)
55
55
• Perturbation theory (next semester)
• Variational theory (next semester)
• Ignore the electron—electron repulsion (good for qualitative work only)
7.2. The Pauli Exclusion Principle
Electron are fundamentally indistinguishable. They can not truly be la-belled.
All physical properties of a system where we have labelled the electrons as, say, 1
and 2 must be exactly the same as when the electrons are labelled 2 and 1.
Now, only |ψ|2 is directly measurable–not ψ itself.
All this implies that
ψ(1, 2) =
⎧⎪⎨⎪⎩+ψ(2, 1), symmetric
or
−ψ(2, 1) antisymmetric
(7.2)
The Pauli exclusion principle states: The total wavefunctions for fermions(e.g., electrons) must be antisymmetric under the exchange of indistinguishable
fermions.
Note: a similar statement exists for bosons (e.g., photons): The total wavefunction
for bosons must be symmetric under exchange of indistinguishable bosons.
Let us consider the two electron atom, helium
56
The total wavefunction is
Ψ = ψ(1, 2)χ(1, 2) (7.3)
Since a complete solution for helium is not possible we must use approximate
wavefunctions. Since we are doing this, we may as well simplify matters and use
product state wavefunctions (products of the hydrogenic wavefunctions).
Ψ = ψ(1)ψ(2)| z spatial part
χ(1)χ(2)| z spin part
, (7.4)
where the single particle wavefunctions are that of the hydrogenic system.
The Pauli exclusion principle implies that if the spatial part is even with respect
to exchange then the spin part must be odd. Likewise if the spatial part is odd
then the spin part must be even.
Now let’s blindly list all possibilities for the ground state wave function of helium
Ψa = ψ1s(1)α(1)ψ1s(2)α(2) (7.5)
Ψb = ψ1s(1)α(1)ψ1s(2)β(2)
Ψc = ψ1s(1)β(1)ψ1s(2)α(2)
Ψd = ψ1s(1)β(1)ψ1s(2)β(2)
These appear to be four reasonable ground state wavefunctions which would im-
ply a four-fold degeneracy. However considering the symmetry with respect to
exchange we see the following
• Ψa has symmetric spatial and spin parts and is there for symmetric. It must
be excluded.
• Similarly for Ψd.
• Ψb and Ψc have symmetric spatial parts, but the spin part is neither sym-
metric or antisymmetric. So, one must make an antisymmetric linear com-
bination of the spin parts.
57
The appropriate linear combination is
α(1)β(2)− α(2)β(1). (7.6)
So the ground state wave function for helium is
Ψg = ψ1s(1)ψ1s(2) [α(1)β(2)− α(2)β(1)] . (7.7)
Consequences of the Pauli exclusion principle
• No two electrons can have the same five quantum numbers
• Electrons occupying that same subshell must have opposite spins
7.3. Many Electron Atoms
The remaining atoms on the periodic table are handled in a manner similar to
helium.
Namely the wavefunction is product state that must be antisymmeterized in ac-
cordance with the Pauli exclusion principle.
The product wavefunction for the ground state is determined by applying the
aufbau principle. The aufbau principle states that the ground state wavefunction
is built-up of hydrogenic wavefunctions
To arrive at an antisymmetric wavefunction we construct the Slater determinant :
Ψ =
¯¯¯ψ1s(1)α(1) ψ1s(1)β(1) · · · ψn(1)α(1) ψn(1)β(1)
ψ1s(2)α(2) ψ1s(2)β(2) · · · ψn(2)α(2) ψn(2)β(2)...
......
......
ψ1s(N)α(N) ψ1s(N)β(N) ψn(N)α(N) ψn(N)β(N)
¯¯¯ (7.8)
58
The reason one can be sure that this wavefunction is the antisymmeterized is that
we know from linear algebra that the determinant is antisymmetric under exchange
of rows (corresponds to exchanging two electrons). It is also antisymmetric under
exchange of columns.
Another property of the determinant is that if two rows are the same (corresponds
to two electrons in the same state) the determinant is zero. This agrees with the
Puli exclusion principle.
As an example consider lithium:
• There are three electrons so we need three hydrogenic wavefunctions: ψ1sα,ψ1sβ, and ψ2sα (or ψ2sβ).
• We construct the Slater determinant as
Ψ1 =
¯¯ ψ1s(1)α(1) ψ1s(1)β(1) ψ2s(1)α(1)
ψ1s(2)α(2) ψ1s(2)β(2) ψ2s(2)α(2)
ψ1s(3)α(3) ψ1s(3)β(3) ψ2s(3)α(3)
¯¯ (7.9)
or
Ψ2 =
¯¯ ψ1s(1)α(1) ψ1s(1)β(1) ψ2s(1)β(1)
ψ1s(2)α(2) ψ1s(2)β(2) ψ2s(2)β(2)
ψ1s(3)α(3) ψ1s(3)β(3) ψ2s(3)β(3)
¯¯ (7.10)
• The short hand notation for these states is (1s)2(2s)1
7.3.1. The Total Hamiltonian
The total Hamiltonian for a many electron (ignoring spin-orbit coupling which
will be discussed next semester) atom is
H =NXi=1
"−~22me∇2i −
Ze2
4π 0ri+Xj>i
e2
4π 0rij
#(7.11)
59
8. Diatomic Molecules and the Born
Oppenheimer Approximation
Now that we have applied quantum mechanics to atoms, we are able to begin the
discussion of molecules.
This chapter will be limited to diatomic molecules.
8.1. Molecular Energy
A diatomic molecule with n electrons requires that 3n+6 coordinates be specified.
Three of these describe the center of mass position.
3n of these describe the position of the n electrons.
This leaves three degrees of freedom (R, θ, φ) which describe the position of the
nuclei relative to the center of mass. R determines the internuclear separation
and θ and φ determine the orientation.
60
60
8.1.1. The Hamiltonian
In the center of mass coordinates the Hamiltonian for a diatomic molecule is
H = TN + Te + VNN + VNe + Vee. (8.1)
TN is the nuclear kinetic energy operator and is given by
TN = −~2
2μ∇2N = −
~2
2μR2∂
∂RR2
∂
∂R+~2
2μJ2, (8.2)
where J is angular momentum operator for molecular rotation and μ = m1m2
m1+m2is
the reduced mass of the diatomic molecule.
Te =P
i− ~22me∇2ei is the kinetic energy operator for the electrons.
VNN =ZAZBee
2
4π 0Ris the nuclear—nuclear potential energy operator.
VNe = −P
i
hZAe
2
4π 0rAi+ ZBe
2
4π 0rBi
iis the nuclear—electron potential energy operator.
Vee =P
i>je2
4π 0rjiis the electron—electron potential energy operator.
61
8.1.2. The Born—Oppenheimer Approximation
The Born—Oppenheimer approximation: The nuclei move much slower thanthe electrons. (classical picture)
We put the Born—Oppenheimer approximation to work by first defining an effec-
tive Hamiltonian
Heff = Te + VNN + VNe + Vee. (8.3)
The approximation comes in by treating R as a parameter rather than an operator
(or variable). So one writes
Heffψe(R, ri) = Ee(R)ψe(R, ri). (8.4)
ψe is the so-called electronic wavefunction.
Now the Schrödinger equation for the diatomic molecule is³TN + Heff
´ψ(R, ri) = Eψ(R, ri). (8.5)
Since the Hamiltonian is a sum of two terms, one can write the wavefunction
ψ(R, ri) as a product wavefunction
ψ = ψNψe, (8.6)
where ψN is the so-called nuclear wavefunction.
Substituting the product wavefunction into the Schrödinger equation gives³TN + Heff
´ψNψe = EψNψe (8.7)³
TN +Ee(R)´ψNψe/ = EψNψe/³
TN +Ee(R)´ψN = EψN .
62
The last equation is exactly like a Schrödinger equation with a potential equal to
Ee(R).
One now models Ee(R) or determines it experimentally.
8.2. Molecular Vibrations
As stated earlier R is the internuclear separation and θ and φ determine the
orientation. Consequently, R is the variable involved with vibration whereas θ
and φ are involved with rotation.
Considering only the R part of the Hamiltonian (under the Born—Oppenheimer
approximation), we have∙− ~
2
2μ
∂2
∂R2+Ee(R)
¸ψvib = Evibψvib. (8.8)
It is convenient at this point to expand Ee(R) in a Taylor series about the equi-
librium position, Req:
Ee(R) = E0 +
µ∂E
∂R
¶Req
(R−Req) +1
2!
µ∂2E
∂R2
¶Req
(R−Req)2 + · · · . (8.9)
Now E0 is just a constant which, by choice of the zero of energy, can be set to an
arbitrary value.
Since we are at a minimum,¡∂E∂R
¢Req
must be zero, so the linear term vanishes.
One defines³∂2E∂R2
´Req≡ ke as the force constant.
The remaining terms in the expansion can collective be defined as O[(R−Req)3] ≡Vanh, the anharmonic potential.
63
As a first approximation we can neglect the anharmonicity. With this, the Schrödinger
equation becomes∙− ~
2
2μ
∂2
∂R2+1
2ke(R−Req)
2
¸ψvib = Evibψvib. (8.10)
If we let x = (R−Req) this becomes∙− ~
2
2μ
∂2
∂x2+1
2kex
2
¸ψvib = Evibψvib, (8.11)
which is exactly the harmonic oscillator equation. Hence
ψvib,n = AnHn(√αx)e−αx
2/2, (8.12)
where α ≡q
keμ~ .
And
Evib,n = hcωe(n+1
2), (8.13)
where ωe ≡ 12π
qkeμ.
8.2.1. The Morse Oscillator
Neglecting anharmonicity and using the harmonic oscillator approximation works
well for low energies. However, it is a poor model for high energies.
For high energies we need a more realistic potential–one that will allow of bond
dissociation.
The Morse potential
Ee(R) = De[1− e−β(R−Req )]2, (8.14)
64
where De is the well depth and β = 2πcωe
qμ2De
is the Morse parameter. Note:
this expression for the Morse potential has the zero of energy at the bottom of
the well (i.e. R = Req, ;Ee(Req) = 0).
The Morse Potential can also be written as
Ee(R) = De[e−2β(R−Req ) − 2e−β(R−Req )]. (8.15)
Now the zero of energy is the dissociated state (i.e. R→∞, ;Ee(R→∞) = 0).
We approach this quantum mechanical problem exactly like all the other.
The Schrödinger equation is∙− ~
2
2μ
∂2
∂R2+De[1− e−β(R−Req )]2
¸ψvib = Evibψvib (8.16)
This is another differential equation that is difficult to solve.
As it turns out, this Schrödinger equation can be transformed into a one of a broad
class of known differential equations called confluent hypergeometric equations–
the solutions of which are the confluent hypergeometric functions, 1F1.
Doing this yields the wavefunctions of the form
ψvib,n(z) = zApne−z1F1(−n, 1 + 2Apn, 2z), (8.17)
z =
√2Deμ
βhe−βx,
A =
√2μ
βh,
pn =pDe +
−12− n
A
and energy levels of the form
Evib,n = −De + hcωe(n+1
2)− hcωexe(n+
1
2)2, (8.18)
65
where ωexe together is the anharmonicity constant, with xe =hcωe4De
.
∗ ∗ ∗ See Handout ∗ ∗∗
8.2.2. Vibrational Spectroscopy
Infrared (IR) and Raman spectroscopy are the two most widely used techniques
to probe vibrational levels.
The spectral peaks appear at v = 4Ehc(in units of wavenumbers, cm−1).
The transition from the n = 0 to the n = 1 state is called the fundamental
transition.
Transitions from n = 0 to n = 2, 3, 4 · · · are called overtone transitions.
Transitions from n = 1 to 2, 3, 4 · · · , n = 2 to 3, 4, 5 · · · , etc. are called hottransitions (or hot bands)
Since the energy levels depend on mass, isotopes will have a different transition
energy and hence appear in a different place in the spectrum. Heavier isotopes
have lower transition energies.
66
9. Molecular Orbital Theory and
Symmetry
9.1. Molecular Orbital Theory
One of the most important concepts in all of chemistry is the chemical bond.
In freshman chemistry we learn of one model for chemical bonding–VSEPR (va-
lence shell electron-pair repulsion) theory, where hybridized atomic orbitals deter-
mine the bonding geometry of a given molecule.
We are now prepared to discuss a bonding theory that is more rigorously based
in quantum mechanics.
Basically we will treat the molecules in the same way as all our other quantum
mechanical problems (e.g., particle in a box, harmonic oscillator, etc.)
As you might expect, it is not possible to obtain the exact wavefunctions and
energy levels so, we must settle for approximate solutions.
As a first example, let us consider the molecular hydrogen ion H+2 .
The Hamiltonianfor H+2 is
H = TN + Tel + VNel + VNN (9.1)
67
67
We use the Born-Oppenheimer approximation and treat the nuclear coordinates
as a parameters rather than as variables. So we only worry about parts of the
Hamiltonian that deal with the electron.
The effective Hamiltonian becomes
H = Tel + VNel (9.2)
=−~22me∇2 − e2
4π 0rA− e2
4π 0rB.
The eigenfunctions of this Hamiltonian are called molecular orbitals.
The molecular orbitals are the analogues of the atomic orbitals.
• Atomic orbitals: Hydrogen is the prototype and all other atomic orbitalsare built from the hydrogen atomic orbitals.
• Molecular orbitals: The hydrogen molecular ion is the prototype and allother molecular orbitals are built from the hydrogen molecular ion molecular
orbitals.
There is one significant difference between the above, which is the hydrogen atomic
orbitals are exact whereas the hydrogen molecular ion molecular orbitals are not
exact.
In fact, we shall see that these molecular orbitals are constructed as linear com-
binations of atomic orbitals.
9.2. Symmetry
Let the atoms of the hydrogen molecular ion lie on the z-axis of the center of mass
coordinate system.
68
Inversion symmetry
• The potential field of the hydrogen molecular ion is cylindrically symmetricabout the z-axis.
• Because of the symmetry the electron density at (x, y, z) must equal theelectron density at (−x,−y,−z).
• The above symmetry therefore requires that the molecular orbitals be eigen-functions of the inversion operator, ı. That is
ıψ(x, y, z) = ψ(−x,−y,−z) = aψ(x, y, z). (9.3)
• Moreover the eigenvalue a can be either +1 or −1.
• If a = +1 the molecular wavefunction is even with respect to inversion andis called gerade and labelled with a “g”: ıψg = ψg
• If a = −1 the molecular wavefunction is odd with respect to inversion andis called ungerade and labelled with a “u”: ıψu = −ψu
• The terms gerade and ungerade apply only to systems that posses inversionsymmetry.
Cylindrical symmetry
69
• The cylindrical symmetry implies that the potential energy can not dependon the φ.
• The molecular wavefunction is described by an eigenvalue λ = 0,±1,±2, . . .
— We use λ to label the molecular orbitals as shown in the table
λ 0 ±1 ±2 · · ·label σ π δ · · ·
Mirror plane symmetry
70
• There is also a symmetry about the x-y plane called horizontal mirror planesymmetry: operator σh.
• Thus the molecular wavefunction must be an eigenfunction of σh with eigen-value ±1.
— If the eigenvalue is +1 (even with respect to σh) the molecular orbitalis called a bonding orbital.
— If the eigenvalue is −1 (odd with respect to σh) the molecular orbitalis called an antibonding orbital.
• There are also vertical mirror plane symmetries, but we will put that dis-cussion off for the time being.
71
10. Molecular Orbital Diagrams
10.1. LCAO–Linear Combinations of Atomic Orbitals
Now that we know what symmetry the molecular orbitals must posses, we need
to find some useful approximations for them.
Useful can mean qualitatively useful or quantitatively useful.
Unfortunately we can’t have both.
We will discuss the approximation which models the molecular orbitals as linear
combinations of atomic orbitals (LCAO).
LCAO is qualitatively very useful but it lacks quantitative precision.
Let us again consider the hydrogen molecular ion H+2 : let one H atom be labelled
A and the other labelled B.
Linear combination of the 1s atomic orbital from each H atom is used for the
molecular orbital of H+2 :
(1sA) = ke−rA/a0 (10.1)
and
(1sB) = ke−rB/a0 (10.2)
72
72
We construct two molecular orbitals as
Φ+ = C+(1sA + 1sB) (10.3)
and
Φ− = C−(1sA − 1sB) (10.4)
The normalization condition is ZΦ±Φ±dΩ = 1 (10.5)
As can be seen from the above figure, Φ+ represents a situation in which the
electron density is concentrated between the nuclei and thus represents a bonding
orbital.
Conversely Φ− represents a situation in which the electron density is very low
between the nuclei and thus represents an antibonding orbital
10.1.1. Classification of Molecular Orbitals
With atoms we classified atomic orbitals according to angular momentum.
For molecular orbitals we shall also classify them according to angular momentum.
But we shall also classify them according to their inversion symmetry and wether
or not they are bonding or antibonding.
73
The classification according to angular momentum is as follows.
λ 0 ±1 ±2 · · ·orbital symbol σ π δ · · ·
Atomic orbitals with m = 0 form σ type molecular orbitals, e.g., s⇒ σ, pz ⇒ σ.
Those with m = ±1 form π type molecular orbitals, e.g., px ⇒ π etc.
The classification according to inversion symmetry is simply a subscript “g” or
“u”. For example, σg or σu etc.
The classification according to bonding or antibonding is an asterisk is used to
denote antibonding. For example, σg is a bonding orbital and σ∗u is an antibonding
orbital.
10.2. The Hydrogen Molecule
Let us now consider the hydrogen molecule. This molecules is a homonuclear
diatomic with two electrons.
If the two atoms are infinitely far apart. The ground state of the system would
consist of two separate hydrogen molecules in their ground atomic states: (1s)1
74
As the atom are brought closer together, their respective s orbitals begin to over-
lap.
It is now more appropriate to speak in terms of molecular orbitals, so one forms
linear combinations of the atomic orbitals.
There are two acceptable linear combinations. These are
σg = 1sA + 1sB (10.6)
and
σ∗u = 1sA − 1sB. (10.7)
75
It can be shown mathematically that the energy level associated with σg is lower
than σ∗u.
We can intuit this qualitatively however since the σ∗u orbital must have a node
whereas the σg does not.
It is also to be expected since we know H2 is a stable molecule.
10.3. Molecular Orbital Diagrams
The energy levels associated with the molecular orbitals are drawn schematically
is what is called a molecular orbital diagram.
The molecular orbital diagram for H2 is shown below
Molecular orbital diagrams can be drawn for any molecule. Some get very compli-
cated. We will focus on the second row homonuclear diatomics and some simple
heteronuclear diatomics.
76
The molecular orbital diagrams for the second row homonuclear diatomics are
rather simple.
∗ ∗ ∗ See Supplement ∗ ∗∗
The supplement that follows this section contains examples for each of the second
row diatomics.
Heteronuclear diatomics are some what more complicated since there is a disparity
in the energy levels of the atomic orbitals for the separated atoms. This disparity
is not present for homonuclear diatomics.
A consequence of this energy level disparity is that molecular orbitals may be
formed from nonidentical atomic orbitals. For example a high lying 1s orbital
may combine with a low lying 2s orbital to form a σ molecular orbital.
The supplement that follows this section contains some examples of heteronuclear
diatomics.
Bond order
• One important property that can be predicted from the molecular orbital
diagrams is bond order.
• Bond order is defined as
BO =1
2(# of bonding electrons−# of antibonding electrons) (10.8)
• Examples follow in the supplement.
77
10.4. The Complete Molecular Hamiltonian and Wavefunc-
tion
We have discussed molecular vibrations which under the Born-Oppenheimer ap-
proximation are governed by the vibrational Hamiltonian and described by the
vibrational wavefunction.
Likewise we have discussed molecular orbitals which are the electronic wavefunc-
tions.
Next semester we will discuss molecular rotations and just like for vibrations
and electronic transitions they are governed by the rotational Hamiltonian and
described by the rotational wavefunction.
We can succinctly express the Schrödinger equation for a molecule as follows.
(Next semester will we look at the details of this for polyatomic molecules)
HmolΨmol = EmolΨmol (10.9)³Hele + Hvib + Hrot
´ψeleψvibψrot = (Eele +Evib +Erot)ψeleψvibψrot
78
11. An Aside: Light Scattering–Why
the Sky is Blue
This chapter addresses the topic of light scattering from two different perspectives.
• Classical electrodynamics
• Classical statistical mechanics
Since this is not a course on electrodynamics, we have to take several key results
from that theory on faith.
11.1. The Classical Electrodynamics Treatment of Light Scat-
tering
As usual we work under the electric dipole approximation and only focus on the
interaction of the electric field part of light with a dipole.
When the light interacts with the molecule an electric dipole is induced according
to
μ = αE, (11.1)
where α is the polarizability of the molecule describing the “flexibility” of its
electron cloud.
79
79
For light, the electric field part is
E(t) = E0 cosωt. (11.2)
The polarizability also depends on the positions of nuclei to some degree. That
is, there is a vibrational (and rotational) contribution to the polarizability:
α(t) = α0 + α1 cosωvt (11.3)
(here for simplicity we assume only one vibrational mode).
Thus the light—matter interaction is described as
μ(t) = α(t)E(t) = (α0 + α1 cosωvt)E0 cosωt (11.4)
= α0E0 cosωt+ α1E0 cosωvt cosωt
= α0E0 cosωt| z Rayleigh
+α1E02
⎡⎣cos(ω − ωv)t| z Stokes Raman
+ cos(ω + ωv)t| z AntiStokes Raman
⎤⎦where a trig identity was used in the last step.
According to classical electrodynamics an oscillating dipole emits an electromag-
netic field at the oscillation frequency.
In this case we see the dipole oscillates at three distinct frequencies: ω, ω − ωv
and ω + ωv as part of three terms in the above expression.
The first term corresponds to Rayleigh scattering where the scattered light is at
the same frequency as the incident light.
The second term corresponds to Stokes Raman scattering where the scattered
light is shifted to the red of the incident frequency.
80
The third term corresponds to anti-Stokes Raman scattering where the scattered
light is shifted to the blue of the incident frequency.
Classical electrodynamics can describe exactly how the oscillating electric dipole
emits electromagnetic radiation. It can be shown that the emitted intensity is
I =ω4
3c3μ20, (11.5)
where μ0 = α0E0 for the case of Rayleigh scattering and μ0 = α1E0/2 for the case
of Raman scattering.
To explicitly derive this expression we would need a fair bit of electrodynamics
and so the derivation is not shown here.
The important point to note is that I ∝ ω4 or alternatively I ∝ 1/λ4. There is avery strong dependence on frequency (or wavelength).
This quartic scattering dependence is, in fact, the reason why the sky is blue (from
the point of view of classical electrodynamics) and is called the Rayleigh scattering
law.
11.2. The Blue Sky
The spectrum of visible light from the sun incident on the outer atmosphere is
essentially flat as shown below.
81
We just learned that light scatters as it traverses the atmosphere according to
Rayleigh’s scattering law: I(λ) ∝ 1/λ4.
The following figures illustrate why Rayleigh scattering implies that the sky is
blue.
11.2.1. Sunsets
We have focused on a blue sky, but red sunsets occur for the same reason–
Rayleigh scattering.
82
If we look directly at the sun during a sunset (or sunrise) it appears red because
most of the blue light has scattered in other directions.
This more pronounced at dawn or dusk since the light must traverse more of the
atmosphere at those times then at noonday at which time the sun appears yellow
in color.
11.2.2. White Clouds
Wemight expect that clouds should be highly colored since they consist of droplets
of water which scatter light very effectively.
83
The key difference between light scattering by clouds versus by the atmosphere is
the size of the scatterer.
The water droplets are much larger than the wavelenght of the light–quite the
opposite case as above.
In this limit an entirely different analysis is made–one does not have Rayleigh
scattering but instead has a process called Mie scattering.
In some contexts, particularly in liquid suspensions, Mie scattering is referred to
as Tyndall scattering
84
Key Equations for Exam 2
Listed here are some of the key equations for Exam 2. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• The wavefunctions for the hydrogenic system are
ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ) (11.6)
• The radial part is.
Rnl(σ) = Anl
µ2σ
n
¶l
e−σ/nL2l+1n+1
µ2σ
n
¶, (11.7)
where the normalization constant, Anl, depends on the n and l quantum
numbers as
Anl = −
sµ2Z
na0
¶3(n− l − 1)!2n[(n+ l)!]3
(11.8)
85
85
• The energy levels for the hydrogenic system are given by
En = −Z2Rn2
(11.9)
• The wavefunctions for the harmonic oscillator are
ψn(y) = AnHn(y)e− y2
2 , y =
µkm
~2
¶ 14
x, An =1p
2nn!√π, (11.10)
where An is the normalization constant for the nth eigenfunction and Hn(y)
are the Hermite polynomials.
• The energy levels are
En = (n+1
2)~ω, ω =
rk
m(11.11)
• The Morse potential is
Ee(R) = De[1− e−β(R−Req )]2, (11.12)
where De is the well depth and β = 2πcωe
qμ2De
is the Morse parameter.
Note: this expression for the Morse potential has the zero of energy at the
bottom of the well (i.e. R = Req, ;Ee(Req) = 0).
• The Morse Potential can also be written as
Ee(R) = De[e−2β(R−Req ) − 2e−β(R−Req )]. (11.13)
Now the zero of energy is the dissociated state (i.e. R→∞, ;Ee(R→∞) =0).
• The energy levels for the Morse oscillator are of the form
Evib,n = −De + hcωe(n+1
2)− hcωexe(n+
1
2)2, (11.14)
where ωexe together is the anharmonicity constant, with xe =hcωe4De
.
86
• Bond order is defined as
BO =1
2(# of bonding electrons−# of antibonding electrons) (11.15)
• The Rayleigh scattering law is
I(λ) ∝ 1/λ4 ∝ ω4 (11.16)
87
Part III
Statistical Mechanics and TheLaws of Thermodynamics
88
88
12. Rudiments of Statistical
Mechanics
When we study simple systems like a single molecule, we use a very detailed
theory, quantum mechanics.
However, most of the time in the real world we are dealing with macroscopic
systems, say, at least 100 million molecules.
It is simply impossible, even with the fastest computers, to write down and solve
the Schrödinger equation for those 100 million molecules, but often Avogadro’s
number of molecules.
So we need a less detailed theory called statistical mechanics, which allows one to
handle macroscopic sized systems without losing to much of the rigor.
12.1. Statistics and Entropy
Probability and statistics is at the heart of statistical mechanics.
We will need some definitions
• Ensemble: A large collection of equivalent macroscopic systems. The sys-tems are the same except that each one is in a different so-called microstate.
89
89
• Microstate: The single particular state of one member of the ensemble givenby listing the individual states of each of the microscopic systems in the
macroscopic state.
• Configuration: The collection of all equivalent microstates. The number ofpossible configurations is defined as W.
Boltzmann developed an equation to connect the microscopic properties of an
ensemble to the macroscopic properties. The Boltzmann equation is
S = k lnW (12.1)
Where S is entropy and k is Boltzmann’s constant.
12.1.1. Combinations and Permutations
Consider a random system that when measured can appear in one of two outcomes
(e.g., flipping coins).
One valuable piece of statistical information about system is knowing how many
different ways the system appears p times in, say, outcome 1 after N measure-
ments.
This is given by the mathematical formula for combinations
C(N, p) =N !
p!(N − p)!. (12.2)
The number C(N, p) is also called the binomial coefficient because it gives the
coefficient for the pth order term in the expansion
(1 + x)N =NXp=0
C(N, p)xp. (12.3)
90
This formula will allow us to derive a normalization constant so that we can obtain
the probability of obtaining p measurements of state 1.
Set x = 1 in the above. This gives
(1 + 1)N =NXp=0
C(N, p)(1)p (12.4)
2N =NXp=0
C(N, p).
So the probability of any one outcome of N measurements is
P (N, p) =1
2NC(N, p) =
1
2NN !
p!(N − p)!(12.5)
For combinations we did not care what order the results of the measurements
occurred.
Sometimes the order is important.
So rather than a particular combination, we are interested in a particular permu-
tation. This is given by
W (N, Ni) =N !
N1!N2!N3! · · ·(12.6)
where N is the total number of measurements and Ni is the number of indistin-
guishable results of type i.
∗ ∗ ∗ See Examples on Handout ∗ ∗∗
For both combinations and permutations we need to evaluated factorials.
91
This is no problem for small numbers, but when we consider macroscopic systems
(1020 or so molecules) no calculator can handle factorials of such large numbers.
Sterlings Approximation:
• In place of evaluating factorials of large number one can use Sterling’s ap-proximation to approximate the value of the factorial.
• Sterling’s approximation is
ln(N !) ' N lnN −N (12.7)
12.2. Fluctuations
When we list the macroscopic properties of a material such as a beaker of benzene
or the air of the atmosphere, we speak of the average value of the property.
Macroscopic equilibrium is a dynamic rather than static equilibrium. Conse-
quently, the value of a certain property fluctuates about the average value. Often
this fluctuation is not important, but sometimes it is important.
The fluctuation about an average value for any observable property O is described
by the variance which is defined as
σ2O ≡ O2 − O2. (12.8)
σO is consider the range of the observable property.
It can be shown thatσOO≈ 1√
N, (12.9)
where N is the number of particles. So for example if N = 1024 then 1√N= 10−12
92
For ensembles having large numbers of particles measured values of a property are
extremely sharply peaked about the average value.
93
13. The Boltzmann Distribution
Consider a isolated system of N molecules that has the set i energy levelsassociated with it.
Since the system is isolated the total energy, E, and the total number of particles
will be constant.
The total energy is given by
E =Xi
Ni i, (13.1)
where Ni is the number of particles in energy state i.
The total number of particles is, of course,
N =Xi
Ni (13.2)
The number of configurations for the system is then given by the number of
distinct permutations of the system
W =N !
N1!N2! · · ·. (13.3)
A system in equilibrium always tries to maximize entropy and minimize energy
and so the equilibrium configuration is a compromise between these two cases.
94
94
For the moment let us relax the isolation constraint.
Maximizing entropy corresponds to maximizing W (via S = k lnW ). This would
be the situation in which every particle was in a different energy state. That is
all Ni = 1 or 0.
Minimizing energy would be the case where all the particles are in the ground
state (say 1).
These two situations are contradictory and some compromise must be obtained.
We start by considering our original system–that being one with constant energy,
E and number of particles N
To determine the equilibrium configuration we must find the maximumW subject
to the constraint of constant energy and constant number of particles.
This is done using the mathematical technique of Lagrange multipliers (page 951
of your calc book).
We will not discuss this method in detail and consequently we cannot derive the
equilibrium configuration.
The derivation using Lagrange multipliers arrives at the configuration in which
the
Ni = Ngie
−β iPj gje
−β j| z pi
, (13.4)
where β ≡ 1kTand gj denotes the degeneracy of states having energy j.
95
The pi represents the probability of finding the a randomly chosen particle or
system which has energy i. This is the Boltzmann distribution
Pi =gie
−β iPj gje
−β j(13.5)
Since we started with a isolated system, β and hence T are constants. A given
energy E will correspond to a unique temperature T.
The analysis readily generalizes to variable energy i.e., nonisolated systems by
considering T as a variable.
13.1. Partition Functions
We have already come across both the partition functions that we will use in this
class.
The first is W–the number of configurations. This is called the microcanonical
partition function.
This partition function is not very useful to us so we will not discuss it further.
The second partition function is
Q =Xj
gje−βEj (13.6)
and is called the canonical partition function.
96
This was first encountered as the denominator of the Boltzmann distribution and
it is extremely important in statistical mechanics. (Note: the symbol Z is also
often used for the canonical partition function.)
The partition function is to statistical mechanics as the wavefunction is to quan-
tum mechanics. That is, the partition function contains all that can be known
about the ensemble.
We shall see in the next chapter that the partition function will provide a link
between the microscopic (quantum mechanics or classical mechanics) and the
macroscopic (thermodynamics).
In fact we have already seen this in the S = k lnW. But this an inconvenient
connection because, for among other reasons, energy levels and temperature do
not explicitly appear.
There are other partition functions that are useful in different situations but we
will do nothing more than list two important ones here: i) the grand canonical
partition function and ii) the isothermal—isobaric partition function
13.1.1. Relation between the Q and W
When we get to connecting quantummechanics with thermodynamics it will prove
convenient to use Boltzmann’s equation (S = k lnW ) but as was stated earlier it
is not convenient to use the microcanonical partition function (W ).
In the following we give an argument which provides a relation between the par-
tition functions. It is not an exact relation as we derive it, but it is a very good
approximation for large numbers of particles.
97
The microcanonical partition function describes a system at fixed energy E. In
fact W is the number of available states of the ensemble at the particular energy
E. This is essentially the same as the degeneracy of the ensemble gE.
Conversely the canonical partition function describes a system with variable en-
ergy.
However, based on our previous discussion of fluctuations, even though the energy
of the ensemble is allowed to vary, the number of states with energy equal to the
average energy E is overwhelmingly large. That is, almost every state available
to the ensemble has energy E.
We can express these ideas mathematically to come up with a relation between
W and Q.
The canonical partition function is
Q =Xj
gje−β j , (13.7)
but to a good approximation
Q ' gEe−βE. (13.8)
Now since the degeneracy is essentially the microcanonical partition function we
have
Q 'We−βE. (13.9)
So the canonical partition function is a Boltzmann weighted version of the micro-
canonical partition function.
We will soon make use of the Boltmann’s equation in terms of the canonical
98
partition function:
lnQ ' ln(We−βE) = lnW + ln(e−βE) (13.10)
= lnW|zS/k
− E
kT.
so,
S = k lnQ+E
T(13.11)
13.2. The Molecular Partition Function
We ended the previous chapter by stating the total molecular energy (about the
center of mass) as
= ele + vib + rot. (13.12)
This is a consequence of the Born Oppenheimer approximation
If we include the center of mass translational motion this is
= ele + vib + rot + trans (13.13)
The ith total energy level is
i = ele,n + vib,v + rot,J + trans,m. (13.14)
Now if we have a collection of molecules in a macroscopic system. A given con-
figuration (say, configuration j) of that system has total energy Ej.
So the canonical partition function is
Q =Xj
gje−βEj (13.15)
99
But, each Ej is made up of the contributions of all of the molecules:
Ej =al +
bm +
cn + · · · (13.16)
The partition function for the molecule is written as
Q =Xj
gje−βEj =
Xl,m,n···
(gal gbmg
cn · · · )e−β(
al +
bm+
cn+··· ) (13.17)
=Xl
gal e−β a
l| z qmol,a
Xm
game−β a
m| z qmol,b
Xn
gane−β a
n| z qmol,c
· · ·
where the qmol,i are the molecular partition functions.
The total canonical partition function is the product of the molecular partition
functions.
For the case where the molecules are the same then all the qmol,i are the same:
qmol,i = qmol thus
Q =qNmolN !
. (13.18)
This allows us to focus only on a single molecule:
qmol =Xi
gie−β i =
Xn,v,J,m
gele,ngvib,vgrot,Jgtrans,me−β( ele,n+ vib ,v+ rot,J+ trans,m)(13.19)X
n
gele,ne−β ele ,n
| z qele
Xv
gvib,ve−β vib,v
| z qvib
XJ
grot,Je−β rot,J
| z qrot
Xm
gtrans,me−β trans,m
| z qtrans
We now collect below the expression for each of these partition functions. You
will get the chance to derive each of these for your home work
100
The Translational Partition Function
qtrans =V
Λ3(13.20)
where
Λ ≡ h√2πmkT
(13.21)
is the thermal de Broglie wavelength.
The Rotational Partition Function (linear molecules)
We will discuss rotations next semester.
However, the high temperature limit, which works for all gases (of linear molecules)
except H2 is
qrot ≈T
σθr(13.22)
where θr ≡ h2
8π2Ik(I is the moment of inertia) and σ is the so-called symmetry
number in which σ = 1 for unsymmetrical molecules and σ = 2 for symmetrical
molecules.
The Vibrational Partition Function
qvib =e−
12β~ω
1− e−β~ω=
1
2 sinh 12β~ω
(13.23)
Note this is for the harmonic oscillator. At temperatures well below the dissocia-
tion energy this is a very good approximation. (You will derive this as a homework
problem.)
The Electronic Partition Function
There is usually only a very few electronic states of interest. Only at exceedingly
high temperatures does any state other that the ground state(s) become important
101
so
qele =Xi
gele,ie−β tele ,i ≈ gele,ground (13.24)
102
14. Statistical Thermodynamics
The partition function allows one to calculate ensemble averages which correspond
to macroscopically measurable properties such as internal energy, free energy,
entropy etc.
In this chapter we will obtain expressions for internal energy, U, pressure, P,
entropy, S, and Helmholtz free energy, A. With these quantities in hand we will,
in the subsequent chapters, formally develop thermodynamics with no need to
refer back to the partition function.
Ensemble averages
The ensemble average of any property is given by
O =1
Q
Xi
Oigie−β i . (14.1)
Internal energy
One critical property of an ensemble is the average (internal) energy U.
U ≡ E =1
Q
Xi
igie−β i . (14.2)
Let us look closer at the above expression. Recall that
Q =Xi
gie−β i . (14.3)
103
103
Now take the derivative of Q with respect to β givesµ∂Q
∂β
¶n,V
=
̶
∂β
"Xi
gie−β i
#!n,V
=Xi
gi
µ∂e−β i
∂β
¶n,V
(14.4)
= −Xi
gi ie−β i
By comparing this to the expression for U, we see
U = − 1Q
µ∂Q
∂β
¶n,V
= −µ∂ lnQ
∂β
¶n,V
, (14.5)
where we used the identity 1y∂y∂x= ∂ ln y
∂x.
Pressure
Another important property is pressure.
When the ensemble is in the particular state i, d i = −pidV . So at constanttemperature and number of particles
pi = −µ∂ i
∂V
¶n,β
(14.6)
Thus the ensemble average pressure is given by
P = p = − 1Q
Xi
gi
µ∂ i
∂V
¶n,β
e−β i . (14.7)
Multiplying by β/β we get
P = − 1
βQ
Xi
gi
µ∂ i
∂V
¶n,β
βe−β i . (14.8)
Using the chain rule in reverse, i.e.,
∂e−β i
∂V=
−βe−β iz | µ∂e−β i
∂ i
¶µ∂ i
∂V
¶= −
µ∂ i
∂V
¶βe−β i (14.9)
104
we proceed as
P =1
βQ
Xi
gi
µ∂e−β i
∂V
¶n,β
=1
βQ
̶
∂V
Xi
gie−β i
!n,β
(14.10)
=1
βQ
µ∂Q
∂V
¶n,β
=1
β
µ∂ lnQ
∂V
¶n,β
.
Entropy
We have already obtained the expression for entropy. It is
S =U
T+ k lnQ (14.11)
= −kβµ∂ lnQ
∂β
¶n,V
+ k lnQ
105
Helmholtz Free Energy
Free energy is the energy contained in the system which is available to do work.
That is, it is the energy of the system minus the energy that is “tied-up” in the
random (unusable) thermal motion of the particle in the system: A ≡ U − TS
Free energy is probably the key concept in thermodynamics and so we will discuss
it in much greater detail later. We will make the distinction between the Helmholtz
free energy and the more familiar Gibb’s free energy (G) later as well.
The Helmholtz free energy has the most direct relation to the partition function
as can be seen from
A ≡ U − TS = −µ∂ lnQ
∂β
¶n,V
+ kTβ
µ∂ lnQ
∂β
¶n,V
− kT lnQ (14.12)
= −kT lnQ
Any thermodynamic property can now be obtained from the above functions as
we shall see in the following chapters.
106
15. Work
We now begin the study of thermodynamics.
Thermodynamics is a theory describing the most general properties of macroscopic
systems at equilibrium and the process of transferring between equilibrium states.
Thermodynamics is completely independent of the microscopic structure of the
system.
15.1. Properties of Partial Derivatives
Of critical importance in mastering thermodynamics is to become proficient with
partial derivatives.
∗ ∗ ∗ See Handout ∗ ∗∗
15.1.1. Summary of Relations
1. The total derivative of z(x, y):
dz =
µ∂z
∂x
¶y
dx+
µ∂z
∂y
¶x
dy (15.1)
2. The chain rule for partial derivatives:µ∂z
∂x
¶y
=
µ∂z
∂u
¶y
µ∂u
∂x
¶y
(15.2)
107
107
3. The reciprocal rule: µ∂z
∂x
¶y
µ∂x
∂z
¶y
= 1 (15.3)
4. The cyclic rule: µ∂z
∂x
¶y
= −µ∂z
∂y
¶x
µ∂y
∂x
¶z
(15.4)
5. Finally µ∂z
∂x
¶u
=
µ∂z
∂x
¶y
+
µ∂z
∂y
¶x
µ∂y
∂x
¶u
(15.5)
15.2. Definitions
System: a collection of particles
Macroscopic systems: Systems containing a large number of particles.
Microscopic systems: Systems containing a small number of particles.
Environment : Everything not included in the system (or set of systems)
Note that the distinction between the system and the environment is arbitrary
and is chosen as a matter of convenience.
15.2.1. Types of Systems
Isolated system: A system that cannot exchange matter or energy with its envi-
ronment.
Closed system: A system that cannot exchange matter with its environment but
may exchange energy.
108
Open system: A system that may exchange matter and energy with its environ-
ment.
Adiabatic system: A closed system that also can not exchange heat energy with
its environment.
15.2.2. System Parameters
Extensive parameters (or properties): properties that depend on the amount ofmatter.
• For example, volume, mass, heat capacity.
Intensive parameters (or properties): properties that are independent of theamount of matter.
• For example, temperature, pressure, density.
Extensive properties can be “converted” to intensive properties through ratios:
Extensive propertyExtensive property
→ Intensive property. (15.6)
For example massvolume = density,
volumemoles = molar volume,
heat capacitymass = specific heat.
15.3. Work and Heat
A system may exchange energy with its environment or another system in the
form of work or heat.
• Work is exchanged if external parameters are changed during the process.
• Heat is exchanged if only internal parameters are changed during the process.
109
Convention
Work, w, is positive (w > 0) if work is done on the system.
Work is negative (w < 0) if work is done by the system.
Heat, q, is positive (q > 0) if heat is absorbed by the system.
Heat is negative (q < 0) if heat is released from the system.
15.3.1. Generalized Forces and Displacements
In physics you learned that an infinitesimal change in work is given by the product
of force, F , times and infinitesimal change in position, dx:
dw = Fdx. (15.7)
For thermodynamics, we need a more general definition if infinitesimal work.
Any given external parameter, A may be considered as a ‘generalized force’ which
is coupled to a particular internal parameter, a, which acts as ‘generalized dis-
placement.’
Note that the generalized force need not have units of force (e.g., Newtons) and
the generalized displacement need not have units of position (e.g., meters), but
the product of the two must have units of energy (e.g., Joules).
The infinitesimal amount of work done on the system is then given by
dw = Ada, (15.8)
or more generally as
dw =Xi
Aidai (15.9)
110
if more than one set of parameters change.
The following table gives some examples of generalized forces and displacementsGeneralized Force, A Generalized Displacement, a Contribution to dw
Pressure, −P Volume, dV −PdVStress, σ Strain, dε σdε
Surface tension, γ Surface area, dA γdAVoltage, E Charge, dQ EdQ
Magnetic Field, H Magnetization, dM HdMChemical Potential, μ Moles, dn μdn
Gravity, mg Height, dh mgdh
15.3.2. PV work
In principle all work is interchangeable so that without loss of generality we will
develop the formal aspects of thermodynamics assuming all work is due to changes
in volume under a given pressure. That is
dw = −PdV, (15.10)
this is called PV work.
When we get to applications of thermodynamics we will then be concerned with
the various forms of work like those shown in the table above.
Expanding Gases
Consider the work done by a gas expanding in piston from volume V1 to V2 against
some constant external pressure P = Pex (see figure)
111
The force exerted on a gas by a piston is equal to the external pressure times the
area of the piston: F = PexA⇒ Pex = F/A.
Recall from physics that work is the (path) integral over force: w = −R x2x1
Fdx.
This can be manipulated as
w = −Z x2
x1
Fdx = −Z x2
x1
F
A|zPex
Adx|zdV
= −Z V2
V1
PexdV (15.11)
If Pex is independent of V then
w = −Z V2
V1
PexdV = −PexZ V2
V1
dV = −Pex4V (15.12)
112
16. Maximum Work and Reversible
changes
Now that we have learned about PV work we will consider the situation where
the system does the maximum amount of work possible.
16.1. MaximalWork: Reversible versus Irreversible changes
The value of w depends on Pex during the entire expansion.
In the figure
wA = −Z V2
V1
PatmdV = −Patm (V2 − V1) (16.1)
113
113
and
wB = w1 + w2, (16.2)
where
w1 = −Z Vi
V1
Patm+2WdV = −Patm+2W (Vi − V1) (16.3)
and
w2 = −Z V2
Vi
PatmdV = −Patm (V2 − Vi) (16.4)
Hence it is clear that |wB| > |wA| .
Now consider case in the figure below
The expansion is reversible. That is, there is always an intermediate equilibriumthroughout the expansion. Namely Pgas = Pex. So,
wrev = −Z V2
V1
PgasdV (16.5)
This is the limiting case of path B in the previous figure. Thus wrev is the maxi-
mum possible work that can be done in an expansion. wrev = wmax.
114
16.2. Heat Capacity
Temperature and heat are different.
Temperature is not the amount of heat.
Temperature is an intensive property and heat is an extensive property.
However, heat is related to temperature through the heat capacity
C(T ) =dq
dT(16.6)
n.b., heat capacity is a function of T ; it is not a constant.
From this equation
dq = C(t)dT, (16.7)
That is, when the temperature of a substance having a heat capacity C(t) is
changed by dT, dq amount of heat energy is transferred.
The heat capacity also depends on the conditions during the temperature change,
e.g., CV (T ) =¡dqdT
¢Vand CP (T ) =
¡dqdT
¢Pare not the same
Heat capacity is an extensive property. To make an intensive property
1. divide by the number of moles to get molar heat capacity
CVm(T ) =1
n
µdq
dT
¶V
(16.8)
2. divide by mass to get specific heat
cV =1
m
µdq
dT
¶V
(16.9)
We will discuss heat capacity more later.
115
16.3. Equations of State
The macroscopic properties of matter are related to one another via a phenom-
enological equation of state.
The state of a pure, homogeneous material (in the absence of external fields) is
given by the values of any two intensive properties.
(More complicated systems require more than two independent variables, but
behave in the same way as the more simple pure system, so we will focus our
development of thermodynamics on simple systems.)
The functional dependence of any property on the two independent variables is
an equation of state. e.g., T , P independent then heat capacity is a function of T
and P , C(T, P ).
16.3.1. Example 1: The Ideal Gas Law
The equation of state for volume of an ideal gas is
PV = nRT, (16.10)
where R is the gas constant (8.315 J K−1 mol−1) and n is the number of moles.
The ideal gas equation of state can be expressed in terms of intensive variables
only
PVm = RT, (16.11)
where Vm = Vn.
The equation of state can also be expressed in terms of density ρ = mV(and molar
mass m/n)
ρ =mP
nRT=
MP
RT. (16.12)
116
16.3.2. Example 2: The van der Waals Equation of State
A more realistic equation of state was presented by van der Waals:
P =nRT
V − nb− n2a
V 2. (16.13)
The parameter a attempts to account for the attractive forces among the particles
The parameter b attempts to account for the repulsive forces among the particles
b originates from hard sphere collisions (see figure):
117
In term of intensive variables
P =RT
Vm − b− a
V 2m
. (16.14)
16.3.3. Other Equations of State
The van der Waals equation of state is not the only one that has been proposed.
Some other equations of state are
• BerthelotP =
nRT
V − nb− n2a
TV 2=
RT
Vm − b− a
TV 2m
(16.15)
• DietericiP =
nRTe−anRTV
V − nb=
RTe−a
RTVm
Vm − b(16.16)
• Redlich-Kwang
P =nRT
V − nb− n2a√
TV (V − nb)=
RT
Vm − b− a√
TVm (Vm − b)(16.17)
118
17. The Zeroth and First Laws of
Thermodynamics
Over the course of the next two lectures we will discuss the four core laws of
thermodyanmics.
Today we will cover the zeroth and first laws, which deal with temperature and
total energy respectively.
Next time we will cover the second and third laws which both deal with entropy.
17.1. Temperature and the Zeroth Law of Thermodynamics
Temperature tells us the direction of thermal energy (heat) flow.
• Heat flows from high T to Low T.
Temperature scales
• Celsius: A relative scale based on water (T = 0C for melting ice and
T = 100C for boiling water)
• Kelvin: An absolute temperature scale based on the ideal gas law. Thetemperature at which (for fixed V and n) the pressure is zero is defined as
T = 0 K
• T (Kelvin) = T (Celsius) + 273.15
119
119
Standard conditions
• standard temperature and pressure (STP): T = 273.15 K and P = 1 atm.
(Vm(STP) = 22.414 L/mol)
• standard ambient temperature and pressure (SATP): T = 298.15 K and
P = 1 bar. (Vm(SATP) = 24.789 L/mol)
Diathermic wall : A wall that allows heat to flow through it.
Adiabatic wall : A wall the does not allow heat to flow through it.
Thermal equilibrium: If two systems are in contact along a diathermic wall and
no heat flows across the wall, then the systems are in thermal equilibrium.
The zeroth law of thermodynamics
• Mathematical statement:
If TA = TB and TB = TC , then TA = TC (17.1)
This the mathematical statement of transitivity
• Verbal statement: If system A is in thermal equilibrium with system B
and system B is in thermal equilibrium system C then system A is also in
thermal equilibrium with system C.
The zeroth law implies that if an arbitrary system, C, is chosen as a thermometer
then it will read the same temperature when it is in thermal contact along a
diathermic wall with system A as when it is in thermal contact along a diathermic
wall with system B.
120
17.2. The First Law of Thermodynamics
Definitions:
• State: the state of a system is defined by specifying a minimum number inintensive variables
• State Function: A function of the chosen independent variables that de-scribes a property of the state (e.g., V (T, P )). The value of the state func-
tion depends only on that given state and on no other possible state of the
system.
17.2.1. The internal energy state function
For characterizing the change in energy of a system, one is concerned with the
work done on the system (w) and the heat supplied to the system (q). The energy
of a system is called the internal energy (U) of the system.
The first law of thermodynamics:
• Mathematical statement:4U = q + w (17.2)
or in differential form
dU = dq + dw (17.3)
• Verbal statement: The change in internal energy of a system is equal to theamount of work done on the system plus the amount of heat provided to the
system.
So for a system where all the work is PV work the first law becomes
4U = q −Z V2
V1
PexdV (17.4)
121
in differential form this is
dU = dq − PexdV (17.5)
Although U can be expressed as a function of any two state variables, the most
convenient at this time are V and T. U → U(T, V ).
The total differential of U(T, V ) is
dU =
µ∂U
∂T
¶V
dT +
µ∂U
∂V
¶T
dV (17.6)
Consider adding heat at a constant volume then
dU =
µ∂U
∂T
¶V
dT +
µ∂U
∂V
¶T
dV = dq − PexdV. (17.7)
So, µ∂U
∂T
¶V
dT = dq =⇒µ∂U
∂T
¶V
=dq
dT= CV (17.8)
Hence the slope¡∂U∂T
¢Vis the heat capacity.
The other slope,¡∂U∂V
¢T, is called the internal pressure (it has no standard symbol).
A useful relation (derivation to come) isµ∂U
∂V
¶T
= T
µ∂P
∂T
¶V
− P (17.9)
Example: A van der Waals gas
P =nRT
V − nb− n2a
V 2⇒µ∂P
∂T
¶V
=nR
V − nb(17.10)
122
so the useful relation becomesµ∂U
∂V
¶T
= TnR
V − nb− P =
nRT
V − nb− nRT
V − nb+
n2a
V 2
= +n2a
V 2(17.11)
The equation of state for U : Express U in terms of T, V, and P.
Start with the total differential of U
dU =
µ∂U
∂T
¶V
dT +
µ∂U
∂V
¶T
dV (17.12)
but¡∂U∂T
¢V= CV and
¡∂U∂V
¢T= T
¡∂P∂T
¢V− P (useful relation). Hence
dU = CV dT +
∙T
µ∂P
∂T
¶V
− P
¸dV (17.13)
is the equation of state for U.
A useful approximation is 4U = CV4T which is valid for
i) heat capacity nearly constant over 4T and with no phase transitions.
ii) ideal gas or at constant volume.
123
18. The Second and Third Laws of
Thermodynamics
18.1. Entropy and the Second Law of Thermodynamics
We learned from statistical mechanics that entropy, S, is a measure of the disorder
of the system and is expressed via Boltzmann’s equation S = k lnW (where W is
the micocanonical partition function)
We expressed Boltzmann’s law in terms of the more convenient canonical partition
function as
S =E
T+ k lnQ. (18.1)
Now, the average energy of the system E is in fact what we call internal energy:
U ≡ E.
Furthermore we derived the simple relation between the Helmholtz free energy
and the canonical partition function as A = −kT lnQ.
Hence,
S =U
T− A
T=1
T(U −A). (18.2)
Since U, A, and T are state functions, S is also a state function .
So we may write
dS =1
T(dU − dA) (18.3)
124
124
for an isothermal process.
Recall the definition of Helmholtz free energy–the energy of the system available
to do work.
We learned previously that the maximum amount of work one can extract from
the system is the work done during a reversible process. Hence dA = dwrev.
For now let us limit the discussion to reversible processes. Then
dS =1
T(dU − dwrev) =
1
T(dqrev + dw/ rev − dw/ rev ) (18.4)
=dqrevT
. (Reversible process)
Note: An alternative approach to thermodynamics which makes no reference to
molecules or statistical mechanics is to simply begin by defining entropy as dS ≡dqrevT
The principle of Clausius
• “The entropy of an isolated system will always increase in a spontaneous
process”
• Mathematical statement: (dS)U,V ≥ 0
For a general process: dU = dq − PexdV
For a reversible process Pex = P and dq = TdS so dU = TdS − PdV
125
Since U, S, T, P, and V are state functions, dU = TdS − PdV holds for any
process, but in general, TdS is not heat and −PdV is not work. (see figure)
TdS is heat and −PdV is work only for reversible processes.
For some dU,
dq − PexdV = TdS − PdV ⇒ TdS = dq − PexdV + PdV (18.5)
TdS = dq + (P − Pex) dV
• Case i) Pex > P then (spontaneous) dV is negative so (P−Pex)dV is positive.
• Case ii) P > Pex then (spontaneous) dV is positive so (P−Pex)dV is positive.
• Case iii) P = Pex then (spontaneous) dV is zero so (P − Pex)dV is zero.
Thus for any spontaneous process TdS ≥ dq.
This is a mathematical statement of the second law of thermodynamics
126
18.1.1. Statements of the Second Law
Unlike the first law, the second law has a number of equivalent statements
1. A cyclic process must transfer heat from a hot to cold reservoir if it is to
convert heat into work.
2. Work must be done to transfer heat from a cold to a hot reservoir.
3. A useful perpetual motion machine does not exist.
4. The entropy of the universe is increasing
5. Spontaneous processes are irreversible in character.
6. The entropy of an isolated system will always increase in a spontaneous
process (the principle of Clausius)
18.2. The Third Law of Thermodynamics
Consider the first law for a reversible change at constant volume.
dU = dq + dw = dq − PexdV (18.6)
From our earlier discussion of heat capacity dq = CV dT (CV since constant vol-
ume). So,
dU = CV dT (18.7)
but also dU = TdS. So
dS =CV dT
T=⇒ 4S =
Z T2
T1
CV
TdT. (18.8)
127
A very similar derivation can be done for a reversible change at constant pressure
(we can not do it quite yet) to yield
4S =
Z T2
T1
CP
TdT (18.9)
18.2.1. The Third Law
Verbal statement
The third law of thermodynamics permits the absolute measurement of entropy.
To derive the mathematical statement of the third laws we starting with
4S =
Z T2
T1
CP
TdT (18.10)
now let T1 → 0
4S = S2 − S0 =
Z T2
0
CP
TdT (18.11)
Hence the mathematical statement of the third law is
S(T2) =
Z T2
0
CP
TdT + S0 (18.12)
From a macroscopic point of view S0 is arbitrary. However, a microscopic point of
view suggests S0 = 0 for perfect crystals of atoms or of totally symmetric molecules
(e.g., Ar, O2 etc.). S0 6= 0 for imperfect crystals and crystals of asymmetric
molecules (e.g., CO).
Alternative statement of the third law: Absolute zero is unattainable.
Consider the heat capacity near T → 0.
For S0 to have significance CPTmust be finite (not infinite) as T → 0. Thus CP → 0.
128
But CP =dqdT→ 0 implies dT
dq→∞.
In other words, an infinitesimal amount of heat causes an infinite change in tem-
perature.
In view of what we have learned about fluctuations, the ever present random
fluctuations in energy provide the infinitesimal amount of heat and so you can
never reach absolute zero corresponding to an average energy of zero.
18.2.2. Debye’s Law
Heat capacity data only goes down so far. So one needs a theoretical extrapolation
down to T = 0. (Debye)
Postulate: CPm = aT 3. That is at low temperatures heat capacity goes as the
cube of the temperature.
C∗Pm, T∗ are the lowest temperature data points. So, a = C∗Pm/T
∗3.
129
The molar entropy is
Sm(T∗) =
Z T∗
0
CP
TdT
CPm=aT3
=C∗PmT ∗3
Z T∗
0
T 2dT (18.13)
=C∗PmT ∗3
T 3
3
¯T∗0
=C∗Pm3
.
18.3. Times Arrow
Entropy and the second law give a direction to time.
For example, if we see a picture of your PChem book in mint condition and we see
a picture of your PChem book all battered and beaten. We know which picture
was taken first.
The interesting thing is that each molecule in a macroscopic system obeys time
invariant dynamics. Both Newton’s laws and Quantum dynamics (next semester)
are the same if you replace t with −t.
Yet, the behavior of the macrosystem definitely changes if you replace t with −t.
Thus the simple fact that you have an enormous number of particles induces a
perceived asymmetry in time.
130
Key Equations for Exam 3
Listed here are some of the key equations for Exam 3. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• The Boltzmann equation isS = k lnW. (18.14)
• The Boltzmann distribution :
gie−β iP
j gje−β j
(18.15)
• The canonical partition function is
Q =Xj
gje−βEj (18.16)
131
131
• The relation between the partition function and the molecular partitionfunction is
Q =qNmolN !
. (18.17)
• The Translational Partition Function
qtrans =V
Λ3(18.18)
where
Λ ≡ h√2πmkT
(18.19)
is the thermal de Broglie wavelength.
• The Rotational Partition Function (linear molecules) is
qrot ≈T
σθr, (18.20)
where θr ≡ h2
8π2Ik(I is the moment of inertia) and σ is the so-called sym-
metry number in which σ = 1 for unsymmetrical molecules and σ = 2 for
symmetrical molecules
• The Vibrational Partition Function is
qvib =1
2 sinh 12β~ω
. (18.21)
• The ensemble average of any property is given by
O =1
Q
Xi
Oigie−β i . (18.22)
• The relations between the canonical partition function and the thermody-namics variables are
Helmholtz Free Energy A = −kT lnQInternal energy U = − 1
Q
³∂Q∂β
´n,V
= −³∂ lnQ∂β
´n,V
Entropy S = −kβ³∂ lnQ∂β
´n,V+ k lnQ
Pressure P = 1βQ
¡∂Q∂V
¢n,β= 1
β
¡∂ lnQ∂V
¢n,β
132
• PV work is
dw = −PdV. (18.23)
• Heat capacity:dq = C(t)dT. (18.24)
• General forms of the first law:
4U = q + w, (18.25)
in differential form this is
dU = dq − PexdV. (18.26)
Also,
dU = TdS − PdV. (18.27)
• The second lawTdS ≥ dq. (18.28)
• The third lawS(T2) =
Z T2
0
CP
TdT + S0 (18.29)
• Debye’s law for entropy at very low temperatures
Sm(T∗) =
C∗Pm3
, (18.30)
where C∗Pm is the molar heat capacity at the lowest temperature for which
there is data.
133
Part IV
Basics of Thermodynamics
134
134
19. Auxillary Functions and Maxwell
Relations
We have stated that thermodynamics as we are studying it deals with states in
equilibrium or transitions between equilibrium states.
Consequently, the concept of equilibrium plays a key role in much of what we will
discuss for the remainder of the year.
The equilibrium constant for a thermodynamic process,K, (which you are familiar
with from general chemistry) serves are a common point which connects thermo-
dynamics, electrochemistry, and kinetics–topics we will encounter throughout
the year.
19.1. The Other Important State Functions of Thermody-
namics
As was the case in quantum mechanics, here too is energy the key property with
which to work.
So far we have encountered two state functions which characterize the energy of a
macroscopic system–the internal energy and, briefly the Helmholtz free energy.
135
135
From the first law as stated as
dU = TdS − PdV (19.1)
we say that the natural (most convenient) variables for the equation of state for
U are S and V . This is U = U(S, V )
Unfortunately S can not be directly measured and most often P is a more conve-
nient variable than V
Because of this fact, it is handy to define state functions which have different pairs
of natural variables, so that no mater what situation arises we have convenient
equations of state to work with.
The other pairs of natural variables being (S and P ), (T and V ) and (T and P )
The table below lists these state functions
State function Symbol Natural variables Definition Units
Internal Energy U S and V energy
Enthalpy H S and P H ≡ U + PV energy
Helmholtz free energy A T and V A ≡ U − TS energy
Gibbs free energy G T and P G ≡ H − TS energy
We consider each of these functions in turn
19.2. Enthalpy
We want a state function whose natural variables are S and P
Let us try the definition H ≡ U + PV.
136
Now formally
dH = dU + d(PV ) = dU + PdV + V dP, (19.2)
but dU = TdS − PdV, so
dH = TdS − PdV/ + PdV/ + V dP (19.3)
= TdS + V dP.
Hence Enthalpy does indeed have the desired natural variables.
19.2.1. Heuristic definition:
Enthalpy is the total energy of the system minus the pressure volume energy. So
a change in enthalpy is the change in internal energy adjusted for the PV work
done. If the process occurs at constant pressure then the enthalpy change is the
heat given off or taken in.
For example, consider an reversibly expanding gas under constant pressure (dP =
0) and adiabatic (dq = 0) conditions.
The system does work during the expansion; in doing so it must lose energy. Since
the process is adiabatic no heat energy can flow in to compensate for the work
done and the gas cools.
The total internal energy decreases. The enthalpy of the system on the other hand
does not change–it is the internal energy adjusted by an amount of energy equal
to the PV work done by the system. As Freshmen we learn this as 4H = qp.
19.3. Helmholtz Free Energy
Now we want a state function whose natural variables are T and V
137
Let us try the definition A ≡ U − TS.
Formally
dA = dU − d(TS) = dU − TdS − SdT, (19.4)
but dU = TdS − PdV, so
dA = TdS/ − PdV − TdS/ − SdT (19.5)
= −PdV − SdT.
Hence Helmholtz free energy does indeed have the desired natural variables.
19.3.1. Heuristic definition:
As we have said before Helmholtz free energy is the energy of the system which is
available to do work–It is the internal energy minus that energy which is “used
up” by the random thermal motion of the molecules.
19.4. Gibbs Free Energy
Finally we want a state function whose natural variables are T and P
Let us try the definition G ≡ H − TS.
Now formally
dG = dH + d(TS) = dH − TdS − SdT, (19.6)
but from above dH = TdS + V dP, so
dG = TdS/ + V dP − TdS/ − SdT (19.7)
= V dP − SdT.
Hence Gibbs free energy does indeed have the desired natural variables.
138
19.4.1. Heuristic definition:
Gibbs free energy is the energy of the system which is available to do non PV
work–It is the internal minus both that energy which is “used up” by the random
thermal motion of the molecules and used up in doing the PV work.
19.5. Heat Capacity of Gases
19.5.1. The Relationship Between CP and CV
To find how CP and CV are related we begin with
dH = TdS + V dP (19.8)
at constant pressure and reversible conditions
dH = TdS (19.9)
dH = dq
but
dq = CPdT (19.10)
The constant pressure heat capcity can then be expressed in terms of enthalpy as
CP =
µ∂H
∂T
¶P
. (19.11)
So,
CP =
µ∂ (U + PV )
∂T
¶P
=
µ∂U
∂T
¶P
+ P
µ∂V
∂T
¶P
(19.12)
note¡∂U∂T
¢Pis not CV we need
¡∂U∂T
¢V. Use an identity of partial derivativesµ
∂U
∂T
¶P
=
µ∂U
∂T
¶V
+
µ∂U
∂V
¶T
µ∂V
∂T
¶P
(19.13)
139
thus
CP =
µ∂U
∂T
¶V
+
µ∂U
∂V
¶T
µ∂V
∂T
¶P
+ P
µ∂V
∂T
¶P
(19.14)
= CV +
µ∂V
∂T
¶P
∙µ∂U
∂V
¶T
+ P
¸.
Recall the expression for internal pressure¡∂U∂V
¢T= T
¡∂P∂T
¢V− P . Then
CP = CV +
µ∂V
∂T
¶P
∙T
µ∂P
∂T
¶V
− P/ + P/¸
(19.15)
Finally
CP = CV + T
µ∂V
∂T
¶P
µ∂P
∂T
¶V
(19.16)
Example: Ideal gases
1. Ideal gas (equation of state: PV = nRT ): This equation is easily made
explicit in either P or V so we don’t need any of the above replacements
CP = CV + T
µ∂V
∂T
¶P
µ∂P
∂T
¶V
(19.17)
= CV + TnR
P
nR
V=
nRT
PVnR
Thus CP = CV + nR or
CPm = CVm +R (19.18)
19.6. The Maxwell Relations
Summary of thermodynamic relations we’ve seen so far
Definitions and relations:
• H = U + PV
140
• A = U − TS
• G = H − TS
• CV =¡∂U∂T
¢V, CP =
¡∂H∂T
¢P
basic equations Maxwell relations working equations
dU = TdS − PdV¡∂T∂V
¢S= −
¡∂P∂S
¢V
dU = CV dT +£T¡∂P∂T
¢V− P
¤dV
dH = TdS + V dP¡∂T∂P
¢S=¡∂V∂S
¢P
dH = CPdT −£T¡∂V∂T
¢P− V
¤dP
dA = −PdV − SdT¡∂S∂V
¢T= +
¡∂P∂T
¢V
dS = CVTdT +
¡∂P∂T
¢VdV
dG = V dP − SdT¡∂S∂P
¢T= −
¡∂V∂T
¢P
dS = CPTdT −
¡∂V∂T
¢PdP
We will get plenty of practice with derivations based on these equations and on
the properties of partial derivatives. (See handout and Homework)
141
20. Chemical Potential
20.1. Spontaneity of processes
Two factors drive spontaneous processes
1. The tendency to minimize energy
2. The tendency to maximize entropy
Let us begin with Helmholtz free energy
The total differential of A is (A = U − TS)
dA = dU − TdS − SdT = dq − PexdV − TdS − SdT (20.1)
For constant T and V, (dA)T,V = dq − TdS
From the second law, TdS ≥ dq for a spontaneous process, (dA)T,V ≤ 0 for aspontaneous process.
Hence at equilibrium (dA)T,V = 0.
For chemistry it is most often more convenient to use Gibbs free energy
The total differential of G is
dG = dH − TdS − SdT = dq − PexdV + PdV + V dP − TdS − SdT (20.2)
142
142
For constant T and P = Pex, (dG)T,P = dq − TdS
Again from the second law, TdS ≥ dq for a spontaneous process, (dG)T,P ≤ 0 fora spontaneous process.
Hence at equilibrium (dG)T,P = 0.
So free energy provides a measure of the thermodynamic driving force towards
equilibrium.
Note free energy provides no information about how fast a process proceeds toequilibrium.
The free energy functions are the workhorses of applied thermodynamics so we
want to get a feel for them.
Returning to the total differentials of free energy,
dA = dU − TdS − SdT (20.3)
and
dG = dH − TdS − SdT. (20.4)
Expressing dU and dH generally as dU = TdS − PdV and dH = TdS + V dP
(remember that in general TdS cannot be identified with dq and PdV cannot be
identified with −w).
Plugging these into the total differentials of free energy gives
dA = −SdT − PdV (20.5)
and
dG = −SdT + V dP (20.6)
143
These expressions are quite general, but i) only PV work and ii) closed systems.
The total differential of A is also
dA = dq + dw − TdS − SdT. (20.7)
For a reversible process dq = TdS and work is maximal.
Hence (dA)T = dwmax =⇒ (4A)T = wmax. As we have stated in words a number
of times before.
The total differential of G is also
dG = dq + dw + PdV + V dP − TdS − SdT. (20.8)
In general dw = dw0 − PexdV where dw0 is the non-PV work.
The total differential of G becomes
dG = dq + dw0 − PexdV + PdV + V dP − TdS − SdT. (20.9)
For constant T and P = Pex, (dG)T,P = dq + w0 − TdS.
For reversible processes q = TdS and this becomes
(dG)T,P = dw0max =⇒ (4G)T,P = w0max (20.10)
So, as stated earlier, the Gibbs free energy is the energy of the system available
to do non-PV work.
20.2. Chemical potential
What if the amount of substance can change?
144
Extensive properties depend on the amount of “stuff”
For example A(T, V ) now becomes A(T, V, n) and the total differential becomes
dA(T, V, n) =
µ∂A
∂T
¶V,n
dT +
µ∂A
∂V
¶T,n
dV +
µ∂A
∂n
¶V,T
dn (20.11)
Let’s focus on the slope¡∂A∂n
¢V,T.
• This is a measure of the change in Helmholtz free energy of a system (at
constant T and V ) with the change in the amount of material.
• Physically, this is a measure of the potential to change the amount of mate-rial.
• It defines the chemical potential μ ≡¡∂A∂n
¢V,T
.
So we can also write
dA = −SdT − PdV + μdn (20.12)
What about the relation of the chemical potential to Gibbs’ free energy?
G = H − TS = U − TS| z =A
+ PV = A+ PV so,
dG = dA+ PdV + V dP (20.13)
= −SdT − P/ dV/ + P/ dV/ + V dP + μdn
= −SdT + V dP + μdn,
but from
dG =
µ∂G
∂T
¶P,n
dT +
µ∂G
∂P
¶T,n
dP +
µ∂G
∂n
¶P,T
dn (20.14)
145
we see that
μ =
µ∂G
∂n
¶P,T
. (20.15)
So, μ is also a measure of the change in Gibbs free energy of a system (at constant
T and P ) with the change in the amount of material and it still has the same
physical meaning.
The Gibbs free energy per mole (Gm) for a pure substance is equal to the chemical
potential. (Gm = μ)
20.3. Activity and the Activity coefficient
When, for example, a solute is dissolved in a solvent, there exist complicated
interactions which cause deviations from ideal behavior.
To account for this one must introduce the concept of activity and the activity
coefficient.
Activity is hard to define in words and indeed it has an awkward mathematical
definition as we will soon see.
The activity coefficient has a more convenient definition which is that it is the
measure of how a particular real system deviates from some reference system
which is usually taken to be ideal.
The mathematical definition of activity ai of some species i is implicitly stated as
limζ→ζª
aig(ζ)
= 1 (20.16)
where g(ζ) is any reference function (e.g., pressure, mole fraction, concentration
etc.), and ζª is the value of ζ at the reference state.
146
This implicit definition is awkward so for convenience one defines the activity
coefficient as the argument of the above limit,
γi ≡aig(ζ)
(20.17)
which we can rearrange as
ai = γig(ζ). (20.18)
The definition of activity implies that γi = 1 at g(ζª) (the reference state)
20.3.1. Reference States
Thermodynamics is founded on the concept of energy which we know to have an
arbitrary scale. That is, we can define are zero of energy any where we want.
Because of this it is always necessary to specify a reference state to which our real
state can be compared.
The choice of this state is completely up to us, but it is often the case that the
reference state is chosen to be some ideal state.
For example, if we are talking about a gas we will mostly likely choose the ideal
gas law in terms of pressure (P = nRT/V ) as our reference function and the
reference state being when P = 0 since we know all gases behave ideally in the
limit of zero pressure.
Let us consider the activity of a real gas for the above reference function and
reference state. Note: the activity of gases as referenced to pressure has the
special name fugacity (fugacity is a special case of activity).
147
Our reference function is very simple: g(ζ) = ζ = P , so
γ =a
P⇒ a = γP. (20.19)
Thus the activity of our real gas is given by the activity coefficient times the
pressure of an ideal gas under the same conditions.
Based on the condition that γ → 1 as we approach the reference state (P = 0
in this case) we see that the activity (or fugacity) of a real gas becomes equal to
pressure for low pressures
20.3.2. Activity and the Chemical Potential
One cannot measure absolute chemical potentials, only relative potentials can be
measured. By convention we chose a standard state and measure relative to that
state.
The deviation of the chemical potential at the state of interest versus at the
reference state is determined by the activity at the current state (the activity at
the reference state is unity by definition).
μi − μªi = RT ln ai. (20.20)
Rather than referencing to the standard state one can also reference to any con-
venient “ideal” state. This ideal state is in turn referenced to the standard state.
For the state of interest
μi = μªi +RT ln ai (20.21)
and for the ideal state
μidi = μªi +RT ln aidi ⇒ μªi = μidi −RT ln aidi . (20.22)
148
Thus,
μi = μidi −RT ln aidi +RT ln ai (20.23)
μi − μidi = +RT ln ai −RT ln aidi
= RT lnaiaidi
Example: Real and ideal gases at constant temperature, but any pressure.Starting from the begining
dμid = dGm = −Sm=0z|dT + VmdP (20.24)
dμid = VmdP
dμid =RT
PdP.
Now we integrate from the reference state to the current state of interestZμª
dμid =
ZPª
RT
PdP. (20.25)
This gives
μid − μª = RT lnP
Pª. (20.26)
The usual standard state is the ideal gas at Pª = 1, so
μid = μª +RT lnP. (20.27)
(Note that as P → 1, μid → μª).
Lets say our gas is not ideal, then at a given pressure
μ = μª +RT ln a. (20.28)
For gases activity is usually called fugacity and given the symbol f , so a = f for
real gasses. Thus
μ = μª +RT ln f. (20.29)
149
Lets say that instead of referencing to the ideal gas at P = 1, we want to reference
to the ideal gas at the current pressure P.
This is easily done by using μª = μid −RT lnP in the above equation for μ,
μ = μid −RT lnP +RT ln f
μ = μid +RT lnf
P.
Example: The barometric equation for an ideal gas.We have an ideal gas so,
μid = μª +RT lnP (20.30)
where we will take the reference state to be at sea level, i.e. Pª = 1 atm.
So at sea level
μid(0) = μª +
=0z | RT ln 1 = μª (20.31)
and at elevation h
μid(h) = μª +RT lnPh (20.32)
The gas fields the gravitational force which gives it a potential energy per mole
of Mgh at height h. We add this energy per mole term to the chemical potential
(which is free energy per mole) thus at equilibrium
μid(0) = μid(h) +Mgh (20.33)
Referencing to the reference state we get
μª/ = μª/+RT lnPh +Mgh (20.34)
RT lnPh = −Mgh
Ph = e−MghRT
The last line is the barometric equation and it shows that pressure is exponentially
decreasing function of altitude.
150
21. Equilibrium
First let us consider the equilibrium A B.
Since A and B are in equilibrium their chemical potentials must be equal
μA = μB (21.1)
Now,
μA = μªA +RT ln aA (21.2)
and
μB = μªB +RT ln aB
So the equilibrium condition becomes
μªA +RT ln aA = μªB +RT ln aB (21.3)
−4μª = μªA − μªB = RT ln aB −RT ln aA
−4μª = RT lnaBaA
Since chemical potential is free energy per mole, if we multiply the above by n
moles we have
−4Gª = nRT lnaBaA
as a consequence of the equilibrium condition.
The quantityaBaAdefines the equilibrium constant, Ka, for this process.
151
151
Say the system A→ B is not in equilibrium then we can not write
μA = μB
but we can write
μA +
4μz | μB − μA = μB (21.4)
Proceeding as above we get
μªA +RT ln aA +4μ = μªB +RT ln aB (21.5)
4μ = μªB − μªA +RT ln aB −+RT ln aA4μ = 4μª +RT ln
aBaA
.
Again multiplying by n gives
4G = 4Gª + nRT lnaBaA
.
If the 4G < 0 then the transition A→ B proceeds spontaneously as written.
Consider a more complicated equilibrium
aA+ bB cC + dD. (21.6)
The equilibrium condition is
aμA + bμB = cμC + dμD. (21.7)
In a manner similar to the above
aμªA+aRT ln aA+ bμªB+ bRT ln aB = cμªC + cRT ln aC +dμªD+dRT ln aD (21.8)
Rearranging gives
≡−4rxnGªz | aμªA + bμªB − cμªC − dμªD = RT ln
acCadD
aaAabB
(21.9)
152
the equilibrium constant is
Ka =acCa
dD
aaAabB
=⇒4Gª = −RT lnKa (21.10)
Note: n is absent in the above since the molar values are implied by the stoi-
chiometry.
21.0.3. Equilibrium constants in terms of KC
Equilibrium constant in terms of a condensed phase concentration:
K 0C =
[C]c [D]d
[A]a [B]b, (21.11)
which is related to Ka by
Ka = K 0C
µγcCγ
dD
γaAγbB
¶. (21.12)
If the reactants are solutes then as the solution is diluted all the activity coefficients
go to unity and K 0C → Ka.
21.0.4. The Partition Coefficient
Up to now we have only considered miscible solutions.
We now consider the problem of determining the equilibrium concentrations of a
solute A in both phases of an immiscible mixture.
153
The equilibrium equation is
Aα Aβ (21.13)
The equilibrium expression for this process is
4Gα→β = 0 = 4Gªα→β − nRT lnKa, (21.14)
where,4Gªα→β ≡ Gª
β −Gªα . The equilibrium constant for this process has a special
name; it is called the partition coefficient, P β/α ≡ Kβ/αpart, for species A in the α—β
mixture.
We can solve for the partition coefficient to yield
P β/α =aβAaαA= e−
4Gªα→β
nRT . (21.15)
For low concentrations
P β/α ' [A]β
[A]α. (21.16)
Knowledge of the partition function is important on the delivery of drugs because,
to enter the body, the drugs must transfer between an aqueous phase and a oil
phase.
154
For most drugs
0 < Po/wpart < 4 (21.17)
Partition coefficient Delivery mechanism
low Po/wpart (likes water) injection
medium Po/wpart oral
high Po/wpart (likes oil) skin patch/ointment
Factors other than the partition coefficient influence the drug delivery choice. For
example, can the drug handle the acidic environment of the stomach?
155
22. Chemical Reactions
Up to now we have only been considering systems in the absence of chemical
reactions. After chemical reactions take place the system is in a final “product”
thermodynamic state that is in general different from the initial “reactant” state.
For any extensive property
• 4rxn(Property) = property of products − property of reactants
• Example
— Reaction: aA+bB= cC+dD
— 4rxnS = cSm,C + dSm,D − aSm,A − bSm,B
22.1. Heats of Reactions
Exothermic reaction: heat is given off to the surroundingsEndothermic reaction: heat is given taken in from the surroundings
At constant pressure (Pex = P
q = 4rxnU − w = 4rxnU − P4rxnV = 4rxnH (22.1)
4rxnH < 0 for Exothermic reactions.
4rxnH > 0 for Endothermic reactions.
156
156
22.1.1. Heats of Formation
Hess’s Law of heat summation: 4rxnH is independent of chemical pathway
Example: C2H2+H2 = C2H4.This direct reaction is not easy but it can be done in steps
C2H2 + 52O2 → 2CO2 +H2O(liq) 4rxnH
ª = −1299.63 kJ2CO2 +2H2O(liq)→C2H4 + 3O2 4rxnH
ª = +1410.97 kJ
H2 + 12O2 →H2O(liq) 4rxnH
ª = −285.83 kJC2H2+H2 = C2H4 4rxnH
ª = −174.49 kJ
The heat of formation 4fHª is the 4rxnH at STP in forming a compound from
its constituent atoms in their natural states.
O2, H2, C(graphite) are examples of atoms in their natural state.
Example: Formation of water
• H2 + 12O2 = H2O not 2H2+O2 = 2H2O
• 4rxnH =P
i νi4fH(i), where νi is the stoichiometric factor of the ith com-
ponent.
Example: H2O(liq)→H2O(gas) at SATPH2 + 1
2O2 = H2O(gas) 4fH
ª = −241.818 kJH2 + 1
2O2 = H2O(liq) 4fH
ª = −285.830 kJH2O(liq)→H2O(gas) 4rxnH
ª = −241.818− (−285.830) = 44.012 kJ
22.1.2. Temperature dependence of the heat of reaction
4rxnH(T2) = 4rxnH(T1) +
Z T2
T1
4rxnCPdT (22.2)
157
22.2. Reversible reactions
Recall the requirement for a spontaneous change: 4G < 0 for constant T and P.
4rxnG = G(products)−G(reactants) =Xi
νiμi, (22.3)
(remember μi = Gm,i for pure substance i).
As we saw before μi can be defined in terms of activity
μi = μªi +RT ln ai. (22.4)
So,
4rxnG =
4rxnGªz | Xi
νiμªi +RT
Xi
νi ln ai. (22.5)
Using the property of logarithms: a lnx + b ln y = ln(xayb) the above expression
becomes
4rxnG = 4rxnGª +RT ln
Yi
aνii (22.6)
4rxnG = 4rxnGª +RT lnQ,
where Q ≡Q
i aνii is the activity quotient.
At equilibrium, 4rxnG = 0 and Q = Ka (Thermodynamic equilibrium constant).
Ka depends on T but is independent of P.
For the reaction aA + bB = cC + dD
Ka =acCa
dD
aaAabB
(22.7)
• Note that the activity of any pure solid or liquid is for all practical purposesequal to 1.
158
• For ideal gases, ai = PiPª =
XiPPª (Pª = 1 bar) This leads to the sometimes
useful relation
KP =P cCP
dD
P aAP
bB
=(PªaC)
c (PªaD)d
(PªaA)a (PªaB)
b= Ka
¡Pª¢c+d−a−b , (22.8)
or more generally KP = Ka (Pª) i νi .
So at equilibrium, 4rxnG = 4rxnGª +RT lnQ becomes
0 = 4rxnGª +RT lnKa ⇒4rxnG
ª = −RT lnKa. (22.9)
22.3. Temperature Dependence of Ka
Starting with G = H − TS or G/T = H/T − S.
From this³∂(G/T )∂(1/T )
´P= H.
Applying this to4rxnG
ª
T=4rxnH
ª
T−4rxnS (22.10)
gives µ∂(4rxnG
ª/T )
∂(1/T )
¶P
= 4rxnHª (22.11)
Using 4rxnGª = −RT lnKa, we getµ
∂ lnKa
∂(1/T )
¶P
ind.=of P
d lnKa
d(1/T )= −4rxnH
ª
R(22.12)
or (using dd(1/T )
= dTd(1/T )
ddT= −T 2 d
dT)
d lnKa
dT=4rxnH
ª
RT 2(22.13)
Integration gives
lnKa(T2) = lnKa(T1) +1
R
Z T2
T1
4rxnHªm
T 2(22.14)
For a reasonably small range T2 − T1 this is well approximated by
lnKa(T2) = lnKa(T1)−4rxnH
ªm
R
µ1
T2− 1
T1
¶(22.15)
159
22.4. Extent of Reaction
There are other equilibrium “constants” that are used in the literature.
• From Pj = XjP , KX = KPP−4υg
• From nj = PjVRT(ideal gas approximation), Kn = KP
¡RTV
¢−4υg
• From concentration Cj =njV=
PjRT
, KC = KP (RT )−4υg
Equilibrium “constants”“constants” expression relation to Ka situation used
Kaactivity(products)activity(reactants) – when an exact answer is needed
KPpartial pressure(products)partial pressure(reactants)
Ka
KγPª−4υg gas reactions
KXmole fraction(products)mole fraction(reactants)
µKa
KγPª−4υg
¶P−4υg when eq. P is known
Knmoles(products)moles(reactants)
µKa
KγPª−4υg
¶¡RTV
¢−4υg when V is known and constant
KCconcentration(products)concentration(reactants)
µKa
KγPª−4υg
¶(RT )−4υg when concentration known
160
23. Ionics
Many chemical processes involve electrolytes and or acids and bases.
To understand these processes we must know something about how ions behave
in solution.
23.1. Ionic Activities
Consider a salt in solution
Mv+Xv− → v+Mz+(aq) + v−X
z−(aq), (23.1)
where v+ (v−) is the number of cations (anions) and z+ (z−) is the charge on the
cation (anion).
The chemical potential for the salt may be written in terms of the chemical po-
tential for each of the ions:
μsalt = v+μ+ + v−μ− (23.2)
To determine the activity we start with
ln aj =μj − μªjRT
, j = + or − (23.3)
and
ln asalt =μsalt − μªsalt
RT. (23.4)
161
161
Substituting the expression for μsalt into this gives
ln asalt =v+μ+ − v−μ− + v+μ
ª+ − v−μ
ª−
RT(23.5)
=v+μ+ − v+μ
ª+
RT| z v+ ln a+
+v−μ− − v−μ
ª−
RT| z v− ln a−
So,
ln asalt = v+ ln a+ + v− ln a− (23.6)
or, alternatively,
asalt = av+av− (23.7)
It is the case that 1 mole of salt behaves like v = v++ v− moles of nonelectrolytes
in terms of the colligative properties. This suggests that the interesting quantity
is μsaltv:
μsaltv=
μªsaltv+RT ln a
1/vsalt. (23.8)
We see that
a1/vsalt = (a
v+av−)1/v ≡ a±. (23.9)
The quantity a± is the mean ionic activity.
23.1.1. Ionic activity coefficients
The activity coefficients for ionic solutions can also be defined via
a+ = γ+m+, a− = γ−m−, (23.10)
where m+ = v+m and m− = v−m.
The mean ionic activity coefficient is
γ± = (γv++ γ
v−− )
1/v. (23.11)
162
The quantities a+, a−, γ+ and γ− cannot be measured individually.
One can use the colligative properties to measure the ionic activity coefficients.
It is convenient to redefine the osmotic coefficient as
φ =−1000 g/kg
vmM1ln a1, (23.12)
where the subscript 1 refers to the solvent.
Similarly freezing point depression is redefined as
θ = vφKfm. (23.13)
So, vφ corresponds to the empirical factor i discussed earlier.
Recall how γ was calculated from the Gibbs-Duhem equation:
ln γ± = −j −Z m
0
j
m0dm0, (23.14)
where j = 1− φ.
23.2. Theory of Electrolytic Solutions
Ionic strength is defined as
I =1
2
Xi
z2imi, (23.15)
where z is the charge of the ion and m its concentration.
Results from Debye—Hückel theory: point charge in a continuumThe Debye—Hückel equation:
ln γ± =−α |z+z−|
√I
1−Ba0√I
, (23.16)
163
where
α =e3
(εkT )3/2
µ2πρ•L
1000
¶1/2, (23.17)
B =8πLe2ρ•
1000εkT, (23.18)
a0 is the radius of closest approach, e is the charge on the electron, ρ• is the
density of the pure solvent, ε is the dielectric constant for the pure solvent and L
is Avogadro’s number.
Notice that the parameters α and B depend only on the solvent.
One important approximation to this equation is to neglect the B term to get the
Debye—Hukel limiting Law (DHLL):
ln γ± = −α |z+z−|√I. (23.19)
This gives the dependence of ln γ± for dilute solutions (m → 0). It is seen that
the DHLL correctly predicts the√m dependence of ln γ±, which is observed ex-
perimentally (recall I = 12
Pi z2imi).
A useful empirical approximation is to set Ba0 = 1 and to add an empirical
correction to get the :
ln γ± =−α |z+z−|
√I
1−√I
+ 2βm
µv2+ + v2−v+ + v−
¶. (23.20)
This equation works well to ionic strengths of about I = 0.1
23.3. Ion Mobility
Current, I is given by the rate of change (in time) of charge, Q:
I =dQ
dt(23.21)
164
(Electrical) work, w, is required to move a change through a potential (or voltage),
ε :
w = −εQ (23.22)
Power is given by the product of the voltage and the current:
p = −εI (23.23)
Resistance is given by the ratio of the voltage to current:
R =ε
I
Conductance is the inverse of the resistance (R−1).
Some relevant constants
• charge of an electron e = 1.602177× 10−19 C.
• Faraday’s constant F = Le = 96485 C/mol (Avogadro’s number of electrons)
23.3.1. Ion mobility
165
The total current passing through an ionic solution is determined by the sum of
the current carried by the cations and by the anions
I = I+ + I− (23.24)
Now
Ii =dQi
dt= |zi| e
dNi
dt, (23.25)
where i = +,−.
For uniform ion velocity (vi) the number of ions arriving at the electrode during
any given time interval 4t is
4Ni =Ni
VAvi4t =⇒ dNi
dt=
Ni
VAvi (23.26)
so
Ii = |zi| eNi
VAvi (23.27)
Recall Coulomb’s law
Fi = zieE, (in vacuum) (23.28)
where E is the electric field, E = dεdx.
Also recall Newton’s law
Fi = mai = mdvidt= zieE. (23.29)
The moving ions experience a viscous drag f that is proportional to their velocities.
So the total force on the ions is a sum of the Coulomb force and the viscous drag
Fi = zieE − fvi (in solution). (23.30)
The ions quickly reach terminal velocity, i.e., the viscous drag equals the Coulomb
force. Hence Fi = 0.
zieE = fvi =⇒ vi =zieE
f. (23.31)
The drag f has three basic origins.
166
1. Stoke’s Law type force
• “spherical” ion moving through a continuous medium
• this contribution is independent of the other ions
2. Electrophoretic effect.
• oppositely charged ions “pull” at each other
3. Relaxation effects
• solvation shell must re-adjust as ion moves. a “dressed” ion.
167
A more fundamental quantity than ion velocity is the ion mobility, ui which is the
ion’s velocity per field,
ui =viE. (23.32)
For the case for parallel plate capacitors E = εl, where l is the separation of the
plates. So,
ui =vil
ε. (23.33)
Here the current carried by ion i is
Ii = |zi| eNi
VAuiε
l. (23.34)
Suppose a salt has a degree of dissociation α (α = 1 for strong electrolytes) to
produce ν+ cations and ν− anions, then each mole of salt gives: N+ = αν+Ln
and N− = αν−Ln.
The current then becomes
Ii = |zi| eανiLn
VAuiε
lF=Le= ανin |zi|uiAF
ε
V l(23.35)
It is of interest to determine the ratio of the current carried by the cation versus
the anion.
I+I−=
α/ ν+n/ |z+|u+A/ F/ εV l/
α/ ν−n/ |z−|u−A/ F/ εV l/=
=1z | ν+ |z+|ν− |z−|
u+u−
=u+u−
(23.36)
Thus the ratio of the currents is determined by simply the ratio of the mobilities.
168
24. Thermodynamics of Solvation
An extremely important application of thermodynamics is to that of ion solvation.
Solvation describes how a solute dissolves in a solvent.
We will focus on ions in solution.
As a basic treatment of solvation we shall consider the solvent as a non-structural
continuum and the ion as a charged particle.
Of course this is an approximation and numerous statistical mechanical models
for solvents which incorporate a more realistic structure can be used, but we will
stick with this simple thermodynamic model.
The way to investigate the ion—solvent interaction upon solvation from a thermo-
dynamics point of view is to consider the change in the properties of the ion in a
vacuum versus the ion in solution.
Primarily we will determine 4Gv→s ≡ Gion in solv. −Gion in vac.
Since Gibbs free energy corresponds to non-PV work, 4Gv→s can be determined
by calculating the reversible work done in transferring an ion into the bulk of the
solvent.
169
169
24.1. The Born Model
The Born model is a simple solvation model in which the ions are taken to be
charged spheres and the solvent is take to be a continuum with dielectric constant
εs
170
4Gv→s for the Born model is obtained by considering the following contribution
to the work of ion transfer from the vacuum state to the solvated state (see figure)
• Begin with the state in which the charged sphere (the ion) is in a vacuum.
• Determine the work, wdis, done in discharging the sphere.
• Assume the uncharged sphere can pass from the (neutral) vacuum to the
neutral solvent without doing any work, wtr = 0. (This is an approximation).
• Determine the work, wch, done in charging the sphere which is now in thesolvent.
171
So,
4Gv→s = wdis + wtr + wch = wdis + wch (24.1)
Work done in discharging the sphere:
The act of discharging a sphere involves bringing out to infinity from the surface
infinitesimal amounts of charge.
The work done is discharging is some what complicated since as one removes the
charge the work done in removing more charge changes according to the amount
of charge currently on the sphere.
172
This is expressed mathematically as
wdis =
Z 0
ze
Z ∞
ri
σ
4π 0r2drdσ (24.2)
=
Z 0
ze
σ
4π 0ridσ
= − (ze)2
8π 0ri,
where z is the oxidation state of the ion, e is the charge of the electron, ri is the
radius of the sphere (ion) and 0 is the permittivity of free space.
Work done in charging the sphere:
The only difference in charging the sphere is that the sign of the work will be differ-
ent and that since we are charging in a solvent we must multiply the permittivity
of free space by the dielectric constant of the solvent.
So,
wch = +(ze)2
8π 0εsri(24.3)
24.1.1. Free Energy of Solvation for the Born Model
Combining the above two expression for work gives
4Gv→s = − (ze)2
8π 0ri+
(ze)2
8π 0εsri(24.4)
=(ze)2
8π 0ri
µ1
εs− 1¶
The above expression is 4Gv→s/ion. For n moles of ions (nL = N)
4Gv→s =N (ze)2
8π 0ri
µ1
εs− 1¶
(24.5)
173
The dielectric constant of any solvent is always greater than unity so 1εs− 1 is
always negative hence 4Gv→s < 0. Thus ions always exist more stably in solution
than in a vacuum.
24.1.2. Ion Transfer Between Phases
We can quickly generalize the Born model to describe ion transfer between phases
in a solution of two immiscible phases
Consider an immiscible solution of two phases α and β having dielectric constants
εα and εβ.
Since Gibbs free energy is a state function we can write the change in free energy
for transfer of an ion form the β phase to the α phase as
4Gβ→α =
=−4Gv→βz | 4Gβ→v +4Gv→α (24.6)
= −N (ze)2
8π 0ri
µ1
εβ− 1¶+
N (ze)2
8π 0ri
µ1
εα− 1¶
=N (ze)2
8π 0ri
µ1
εα− 1
εβ
¶The Partition Coefficient
We can now write the partition coefficient for the Born model as
Pα/βi = e
−4Gªβ→α
nRT = e− L(ze)2
8πri 0RT1εα− 1εβ (24.7)
24.1.3. Enthalpy and Entropy of Solvation
We may employ the standard thermodynamic relations which we have derived
earlier to obtain the entropy and enthalpy for the Born model.
174
From µ∂G
∂T
¶P
= −S ⇒µ∂4Gv→s
∂T
¶P
= −4Sv→s, (24.8)
we find entropy to be
4Sv→s = −∂
∂T
"N (ze)2
8π 0ri
µ1
εs− 1¶#
. (24.9)
The only variable in the above equation that has a temperature dependence is the
dielectric constant of the solvent so,
4Sv→s = −N (ze)2
8π 0ri
∂
∂T
µ1
εs
¶=
N (ze)2
8π 0riε2s
∂εs∂T
. (24.10)
Enthalpy is obtained via the relation:
4Hv→s = 4Gv→s + T4Sv→s (24.11)
=N (ze)2
8π 0ri
µ1
εs− 1¶+
N (ze)2 T
8π 0riε2s
∂εs∂T
=N (ze)2
8π 0ri
µ1
εs+
T
ε2s
∂εs∂T− 1¶
24.2. Corrections to the Born Model
The Born model is very valuable because of its simplicity–qualitative statements
about solvation and ion transfer between phases can be made.
Unfortunately however, the Born model does not make quantitatively correct pre-
dictions in many cases.
We simply list here several phenomena that more sophisticated theories of solva-
tion must consider
175
1. The solvophobic effect: a cavity must form in the solvent to accommodate
the ion.
2. Changes in solvent structure: the local environment of the ion has a different
arrangement of solvent molecules than that of the bulk solvent, so the initial
structure of the solvent must breakdown and the new structure must form.
3. Specific interactions: any interaction energy specific to the particular ion-
solvent pair: Hydrogen bonding being the prime example.
4. Annihilation of defects: A small ion may be captured in a micro-cavity
within the solvent releasing the energy of the micro-cavity defect.
176
25. Key Equations for Exam 4
Listed here are some of the key equations for Exam 4. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• Some thermodynamic relations
H = U + PV dH = TdS + V dP
A = U − TS dA = −SdT − PdV
G = H − TS dG = −SdT + V dP
• The chemical potential equation
μi = μªi +RT ln ai (25.1)
• The 4G equation (this should be posted on your refrigerator)
4G = 4Gª +RT lnQ. (25.2)
177
177
At equilibrium 4G = 0 and
4Gª = −RT lnKa (25.3)
• For an ideal gasCPm = Cvm +R (25.4)
• The Debye—Hukel limiting Law (DHLL):
ln γ± = −α |z+z−|√I. (25.5)
• The ratio of the current carried by the cation versus the anion in terms ofion mobility is
I+I−=
u+u−
(25.6)
• The chemical potential equation
μi = μªi +RT ln ai (25.7)
• The 4G equation (this should be posted on your refrigerator)
4G = 4Gª +RT lnQ. (25.8)
At equilibrium 4G = 0 and
4Gª = −RT lnKa (25.9)
• 4G for the Born model:
4Gv→s =N (ze)2
8π 0rs
µ1
εs− 1¶
(25.10)
• 4G for transfer of an ion form the β phase to the α phase,
4Gβ→α =N (ze)2
8π 0ri
µ1
εα− 1
εβ
¶(25.11)
178
Chemistry 352: PhysicalChemistry II
179
179
Part V
Quantum Mechanics andDynamics
180
180
26. Particle in a 3D Box
We now return to quantum mechanics and investigate some of the important
models that we omitted from the first semester.
In particular we will look at the particle in a box in more than one dimension.
We will also solve models which deal with rotations.
26.1. Particle in a Box
Recall that the important ideas from the 1D particle in a box problem were
The potential, V (x), is given by
V (x) =
⎧⎪⎨⎪⎩∞ x ≤ 00 0 < x < a
∞ x ≥ a
. (26.1)
Because of the infinities at x = 0 and x = a, we need to partition the x-axis into
the three regions shown in the figure.
181
181
Now, in region I and III, where the potential is infinite, the particle can never
exist so, ψ must equal zero in these regions.
The particle must be found only in region II.
The Schrödinger equation in region II is (V (x) = 0)
Hψ = Eψ =⇒ −~2
2m
d2ψ(x)
dx2= Eψ, (26.2)
The general solution of this differential equation is
ψ(x) = A sin kx+B cos kx, (26.3)
where k =q
2mE~2 .
Now ψ must be continuous for all x. Therefore it must satisfy the boundary
conditions (b.c.): ψ(0) = 0 and ψ(a) = 0.
From the ψ(0) = 0 b.c. we see that the constant B must be zero because
cos kx|x=0 = 1.So we are left with ψ(x) = A sin kx for our wavefunction.
182
The second b.c., ψ(a) = 0, places certain restrictions on k.
In particular,
kn =nπ
a, n = 1, 2, 3, · · · . (26.4)
The values of k are quantized. So, now we have
ψn(x) = A sinnπx
a. (26.5)
The constant A is the normalization constant.
Solving for A gives
A =
r2
a. (26.6)
Thus our normalized wavefunctions for a particle in a box are (in region II)
ψn(x) =
r2
asin
nπx
a. (26.7)
We found the energy levels to be
En =n2π2~2
2ma2~= h
2π=n2h2
8ma2. (26.8)
26.2. The 3D Particle in a Box Problem
We now consider the three dimensional version of the problem.
The potential is now
V (x, y, z) =
(0, 0 < x < a, 0 < y < b, 0 < z < c
∞, else. (26.9)
183
Now the Schrödinger equation is
Hψ = Eψ ⇒ −~22m∇2ψ = Eψ
⇒ −~22m
µ∂2ψ
∂x2+
∂2ψ
∂y2+
∂2ψ
∂z2
¶= Eψ. (26.10)
It is generally true that when the Hamiltonian is a sum of independent terms, we
can write the wavefunction as a product of wavefunctions
ψ(x, y, z) = ψx(x)ψy(y)ψz(z). (26.11)
This lets us perform a mathematical trick which is sometimes useful in solving
partial differential equations.
Subbing the product wavefunction into the Schrödinger equation we get
−~22m
µ∂2ψxψyψz
∂x2+
∂2ψxψyψz
∂y2+
∂2ψxψyψz
∂z2
¶= Eψxψyψz (26.12)
−~22m
µψyψz∂
2ψx
∂x2+
ψxψz∂2ψy
∂y2+
ψxψy∂2ψz
∂z2
¶= Eψxψyψz.
We now divide both sides by ψxψyψz to get
−~22m
µ1
ψx
∂2ψx
∂x2+1
ψy
∂2ψy
∂y2+1
ψz
∂2ψz
∂z2
¶= E. (26.13)
This equation is now of the form
f(x) + g(y) + h(z) = C, (26.14)
where C is a constant.
If we take the derivative with respect to x we get
d
dx
→
→f(x) + g(y) + h(z) = C,
df(x)
dx+
dg(y)
dx+
dh(z)
dx=
dC
dx,
df(x)
dx= 0, (26.15)
184
So, f(x) is a constant. Similarly for g(y) and h(z)
Applying this to our Schrödinger equation means that we have converted our
partial differential equation into three independent ordinary differential equations,
−~22m
1
ψx
d2ψx
dx2= Ex =⇒
−~22m
d2ψx
dx2= Exψx (26.16)
−~22m
1
ψy
d2ψy
dy2= Ey =⇒
−~22m
d2ψy
dy2= Eyψy
−~22m
1
ψz
d2ψz
dz2= Ez =⇒
−~22m
d2ψz
dz2= Ezψz
which we recognize as the 1D particle in a box equations.
Hence we immediately have
ψx =
r2
asin
nxπx
a, (26.17)
ψy =
r2
bsin
nyπy
b,
ψz =
r2
csin
nzπz
c
and
Ex,nx =n2xh
2
8ma2, (26.18)
Ey,ny =n2yh
2
8mb2,
Ez,nz =n2zh
2
8mc2.
The total wavefunction is
ψ =2√2√
abcsin
nxπx
asin
nyπy
bsin
nzπz
c(26.19)
and the total energy is
E = Ex,nx +Ey,ny +Ez,nz . (26.20)
185
DegeneracyThe 3D particle in a box model brings up the concept of degeneracy.
When n(> 1) states have the same total energy they are said to be n-fold degen-
erate.
Let the 3D box be a cube (a = b = c) then the states
(nx = 2, ny = 1, nz = 1), (26.21)
(nx = 1, ny = 2, nz = 1),
(nx = 1, ny = 1, nz = 2)
have the same total energy and thus are degenerate.
186
27. Operators
27.1. Operator Algebra
We now take a mathematical excursion and discuss the algebra of operators.
Definitions
• Function: A function, say f , describes how a dependent variable, say y, is
related to an independent variable, say x: y = f(x)
— e.g., y = x2, y = sinx, etc.
• Operator: An operator, say O, transforms a function, say f , into another
function, say g: Of(x) = g(x).
• Algebra: An algebra is a specific collection of rules applied to a set of objectsand a particular operation
— Rules
∗ Transitivity∗ Associativity∗ Existence of an identity∗ Existence of an inverse
— e.g., Addition on the set of real numbers, Multiplication on the set ofreal numbers
187
187
— Note: Commutivity is not a requirement of an algebra
∗ example 1: multiplication on the set of real number is commutive:ab = ba
∗ example 2: multiplication on the set of n× n matrices is not com-
mutive: ab 6= ba in general. e.g.,"1 0
2 1
#"3 1
1 1
#=
"3 1
7 3
#(27.1)
but "3 1
1 1
#"1 0
2 1
#=
"5 1
3 1
#6="3 1
7 3
#(27.2)
Algebraic rules for operators
1. Equality:
if α = β, then αf(x) = g(x) = βf(x) (27.3)
2. Addition:
if αf(x) = g(x) and βf(x) = h(x), (27.4)
then (α+ β)f(x) = αf(x) + βf(x) = g(x) + h(x)
3. Multiplication:
αβf(x) = α³βf(x)
´(27.5)
βαf(x) = β (αf(x)) ,
but in general αβf(x) 6= βαf(x).
4. Inverse:
if αf(x) = g(x) and βg(x) = f(x) (27.6)
then β = α−1 and is said to be α inverse
188
Linear operators:
• A special and important class of operators
• They obey all of the above properties in addition to
— α (f(x) + g(x)) = αf(x) + αg(x), and
— α(λf(x)) = λαf(x), where λ is a complex number.
Hermitian operators:
• A special class of linear operators
• All observables in quantum mechanics are associated with Hermitian oper-
ators
• The eigenvalues of Hermitian operators are real
Some important operators
1. • x: xf(x) = xf(x)
• d: df(x) = ddxf(x)
• d2: d2f(x) = d³df(x)
´= d
¡ddxf(x)
¢= d
dx
¡ddxf(x)
¢= d2
dx2f(x)
• ı: ıf(x, y, z) = f(−x,−y,−z)
• ∇: ∇f(x, y, z) =³
∂∂xex +
∂∂yey +
∂∂zez´f(x, y, z)
• ∇2: ∇2f(x, y, z) =³
∂2
∂x2+ ∂2
∂y2+ ∂2
∂z2
´f(x, y, z)
Commutators:
We have seen that in general αβ 6= βα. This leads to the construction of the
commutator, [, ]: hα, β
i≡ αβ − βα. (27.7)
189
If αβ = βα, thenhα, β
i= 0 and α and β are said to commute with one another.
The eigenvalue equation:
If αf(x) = g(x) and g(x) = af(x), then the operator equation, αf(x) = g(x)
becomes the eigenvalue equation
αf(x) = af(x). (27.8)
The eigenvalue equation is of fundamental importance in quantum theory. We
shall see that eigenvalues of certain operator can be identified as experimental
observables.
Commuting operators and simultaneous sets of eigenfunctions.
If αf(x) = af(x) and β and α commute, then βf(x) = bf(x).
The proof goes as follows: On the one hand,
β (αf) = β (af) = aβf (27.9)
because f is an eigenfunction of α.
On the other hand,
β (αf) = α³βf´
(27.10)
because β and α commute.
Thus
α³βf´= a
³βf´. (27.11)
which states that βf is an eigenfunction of α with eigenvalue a. The only way for
this to be true is if βf = bf.
190
27.2. Orthogonality, Completeness, and the Superposition
Principle
Theorem 1: The eigenfunctions of a Hermitian operator corresponding to differ-ent eigenvalues are orthogonal:Z
spaceψ∗jψk = 0, j 6= k. (27.12)
Theorem 2: The eigenfunctions of a Hermitian operator form a complete set
Corollary (the superposition principle): Any arbitrary function ψ in the
space of eigenfunctions ϕi can be written as a superposition of these eigenfunc-
tions:
ψ =Xi
aiϕi (27.13)
191
28. Angular Momentum
We will encounter several different types of angular momenta, but fortunately
they are all described by a single theory
Before starting with the quantum mechanical treatment of angular momentum,
we first review the classical treatment.
28.1. Classical Theory of Angular Momentum
The classical angular momentum, L, is given by
L = x× p (28.1)
The vector cross-product can be computed by finding the following determinant:
L =
¯¯ ex ey ez
x y z
px py pz
¯¯ = Lxz |
(ypz − zpy)ex +
Lyz | (zpx − xpz)ey +
Lzz | (xpy − ypx)ez (28.2)
Hence,
Lx = (ypz − zpy) , (28.3)
Ly = (zpx − xpz) , (28.4)
Lz = (xpy − ypx) . (28.5)
Another quantity that we will find useful is
L2 = L · L = L2x + L2y + L2z (28.6)
192
192
28.2. Quantum theory of Angular Momentum
So, in accordance with postulate II, we replace the classical variables with their
operators. That is,
Lx = (ypz − zpy) =~i
µy∂
∂z− z
∂
∂y
¶, (28.7)
Ly = (zpx − xpz) =~i
µz∂
∂x− x
∂
∂z
¶, (28.8)
Lz = (xpy − ypx) =~i
µx∂
∂y− y
∂
∂x
¶. (28.9)
Recall the basic commutators. ∙∂
∂u, u
¸= 1, (28.10)∙
∂
∂u, v
¸= 0,
where u, v = x, y, or z and u 6= v.
From these basic commutators one can derivehLx, Ly
i= i~Lz,
hLy, Lz
i= i~Lx,
hLz, Lx
i= i~Ly (28.11)
and hL2, Lx
i=hL2, Ly
i=hL2, Lz
i= 0 (28.12)
It is often convenient to express the angular momentum operators in spherical
polar coordinates as follows.
Lx = i~µsinφ
∂
∂θ+ cot θ cosφ
∂
∂φ
¶, (28.13)
Ly = −i~µcosφ
∂
∂θ− cot θ sinφ ∂
∂φ
¶, (28.14)
193
Lz = −i~∂
∂φ(28.15)
L2 = −~2µ
∂2
∂θ2+ cot θ
∂
∂θ+
1
sin2 θ
∂2
∂φ2
¶(28.16)
28.3. Particle on a Ring
Consider a particle of mass μ confined to move on a ring of radius R.
The moment of inertia is I = μR2
The Hamiltonian is given by
H =L2z2I=−~22I
d2
dφ2(28.17)
(note that we use d rather than ∂ since the problem is one-dimensional).
The Schrödinger equation becomes
−~22I
d2ψ
dφ2= Eψ (28.18)
Notice that this Schrödinger equation is exactly the same form as the particle in
a box. The only difference is the boundary conditions.
The boundary condition for the particle in a box were ψ was zero outside the box.
Now the boundary condition is that ψ(φ) = ψ(φ + 2π). The wavefunction must
by 2π periodic.
The allowable wavefunctions are
ψm(φ) =
⎧⎪⎨⎪⎩A cosmφ
A sinmφ
Aeimφ
, (28.19)
194
m = 0, ±1, ±2, ±3, . . .These wavefunctions are really the “same.” It will be most convenient to use
ψm(φ) = Aeimφ as our wave functions.
Plugging ψm(φ) = Aeimφ into the Schrödinger equation gives
−~22I
d2Aeimφ
dφ2= EmAe
imφ (28.20)
~2m2
2IAeimφ = EmAe
imφ
Therefore the energy levels (the eigenvalues) for a particle in a ring are
Em =~2m2
2I=
m2h2
8π2I. (28.21)
Next we need to find the normalization constant, A.
1 =
Z 2π
0
ψ∗ψdφ (28.22)
1 =
Z 2π
0
A2e−imφeimφdφ
1 = A2Z 2π
0
dφ = 2πA2,
thus
A =
r1
2π. (28.23)
Hence the normalized wavefunctions for a particle on a ring are
ψ =1√2π
eimφ. (28.24)
28.4. General Theory of Angular Momentum
To discuss angular momentum in a more general way it is convenient to define
two so-called ‘ladder’ operators
L+ ≡ Lx + iLy (28.25)
195
and
L− ≡ Lx − iLy (28.26)
We collect here the commutators of L+ and L−:hLz, L+
i= L+ ⇒ L+Lz = LzL+ − L+ (28.27)h
Lz, L−
i= −L− ⇒ L−Lz = LzL− + L− (28.28)
Now, since Lz and L2 commute there must exist a set of simultaneous eigenfunc-
tions ψiLzψi = mψi (28.29)
and
L2ψi = k2ψi (28.30)
Physically, k~ represents the length of the angular momentum vector and m~represents the projection onto the z-axis. (Note: for simplicity in writing we are‘hiding’ the ~ in the wavefunctions.)
On these physical grounds we conclude |m| ≤ k, i.e., k sets an upper and lower
limit on m.
Let’s define the maximum value of m to be a new quantum number l ≡ mmax.
(Thus l ≤ k).
And let’s define the minium value of m to be a new quantum number l0 ≡ mmin.
(Thus −l0 ≤ k)
Now, at least one of the eigenfunctions in the set ψi yields the eigenvalue mmax
(or l) when operated on by Lz. Let’s call that eigenfunction ψl;
Lzψl = lψl. (28.31)
Now we can operate on both sides of this equation with L−:
L−Lzψl = L−lψl (28.32)
196
Using the commutator relation L−Lz = LzL− + L− we get³LzL− + L−
´ψl = lL−ψl (28.33)
LzL−ψl + L−ψl = lL−ψl
Bringing the second term on the left hand side over to the right hand side gives
LzL−ψl = lL−ψl − L−ψl (28.34)
LzL−ψl| z ψl−1
= (l − 1)L−ψl| z ψl−1
We see that L−ψl ≡ ψl−1 is in fact an eigenfunction of Lz (with associated eigen-
value (l − 1)) and is thus a member of ψi .
The eigenfunction ψl−1 has an associated eigenvalue that is one unit less then the
maximum value.
The above procedure can be repeated n times so that Ln−ψl = ψl−n provided n
does not exceed l − l0.
The eigenfunction ψl−n has an associated eigenvalue that is n units less then the
maximum value, i.e.,
Lzψl−n = (l − n)ψl−n. (28.35)
The largest value of n is l − l0. For that case,
Lzψl0 = (l − l + l0)ψl0 = l0ψl0 . (28.36)
Similar behavior is seen for the operator L+, except in the opposite direction–the
eigenvalue is increased by one unit for each action of L+. For example
L+Lzψl0 = L+l0ψl0 (28.37)³
LzL+ − L+´ψl0 = l0L+ψl0
LzL+ψl0 = (l0 + 1)L+ψl0 .
197
The raising and lowering nature of L+ and L− is why they are called ladder
operators.
We can not act with L+ and L− indefinitely since we are limited by l–we reach
the ends of the ladder. This requires that
L−ψl0 = 0 (28.38)
(we can’t go lower than the lowest step) and
L+ψl = 0 (28.39)
(we can’t go higher than the highest step).
Often times the ladder operators appear in tandem either as L−L+ or L+L− so it
is useful list some identities for these products
L−L+ = L2 − L2z − Lz (28.40)
and
L+L− = L2 − L2z + Lz (28.41)
We can use these identities to derive a relation between the quantum numbers k
and l.
We begin with
L−L+ψl = L−³L+ψl
´= 0, (28.42)
but from the first of the above identities
L−L+ψl =³L2 − L2z − Lz
´ψl = (k
2 − l2 − l)ψl (28.43)
Therefore
k2 − l2 − l = 0⇒ k =pl(l + 1). (28.44)
198
We we can also consider
L+L−ψl0 = L+³L−ψl0
´= 0 (28.45)
and
L+L−ψl0 =³L2 − L2z + Lz
´ψl0 = (k
2 − l02 + l0)ψl0 . (28.46)
substituting in the relation we just found for k gives
l(l + 1)− l02 + l0 = 0; (28.47)
simplifying gives
l = −l0 (28.48)
Thus mmax = l, mmin = −l and so m = l, l − 1, l − 2, . . . , −l + 1 ,−l.
This also implies that the number of ‘rungs’ is 2l+1 and that l must be either an
integer or a half-integer.
28.5. Quantum Properties of Angular Momentum
The eigenfunctions of angular momentum are entirely specified by two quantum
numbers l and m: ψlm.
L2ψlm = l(l + 1)ψlm Lzψlm = mψlm (28.49)
If we write out the first of these explicitly in spherical polar coordinates as a
partial differential equation we obtain
∂2ψlm
∂θ2+ cot θ
∂ψlm
∂θ+
1
sin2 θ
∂2ψlm
∂φ2+ l(l + 1)ψlm = 0 (28.50)
The solutions to this partial differential equation are known to be the spherical
harmonic functions
ψlm = Ylm(θ, φ). (28.51)
199
The spherical harmonics are functions of two variables, but they are a product of
a function only of θ and a function only of φ,
ψlm = Ylm(θ, φ) = AP|m|l (θ)eimφ, (28.52)
where the P |m|l (θ) are the Legendra polynomials and A is normalization constant.
Both the spherical harmonics and the Legendra polynomials are tabulated. They
are also built-in functions of Mathematica.
The spherical harmonics (and hence the angular momentum wavefunctions) are
orthonormal; meaning,Z 2π
0
Z π
0
Y ∗l0m0(θ, φ)Ylm(θ, φ) sin θdθdφ =
(1 l0 = l and m0 = m
0 l0 6= l or m0 6= m(28.53)
28.5.1. The rigid rotor
Rotational energyFor general rotation in three dimensions the is
H =~2
2IL2, (28.54)
so the Schrödinger equation is
Hψlm = Elmψlm ⇒~2
2IL2ψlm = Elmψlm ⇒
~2
2Il(l + 1)ψlm = Elmψlm. (28.55)
Thus
Elm =l(l + 1)~2
2I=
l(l + 1)h2
8π2I= El. (28.56)
There is no m dependence for the energy. In other words, the energy levels are
determined only by the value of l.
We know that there are 2l+1 different m values for a particular l value. All 2l+1
of these wavefunctions correspond to the same energy. We say the there is a 2l+1
degeneracy of the energy levels.
200
29. Addition of Angular Momentum
29.1. Spin Angular Momentum
We learned above that l may take on integer or half-integer values.
Systems in which l takes on half-integer values are peculiar.
These systems have no classical analogs.
One example of such a system is the spin of an electron, l = s = 1/2. The values
of m = ms are limited to +1/2 and −1/2.
One peculiarity of this system is that the wavefunctions are 4π periodic (and 2π
antiperiodic):
ψs(θ) = −ψs(θ + 2π) (29.1)
and
ψs(θ) = ψs(θ + 4π). (29.2)
That means that the system has to ‘rotate’ twice (in spin space not coordinate
space) to get back to its original state.
∗ ∗ ∗ See in-class demonstration: the belt trick ∗ ∗∗
201
201
29.2. Addition of Angular Momentum
In atoms the are a number of sources of angular momentum: The l’s and s’s of
each of the electrons.
One measures, however, the total angular momentum, J.
The electrons in many electron atoms couple. The are two main coupling schemes
which account for the total angular momentum of the atom.
1. LS coupling (also called Russell-Saunders coupling)
• works well for low atomic weight atoms (first couple of rows of the
periodic table)
• find the total spin angular momentum S =Ms,max, (Ms =P
imsi)
• find the total orbital angular momentum L = Mmax, (M =P
imi)
• then J = L+ S
2. jj coupling
• applies to higher atomic weight atoms
• find subtotal angular momentum for each electron ji = li + si
• then find total angular momentum by J =P
i ji.
• we will not use this method.
29.2.1. The Addition of Angular Momentum: General Theory
Consider two sources of angular momentum for a system represented by the op-
erators J1 and J2 (J1 and J2 could be L or S angular momentum; we use J when
we speak generally.)
202
The total angular momentum is JT = J1 + J2.
The total z-component of the angular momentum is JzT = Jz1 + Jz2
The last statement implies that the orientation quantum number of the total
system is simple the sum of that for the components
M = m1 +m2 (29.3)
We need to determine the allowed values of the total angular momentum quantum
number J.
The maximum value of J is determined by the maximum value of M by
Jmax =Mmax = m1max +m2max = j1 + j2 (29.4)
This corresponds to a situation in which component angular momentums add in
the most favorable manner
The minimum value of J is determined by the case when the components add in
the least favorable manner. That is,
Jmin = |j1 − j2| . (29.5)
The total angular momentum is quantized is exactly the same manner as any
other angular momentum. Thus the allowed values of J are
J = j1 + j2, j1 + j2 − 1, . . . , |j1 − j2|+ 1, |j1 − j2| . (29.6)
29.2.2. An Example: Two Electrons
The table below shows the total spin angular momentum S for a two electron
system
203
spin state ms1 ms1 MS S
α(1)α(2) 12
12
1 1
β(1)β(2) −12−12−1 1
α(1)β(2) + β(1)α(2) 0 0 0 1
α(1)β(2)− β(1)α(2) 0 0 0 0
Counting states:
The spin degeneracy, gS, of the states is given by 2S + 1. In the above example
the degeneracy is gS = 3 for the S = 1 states and gS = 1 for the S = 0 states.
29.2.3. Term Symbols
We have already seen several term symbols, those being 1S and 3S during our
discussion of helium.
Term symbols are simply shorthand notion used to identify states. Term symbols
are useful for predict and understanding spectroscopic data. So, it is worthwhile
to briefly discuss them.
In general the term symbol is simply notates the total orbital angular momentum
and spin degeneracies of a particular set of states (or a state in the case of a singlet
state).
The orbital degeneracy is given by gL = 2L+ 1.
For historical reasons L values are associated with a letter like the l values of a
hydrogenic system are.
L 0 1 2 3 4 5
symbol S P D F G H.
204
The term symbol for a particular states is constructed from the following general
templategSLJ .
Many electron atoms have term symbols associated with their states.
Rules:
1. All closed shells have zero spin and orbital angular momentums: L = 0,
S = 0. These states are all singlet S states, notated by 1S
2. An electron and a “hole” lead to equivalent term symbols.
• E.g., p1 and p5 have the same term symbol.
3. Hund’s Rule for the ground state only.
1. The ground state will have maximum multiplicity.
2. If several terms have the same multiplicity then ground state will be
that of the largest L.
3. Lowest J value (regular) “electron”, Highest J value (inverted) “hole”
29.2.4. Spin Orbit Coupling
A charge possessing angular momentum has a magnetic dipole associated with it.
An electron has orbital and spin magnetic dipoles.
These dipoles interact with a certain spin—orbit interaction energy ESO.
The spin—orbit Hamiltonian is
HSO = hcA[L · S (29.7)
HSO =hcA
2
³J2 − L2 − S2
´,
205
where A is the spin—orbit coupling constant.
From the Hamiltonian the spin—orbit interaction energy is
ESO =hcA
2[J(J + 1)− L(L+ 1)− S(S + 1)] (29.8)
206
30. Approximation Techniques
As we learned last semester, there are very few models for which we can obtain
an exact solution.
Consequently we must be satisfied with using approximation methods.
Last semester, we always took the simplest approximation to give the qualitative
properties of the unsolvable system.
Now we will consider two important quantitative approximation methods: (i)
perturbation theory and (ii) variational theory
30.1. Perturbation Theory
The basic procedure of perturbation theory
• Find a solvable system that is similar to the system at hand.
• Treat the difference between the two systems as a perturbation to the solv-able system
• Use the solvable system’s wavefunctions as a zeroth order approximation tothe wavefunctions for the unsolvable system.
• These wavefunctions are used to find a first order correction to the energy.
207
207
• The first order energy is then used to make a first order approximation tothe wavefunction.
• The procedure is repeated to get higher and higher order approximations.
This process get algebraically intensive so we will only go as far as listing the first
order energy correction.
The nth state energy in perturbation theory:
En = E(0)n +E(1)
n + . . . , (30.1)
where E(0)n is the nth state energy for the unperturbed (solvable) system and E
(1)n
is the first order correction. This is given by
E(1)n =
Zallspace
ψ(0)∗n H(1)ψ(0)n dx, (30.2)
where H(1) is the first order correction to the Hamiltonian–the perturbation.
Example: the quartic oscillator
• Consider the quartic oscillator described by the potential V (x) = 12kx2+ax4
where a is very small and can be treated as a perturbation.
• The obvious solvable system is the harmonic oscillator:
H = − ~2
2m
d2
dx2+1
2kx2. (30.3)
This has energy levelsEn = ~ω(n+12) and wavefunctionsAnHn(
√αx)e−αx
2/2,
where α =q
km~
• The perturbative part of the Hamiltonian is
H(1) = ax4. (30.4)
208
• For example, the ground state energy correction is then calculation from
E(1)0 =
Z ∞
−∞ψ(0)∗0 H(1)ψ
(0)0 dx (30.5)
=
Z ∞
−∞A0e
−αx2/2ax4A0e−αx2/2dx
= aA20
Z ∞
−∞x4e−αx
2
dx
=3√πaA20
4α52
,
so the first order ground state energy for a quartic oscillator is
E0 '~ω2+3√πaA20
4α52
.
30.2. Variational method
The basic idea behind the variational method is to use a trial wavefunction with
an adjustable parameter. The value of the parameter which minimizes the energy,
Etrial, gives a trial wavefunction which is closest to the real wavefunction.
The basis for this is the variation theorem which states
Etrial ≥ E.
We will not prove this theorem here.
The trial energy is calculated by
Etrial =
Rallspace
ψ∗trialHψtrialdxRallspace
ψ∗trialψtrialdx(30.6)
The trial energy is now a function of the adjustable parameter, p, that we use to
minimize the trial energy by setting
dEtrialdp
= 0 (30.7)
209
and solving for p. (Strictly speaking we should check that we have a minimum
and not a maximum or inflection point, but with reasonably good trial functions
one is pretty safe in having a minimum.)
210
31. The Two Level System and
Quantum Dynamics
Our entire discussion of quantum mechanics thus far had dealt only with time
independent quantum mechanics.
The time variable never appears in any expression.
Obviously there are cases where quantum objects move with time. For example,
firing an electron down a particle accelerator.
We shall finally get to quantum dynamics in this chapter, but first we will discuss
the very important model of the two level system.
31.1. The Two Level System
If the harmonic oscillator is the most important model in all a physics, the two
level system is a close second.
The spin system discussed above is an example of a two level system.
The two level system is inherently quantum mechanical in nature. Unlike the
harmonic oscillator it has no classical analogue.
211
211
Consequently, we can not use our usual procedure of writing down the classical
Hamiltonian and then replacing the variables with their corresponding operators.
The two level system consists of two states ψ1 and ψ2 separated by energy 4 =
2 − 1 as shown below
The states ψ1 and ψ2 are orthonormal:ZTLS
ψ∗jψkdΩ =
(1 j = k
0 j 6= k, (31.1)
whereRTLS dΩ means integration over the two level space (which is really just the
sumP2
i=1).
The states ψ1 and ψ2 are eigenfunctions of the two level Hamiltonian,
H = 1δ1,° + 2δ2,°, (31.2)
where δj,° “projects out” the jth state of the wavefunction being acted on.
212
For example let some arbitrary wavefunction ψ = aψ1 + bψ2, then
Hψ = ( 1δ1,° + 2δ2,°) (aψ1 + bψ2) (31.3)
= 1δ1,° (aψ1 + bψ2) + 2δ2,° (aψ1 + bψ2)
= a 1ψ1 + b 2ψ2
Another orthonormal set of wavefunctions are the so-called ‘left’
ψL =1√2ψ1 +
1√2ψ2 (31.4)
and ‘right’
ψR =1√2ψ1 −
1√2ψ2 (31.5)
states.
We can invert above equations and solve for ψ1 and ψ2 in terms of ψL and ψR
ψ1 =1√2ψL +
1√2ψR (31.6)
and
ψ2 =1√2ψL −
1√2ψR. (31.7)
213
31.2. Quantum Dynamics
So far we have been concerned with the eigenfunctions and eigenvalues (energy
levels) of the various quantum systems that we have discussed.
What has been kept hidden up to now is the fact that the eigenfunctions are really
multiplied by a phase factor of the form .
Ψn(x, t) ≡ ψn(x)e− i~ Ent (31.8)
We can verify this by obtaining the time independent Schrödinger equation from
the more general time dependent
i~∂Ψn(x, t)
∂t= HΨn(x, t) (31.9)
i~∂ψn(x)e
− i~ Ent
∂t= Hψn(x)e
− i~ Ent
i~ψn(x)∂e−
i~ Ent
∂t= Hψn(x)e
− i~ Ent
i~ψn(x)
µ− i
~En
¶e−
i~ Ent = Hψn(x)e
− i~ Ent
Enψn(x)e− i~ Ent = e−
i~ EntHψn(x)
Enψn(x) = Hψn(x) (31.10)
Does this mean the eigenstates are not stationary states? To determine this we
need to calculate the probability of finding the particle in the same eigenstate at
some future time. This is given by
P (x, t) =
¯ZΨ∗n(x, 0)Ψn(x, t)dx
¯2(31.11)
=
¯Zψ∗n(x)ψn(x)e
− i~ Entdx
¯2=
¯e−
i~ Ent
Zψ∗n(x)ψn(x)dx
¯2=
¯e−
i~ Ent(1)
¯2= 1,
214
so no matter what time t we check we will always find the system in the same
eigenstate. Thus the eigenstates are stationary states.
In general the state of the system need not be in one particular eigenstate; it may
be in a superposition of any number of eigenstates.
The “left” and “right” wavefunctions that we saw in the discussion of the two
level system are examples of superposition states.
The phase factor does become important for superposition states.
As an example consider the state
Φ(x, t) =1√2Ψ1(x, t) +
1√2Ψ2(x, t) (31.12)
exposing the phase factors we get
Φ(x, t) =1√2ψ1(x)e
− i~ E1t +
1√2ψ2(x)e
− i~ E2t (31.13)
Let’s now track the probability of finding the particle in the same superposition
state. Similar to before we calculate
P (x, t) =
¯ZΦ∗(x, 0)Φ(x, t)
¯2=
¯Z µ1√2ψ∗1(x) +
1√2ψ∗2(x)
¶µ1√2ψ1(x)e
− i~ E1t +
1√2ψ2(x)e
− i~ E2t
¶¯2=
¯¯12Z Ã
ψ∗1(x)ψ1(x)e− i~ E1t + ψ∗1(x)ψ2(x)e
− i~ E2t
+ψ∗2(x)ψ1(x)e− i~ E1t + ψ∗2(x)ψ2(x)e
− i~ E2t
!dx
¯¯2
. (31.14)
The “cross-terms” (those of the form ψ∗1(x)ψ2(x) and ψ∗2(x)ψ1(x)) are zero when
215
integrated because the eigenfunctions are orthogonal. This leaves
P (x, t) =
¯ZΦ∗(x, 0)Φ(x, t)
¯2(31.15)
=
¯1
2
Z ³ψ∗1(x)ψ1(x)e
− i~ E1t + ψ∗2(x)ψ2(x)e
− i~ E2t
´dx
¯2=
¯1
2
µe−
i~ E1t
Zψ∗1(x)ψ1(x)dx+ e−
i~ E2t
Zψ∗2(x)ψ2(x)dx
¶¯2=
¯1
2
³e−
i~ E1t + e−
i~ E2t
´¯2=1
4
³e+
i~ E1t + e+
i~ E2t
´³e−
i~ E1t + e−
i~ E2t
´=
1
4
³1 + e+
i~ (E1−E2)t + e−
i~ (E1−E2)t + 1
´=1
2
µ1 + cos
(E1 −E2)
~t
¶.
The probability of find in the system in its original superposition states is not one
for all times t.
216
Key Equations for Exam 1
Listed here are some of the key equations for Exam 1. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• The short cut for getting the normalization constant .
N =
sZspace
|ψunnorm(x, y, z)|2 dxdydz. (31.16)
• The normalized wavefunction:
ψnorm =1
Nψunnorm. (31.17)
• How to get the average value for some property,
hαi =Zspace
ψ∗αψdxdydz. (31.18)
217
217
• The Laplacian∇2 =
µ∂2
∂x2+
∂2
∂y2+
∂2
∂z2
¶. (31.19)
• Normalized wavefunctions for the 3D particle in a box,
ψn(x) =2√2√
abcsin
nxπx
asin
nyπy
bsin
nzπz
c. (31.20)
• The energy levels for the 3D particle in a box,
Enx,ny,nz =n2xh
2
8ma2+
n2yh2
8mb2+
n2zh2
8mc2. (31.21)
• Orthonormality: Zspace
ψ∗jψk =
(1, j = k
0, j 6= k. (31.22)
• Superpostion:ψ =
Xi
aiϕi (31.23)
• Commonly used comutators of the angular momentum operators arehLx, Ly
i= i~Lz,
hLy, Lz
i= i~Lx,
hLz, Lx
i= i~Ly (31.24)
and hL2, Lx
i=hL2, Ly
i=hL2, Lz
i= 0. (31.25)
• The energy levels for a particle in a ring are
Em =~2m2
2I=
m2h2
8π2I. (31.26)
• The normalized wavefunctions for a particle on a ring are
ψ =1√2π
eimφ. (31.27)
218
• The eigenfunctions of angular momentum are entirely specified by two quan-tum numbers l and m: ψlm.
L2ψlm = l(l + 1)ψlm Lzψlm = mψlm (31.28)
• The energy levels for the rigid rotor are
El =l(l + 1)~2
2I. (31.29)
• Degeneracy for general angular momentum is
gJ = 2J + 1. (31.30)
• The first order energy correction in pertubation theory is
E(1)n =
Zallspace
ψ(0)∗n H(1)ψ(0)n dx, (31.31)
• The trial energy in variation theory is calculated by
Etrial =
Rallspace
ψ∗trialHψtrialdxRallspace
ψ∗trialψtrialdx(31.32)
• In generalΨn(x, t) ≡ ψn(x)e
− i~ Ent (31.33)
• The left and right superposition states are
ψL =1√2ψ1 +
1√2ψ2 (31.34)
and
ψR =1√2ψ1 −
1√2ψ2 (31.35)
219
Part VI
Symmetry and Spectroscopy
220
220
32. Symmetry and Group Theory
We now take a short break from physical chemistry to discuss ideas from the
mathematical field of group theory.
Inherent to group theory is symmetry.
As far as we are concerned, we will
• determine the symmetry of a particular molecule.
• The types of symmetry it has will determine to which symmetry group itbelongs.
• The mathematical properties of all the possible groups have been workedout
• These mathematical properties translate into a wide variety of variety ofphysical properties including
— Bonding
— Properties of wavefunctions
— Vibrational modes
— Many more applications
221
221
32.1. Symmetry Operators
Any operator that leaves |ψ|2 invariant are symmetry operators for that particularsystem:
O |ψ|2 = |ψ|2 . (32.1)
This implies
Oψ = ±ψ. (32.2)
That is, the eigenvalues for the particular symmetry operator are 1 or −1.
For molecules we will be dealing with point group symmetry operators. These
operators deal with symmetry about the center of mass.
We have seen two such operators in ı and σh.
An example of symmetry operator that is not a point group symmetry operator
would be an operator that performed some sort of translation in space. This type
of operator arrises in the treatment of extended crystal structures.
∗ ∗ ∗ See Handout on Symmetry Elements ∗ ∗∗
32.2. Mathematical Groups
In mathematics the term “group” has special meaning. It is a set of objects and
a single operation, which has the following properties.
1. The group is associative (but not necessarily communative) with respect to
the operation.
2. An identity element exits and is a member of the group
222
3. The “product” of any two members of the group yield a member of the
group.
4. The inverse of every member of the group is also in the group. In other
words, for any member of the group one can find another member of the
group which, upon “multiplication,” yields the identity element.
∗ ∗ ∗ See Handout on Naming Point Groups ∗ ∗∗
∗ ∗ ∗ See Handout on Assigning Point Groups ∗ ∗∗
Associated with a given group is a “multiplication” table.
32.2.1. Example: The C2v Group
The C2v group consists of the symmetry elements E, C2, σv (in-plane) and σ0v
(transverse).
Water is an example of a molecule described by this point group.
The multiplication table for the C2v group is
C2v E C2 σv σ0v
E E C2 σv σ0v
C2 C2 E σ0v σv
σv σv σ0v E C2
σ0v σ0v σv C2 E
32.3. Symmetry of Functions
In the absence of degeneracy, the wavefunctions must be symmetric or antisym-
metric with respect to all elements of the group.
223
Connecting with the C2v group example lets consider the wavefunctions for water.
In this case one can collect the eigenvalues (either +1 or −1) for each of the foursymmetry operators as a four component vector. As it turns out there is four
possible sets of eigenvalues–hence four different vectors:
A1 = (1, 1, 1, 1)
A2 = (1, 1,−1,−1)B1 = (1,−1, 1,−1)B2 = (1,−1,−1, 1).
To see where these four vectors come from, consider the following.
• The first value has to be +1 since the only eigenvalue of E is 1
• The eigenvalue of C2 can be +1 or −1
— When it is +1 the vectors are labelled A
— When it is −1 the vectors are labelled B
• The eigenvalue of σv can be either +1 or −1
— When it is +1 the vectors are labelled with a subscript 1
— When it is −1 the vectors are labelled with a subscript 2
• The eigenvalue of σ0v can be either +1 or −1
• Finally there is a restriction do to the fact that the eigenvalues must obeythe group multiplication table.
— This restriction forces the eigenvalues of σv and σ0v to be the same for
the A type vectors and opposite for the B type vectors.
224
The above considerations leave four vectors.
In fact, there will always be the same number of vectors as symmetry elements.
Altogether, the vectors represent what is call an irreducible representation of the
group.
These vectors make up the :
C2v E C2 σv σ0v
A1 1 1 1 1
A2 1 1 −1 −1B1 1 −1 1 −1B2 1 −1 −1 1
∗ ∗ ∗ See Handout on Character Tables ∗ ∗∗
32.3.1. Direct Products
The direct product of a two vectors is defined as
(x1, x2, x3, . . .)⊗ (y1, y2, y3, . . .) = (x1y1, x2y2, x3y3, . . .) (32.3)
For the example of the C2v group consider
B1 ⊗B2 = (1,−1, 1,−1)⊗ (1,−1,−1, 1)= (1, 1,−1,−1) = A2 (32.4)
32.4. Symmetry Breaking and Crystal Field Splitting
We shall investigate how degeneracies of energy levels are broken as one reduces
the overall symmetry of the system.
225
In doing this we will, for simplicity, consider only proper rotations (Cn). Mirror
symmetry will not be considered (although in real applications one must consider
all symmetry).
First consider a free atom. In this case there is complete rotational symmetry.
Thus the symmetry group is the spherical group (see character table handout.)
This is the group associated with the particle on a sphere model and the angular
part of the hydrogen atom. The vectors are the labeled according to the angular
momentum quantum numbers S, P, D, F, etc.
The degeneracies of these vectors are 1 for S, 3 for P, 5 for D and so on as is
familiar to us already.
Now consider the free atom being placed in a crystal lattice of octahedral sym-
metry. For example placed at the center of a cube which has other atoms at the
centers of each face of the cube.
When moving to octahedral symmetry we now must look at the character table for
such a case–the O group (remember we are considering only proper rotations).
The S vector has the symmetry of a sphere (x2 + y2 + z2) and hence is totally
symmetric. It is also nondegenerate so it will be, of course, nondegenerate in
the octahedral case. It remains totally symmetric so it is now represented by the
vector A1.
The P vector is triply degenerate and has the symmetry of x, y and z as we see
from the character table for the spherical group. In the octahedral crystal the
degeneracy remains in tact and these states are represented by the T1 group.
226
The D vector has a degeneracy of five and the symmetry of 2z2− x2− y2, xz, yz,
xy, x2 − y2. Looking at the table for the O group we see the degeneracy splits:
two states become E type and the remaining three become T2 type.
The F states have a degeneracy of 7 and the symmetry of z3, xz2, yz2, xyz,
z(x2 − y2), x(x2 − 3y2) and y(3x2 − y2). In an octahedral environment the states
split with one becoming A2, three becoming T1 and three becoming T2. This is
not readily apparent from the character tables so one needs to inspect a little
harder to see it (see homework).
The octahedral group is still highly symmetric. Lets say that two atoms on oppo-
site sides of the cube are moved slightly inward. The remaining four atoms remain
in place.
This breaks the octahedral symmetry and the system now assumes D4 symmetry.
Now the A1 vector of the O group becomes the A1 vector of the D4 group. The
triply degenerate T1 vector splits into a A2 state and a doubly degenerate E state.
The E states from the O group become a A1 type state and a B1 type state.
The T1 states from the O group become a A2 type state and a E type state.
The T2 states from the O group become a B2 and a E type state.
227
33. Molecules and Symmetry
From our chapter on diatomic molecules last semester we have learned a great
deal which caries over directly to polyatomic molecules.
So, in this chapter we simply investigate some of the specific details regarding
polyatomic molecules.
33.1. Molecular Vibrations
As for diatomic molecules, it is convenient to work with center of mass coordinates.
With polyatomic molecules one needs to specify the coordinates of N nuclei rather
than just two nuclei.
To do so we begin with the 3N nuclear degrees of freedom.
As for the diatomic case 3 degrees of freedom determine the center of mass motion.
That leaves us with 3N − 3 coordinates to specify.One must now consider two different types of polyatomic molecules: Linear and
Nonlinear.
• For linear molecules there are 2 rotational degrees of freedom
• For nonlinear molecules there are 3 rotational degrees of freedom
This now leaves one with 3N − 5 vibrational degrees of freedom for linear poly-
atomic molecules and 3N − 6 vibrational degrees of freedom for nonlinear mole-
cules.
228
228
33.1.1. Normal Modes
Polyatomic molecules can undergo very complicated vibrational motion.
Regardless of what type of vibrational motion is taking place, however, that mo-
tion is some linear combination of fundamental vibrational motions called normal
modes.
This is analogous to writing an arbitrary wavefunction as a linear combination of
eigenfunctions. One example was the “left” and “right” states of the two level
system.
The number of normal modes equals the number of vibrational degrees of freedom.
At low energies the normal modes are well approximated as harmonic oscillators.
33.1.2. Normal Modes and Group Theory
The symmetry of the normal modes are associated with entries in the character
table of the point group of any particular polyatomic molecule.
Example: Water
The point group symmetry of the water molecule is C2v. The character table is
C2v E C2 σv σ0v
A1 1 1 1 1
A2 1 1 −1 −1B1 1 −1 1 −1B2 1 −1 −1 1
Water has three nuclei and it is nonlinear so it has 3(3)− 6 = 3 normal modes.The three modes are the bending vibration, the symmetric stretching vibration
and the asymmetric stretch.
229
The normal modes are associated with a particular vector (row) of the character
table by considering the action of the each of the symmetry elements on the normal
mode.
For the bending mode, the vibration is complete unchanged by any of the sym-
metry elements. Consequently the bending mode is associated with A1
The same is true for the symmetric stretching mode. It too is associated with A1.
The asymmetric stretch, however, is associated with B1 since C2 and σ0v transform
the mode into its opposite and σv leaves it unchanged.
230
34. Vibrational Spectroscopy and
Group Theory
We now investigate how group theory and, in particular, the character tables can
be used to determine IR and Raman spectra and selection rules for polyatomic
molecules
34.1. IR Spectroscopy
IR absorption is exactly the same as regular electronic absorption except the
frequency of the electromagnetic radiation is much less.
The typical “energies” for IR absorption are from 400 to 4000 cm−1. This is in
the Infrared region of the electromagnetic spectrum.
As for electronic absorption one typically employs the electric dipole approxima-
tion.
The electric dipole approximation
• Molecule is viewed as a collection of charges
• Multipole expansion
monopole+ dipole+ quadrapole+ · · · (34.1)
231
231
• Light—matter interaction is dominated by the light—dipole coupling so theother interactions are ignored.
In order for absorption of the electromagnetic radiation to take place, it must be
able to couple to a changing (oscillating) electric dipole.
The electric dipole is
μ = μxex + μyey + μzez (34.2)
where μx = qx, μy = qy, μz = qz.
The upshot of all this is as far as group theory is concerned is the following
selection rule:
• The vibrational coordinates for an IR active transition must have the samesymmetry as either x, y, or z for the particular group.
Example: Water
Recall that the point group symmetry of the water molecule is C2v.
We now need a column of the character table which we have ignored up to this
point.
The character table is
C2v E C2 σv σ0v Functions
A1 1 1 1 1 z, x2, y2, z2
A2 1 1 −1 −1 xy
B1 1 −1 1 −1 x, xz
B2 1 −1 −1 1 y, yz
The last column describes the symmetry of several important functions for the
point group.
232
Among these functions are x, y, and z.
So we can see immediately that the IR active modes of any molecule having this
point group will be A1, B1, and B2.
The A2 mode is IR forbidden and any vibrations having this symmetry will not
appear in the IR spectrum (or it may appear as a very weak line).
From before we know the modes of water have A1, and B1 symmetry and hence
are all IR active and appear in the IR spectrum
34.2. Raman Spectroscopy
Raman spectroscopy is somewhat different than IR spectroscopy in that vibra-
tional frequencies are measured by way of inelastic scattering of high frequency
(usually visible) light.
The light loses energy to the material in an amount equal to the vibrational energy
of the molecules is the sample.
This lose of energy shows up in the scattered light as a new down shifted frequency
from that of the original input light frequency.
Unlike IR absorption which is based on the electric dipole, Raman scattering is
based on the polarizability of the molecule
Roughly speaking the polarizability of a molecule determines how the electron
density is distorted through interaction with an electromagnetic field.
233
The molecular quantity of interest is the polarizability tensor,↔α.
We will not get into tensors in this course except to say the polarizability tensor
elements are proportional to the quadratic functions, x2, y2, z2, xy, xz, yz, (or
any combinations thereof).
One can now inspect the character table to determine which modes will be Raman
active.
For the example of water, all modes are Raman active
Rule of Mutual exclusion
• Vibrational mode can be both IR and Raman active or inactive
• If, however, the molecule has inversion symmetry (contains ı as a symmetryelement) then no modes will be both IR and Raman active.
234
35. Molecular Rotations
Recall that the three degrees of freedom that described the position of the nuclei
about the center of mass were (R, θ, φ). The R was involved in vibrations. We
now turn our attention to the angular components to describe rotations.
Recall also the Kinetic energy operator for the nuclei in the center of mass coor-
dinates
TN = −~2
2μ∇2N = −
~2
2μR2∂
∂RR2
∂
∂R+~2
2μJ2. (35.1)
We will now be concerned only with the angular part,
−~2
2IJ2. (35.2)
Now, under the Born-Oppenheimer approximation, R is a parameter. For constant
R the rotational energy is given by
Erot =J(J + 1)~2
2μR2=
J(J + 1)h2
8π2I. (35.3)
This is the so-called rigid rotor energy.
It is common to define
Be ≡h
8π2I(35.4)
as the rotational constant. Then
Erot = J(J + 1)hBe (35.5)
with a degeneracy of
gJ = 2J + 1 (35.6)
235
235
35.1. Relaxing the rigid rotor
Of course the rigid rotor is not a perfectly correct model for a diatomic molecule.
There are two corrections we will now make
1. Vibrational state dependence:
• The R value is dependent on the particular vibrational level.
• One defines a rotational interaction constant that depends on the vi-brational level, n.
Bn ≡ Be −µn+
1
2
¶αe, (35.7)
where αe is an empirical rotational—vibrational interaction constant.
2. Centrifugal stretching:
• Rotation tends to stretch the diatomic distance R.
• This is corrected for by the term
−J2(J + 1)2Dc, (35.8)
where
Dc ≡4B3
e
ω2e(35.9)
is the centrifugal stretching constant.
35.2. Rotational Spectroscopy
A rotational transition can occur in the same vibrational level n. This is called a
pure rotational transition. Alternatively, a rotational transition can accompany a
vibrational transition.
In either case the selection rule for the transition is 4J = ±1.
236
It turns out that typical rotational energy gaps are on the order of a few wavenum-
bers or less.
Thermal energy, kT, at room temperature is about 200 cm−1. This means that
at room temperature the many excited rotational states are populated.
∗ ∗ ∗ See Handout ∗ ∗∗
The selection rules and the thermalized states combine to yield a multi-peaked
ro-vibrational spectrum.
∗ ∗ ∗ See Handout ∗ ∗∗
35.3. Rotation of Polyatomic Molecules
There are a few additional details regarding rotations for polyatomic molecules as
compared to diatomics
Of course one could set-up an arbitrary center of mass coordinate system. But
one system is special–the principle axes coordinate system.
The principle axes coordinate system is the one in which the z-axis is taken to be
along the principle symmetry axis.
The total moment of inertia, I = Ixx + Iyy + Izz
The Hamiltonian in the principle axes system is
H =~2
2
"J2xIxx
+J2yIyy
+J2zIzz
#(35.10)
237
There are four classes of polyatomic molecules regarding rotations
1. Linear (e.g., carbon dioxide)
• Izz = 0, Ixx = Iyy
• J2 = J2x + J2y
• The Hamiltonian isH =
~2
2IxxJ2 (35.11)
• The rotational energy is
Erot = hBJ(J + 1), (35.12)
where
B =h
8π2Ixx(35.13)
2. Symmetric tops (e.g., benzene)
• Ixx = Iyy
• J2 = J2x + J2y + J2z
• The Hamiltonian is
H =~2
2
"J2x + J2yIxx
+J2zIzz
#(35.14)
• The rotational energy is
Erot = hBJ(J + 1) + h(A−B)K2, (35.15)
where
A =h
8π2Izz, (35.16)
B =h
8π2Ixx(35.17)
and K is the quantum number describing the projection of the angular
momentum onto the z-axis
238
3. Spherical tops (e.g., methane)
• Ixx = Iyy = Izz
• J2 = J2x + J2y + J2z
• The Hamiltonian isH =
~2
2IxxJ2 (35.18)
• The rotational energy is
Erot = hBJ(J + 1), (35.19)
where
B =h
8π2Ixx(35.20)
4. Asymmetric tops
• Ixx 6= Iyy 6= Izz
• These are more complicated and we will not discuss them in detail
239
36. Electronic Spectroscopy of
Molecules
The electronic spectra of molecules are quite different than that of atoms.
Atomic spectra consist of single sharp lines due to transitions between energy
levels.
Molecular spectra, on the other hand, have numerous lines (bands) due to the
fact that electronic transitions are accompanied by vibrational and rotational
transitions.
36.1. The Structure of the Electronic State
Last semester we saw that under the Born—Oppenheimer approximation we were
able to write the molecular wavefunction as a product of an electronic part and a
nuclear part.
We found that in doing so the electronic energy level, Ee, was parameterized by
the internuclear distance, R.
Ee as a function of R describe the effective potential for the nuclei.
It had a qualitative shape similar to the Morse potential.
240
240
In the figure below the ground and first excited electronic levels (as a function of
R) are shown.
Note: The potential minima are not at the same value of R for each of the
electronic states.
36.1.1. Absorption Spectra
In absorption spectroscopy, light promotes an electron from the ground electronic
state (and usually from the ground vibrational state too) to the excited electronic
state and any of the excited vibrational states of the excited electronic state.
∗ ∗ ∗ See Spectroscopy Supplement p1 ∗ ∗∗
36.1.2. Emission Spectra
In emission spectroscopy, light demotes an electron from the ground vibrational
state of the excited electronic state to any one of a number of excited vibrational
levels in the ground electronic state.
241
∗ ∗ ∗ See Spectroscopy Supplement p2 ∗ ∗∗
36.1.3. Fluorescence Spectra
All during the process of absorption, the process of is taking place.
∗ ∗ ∗ See Spectroscopy Supplement p3 ∗ ∗∗
As seen in the supplement the fluorescence spectrum is shifted to lower energies
(red shifted) from the absorption spectrum.
This is known as the Stokes shift.
The main stream explanation for the stokes shift is as follows
• Light promotes the system from the ground vibrational and ground elec-
tronic state to excited vibrational levels in the excited electronic state.
• The system then very rapidly (on the order of tens to hundreds of fem-
toseconds) relaxes to the ground vibrational state of the excited electronic
state.
• This process is called
• The molecule than emits a photon to drop back down into an excited vibra-tional state of the ground electronic state.
• This requires a lower energy (or “more red”) photon. Hence the Stokes shift.
242
36.2. Franck—Condon activity
We have seen than an electronic tranistion involves not only a change in the
electronic state but also in the vibrational state in general (and in the rotaitonal
state as well, but we will ingore this).
Assuming the electronic transition is allowed one must calculate the probability of
the vibrational transistion as well. This is down by evaulating the Franck—Condon
integral.
36.2.1. The Franck—Condon principle
When the Born—Oppenheimer approximation is applied to spectroscopic transi-
tions, one obtains the Franck—Condon principle.
The Franck—Condon principle states that the nuclei do not move during an elec-
tronic transition.
Physically this means that for a particular transition to be Franck—Condon ac-
tive there must be good overlap of the vibrational wavefunctions involved in the
transition.
Mathematically this means that the strength of a transition fromΨi = ψel,iψvib,i →Ψf = ψel,fψvib,f is given by¯
¯Z
allspace
Ψ∗f μelΨi
¯¯2
=
¯¯Z
elspace
Zvibspace
ψ∗el,fψ∗vib,f μelψel,iψvib,i
¯¯2
, (36.1)
243
where μel is the electronic transition dipole. We can separate the integrals as¯¯Z
elspace
ψ∗el,f μelψel,i
¯¯2
| z if 6=0, allowed
¯¯Z
vibspace
ψ∗vib,fψvib,i
¯¯2
| z Franck—Condon
, (36.2)
244
37. Fourier Transforms
As a spectroscopist it is imperative to have a deep understanding of the relation-
ship between time and frequency.
Spectroscopic data is obtained either in the time domain or in the frequency
domain and one should readily be able to look at data in one domain and know
what is happening in the other domain.
One should be familiar with qualitative aspects of this time—frequency relation,
such as if a signal oscillates in time it will have a peak in it frequency spectrum
at the frequency with which it is oscillating.
Furthermore, if the signal decays rapidly it will have a broad spectrum and, con-
versely, if the signal decays slowly it will have a narrow spectrum. The mathemat-
ics which governs these qualitative statements is Fourier transform theory which
we now review.
37.1. The Fourier transformation
The Fourier transformation, =, of a function f(t) will, in this work, by denoted
by a tilde, f(ω), and is given by
= [f(t)] = f(ω) =
Z ∞
−∞f(t)eiωtdt. (37.1)
245
245
The Fourier transformation is unique and it has a unique inverse, =−1, which isgiven by
=−1hf(ω)
i= f(t) =
1
2π
Z ∞
−∞f(ω)e−iωtdω. (37.2)
The above two relations form the convention used throughout this work.
Other authors use different conventions, so one must take care to know exactly
which convention is being used.
For simplicity the symbol = will be used to represent the Fourier transformationoperation, i.e., = [f(t)] = f(ω).Whereas the symbol =−1 will represent the inverseFourier transformation, i.e., =−1
hf(ω)
i= f(t).
246
Key Equations for Exam 2
Listed here are some of the key equations for Exam 2. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• Vibrational degrees of freedom
— linear: 3N − 5
— not linear: 3N − 6
• The so-called rigid rotor energy is
Erot = J(J + 1)hBe. (37.3)
where
Be ≡h
8π2I(37.4)
is the rotational constant.
247
247
• The degeneracy of the rigid rotor is
gJ = 2J + 1 (37.5)
• Franck—Condon Factor: ¯¯Z
vibspace
ψ∗vib,fψvib,i
¯¯2
(37.6)
• The Fourier transformation is
= [f(t)] = f(ω) =
Z ∞
−∞f(t)eiωtdt. (37.7)
• The inverse Fourier transformation is
=−1hf(ω)
i= f(t) =
1
2π
Z ∞
−∞f(ω)e−iωtdω. (37.8)
248
Part VII
Kinetics and Gases
249
249
38. Physical Kinetics
We now turn our attention to the molecular level and in particular to molecular
motion.
38.1. kinetic theory of gases
A microscopic view of gases
Consider a gas of point mass (m), m is the molecular (or atomic) mass
• Each particle of mass m has velocity v, hence a momentum of p = mv and
a kinetic energy of KE = 12mv · v = 1
2mv2.
• A sample of N molecules is characterized by its number density n∗ = NV.
• From the ideal gas law PV = nRT = NLRT (L is Avogadro’s number):
NV= PL
RT= n∗
Consider the ith particle at position xi = (x, y, z) in coordinate (position) space.
Its velocity is vi = dxidt=¡dxidt, dyidt, dzidt
¢. This can represented in velocity space by
the vector vi = (vxi , vyi , vzi).
The velocities of the particles are characterized by a probability distribution func-
tion for velocities F (vx, vy, vz, t) which is in general a function of time, t.
250
250
The number of particles, NVv , having velocities in a macroscopic volume, Vv, in
velocity space is
NVv = N
ZVv
F (v, t)dv = N
Z Z ZVv
F (vx, vy, vz, t)dvxdvydvz (38.1)
It is more convenient to switch to spherical polar coordinates in velocity space
(v, θ, φ); n.b., v is simply a magnitude (not a vector)–it is the speed.
The probability distribution function then becomes F (v, θ, φ, t)
If we choose the origin of our coordinate system to be at the center of mass of the
gas, then for many cases the velocity distribution will be isotropic–independent
of θ and φ.
F (v, θ, φ, t) = F (v, t). (38.2)
Furthermore, stationary distributions–those independent of time–are often en-
countered.
F (v, θ, φ, t) = F (v, θ, φ). (38.3)
251
We shall consider stationary isotropic distributions F (v). So F (v) represents a
distribution of speeds.
It can be shown from first principles that
F (v) = 4π
µm
2πkbT
¶32
e−mv2
2kbT v2 (38.4)
where kb = 1.380658 × 10−23 is Boltzmann’s constant. This is the Maxwell’sdistribution (of speeds).
38.2. Molecular Collisions
The average speed of a particle can calculated from Maxwell’s distribution:
hvi = v =
Z ∞
0
vF (v)dv =
Z ∞
0
v34π(m
2πkT)32 e−
mv2
2kT dv (38.5)
=
s8kT
πm
µL
L
¶Lk=R=
Lm=M
r8RT
πM
It will be convenient to define number density as n∗ ≡ NVwhere N is the number
of particles. For an ideal gas (V = nRTP), n∗ =
N
n|z=L
PRT= LP
RT.
A simple model for molecular collisions:
252
• Particles are hard spheres of radius σ.
• A Particle moving at v sweeps out a cylinder of radius σ and length v4t =⇒V = πσ2v4t.
∗ ∗ ∗ See Handout ∗ ∗∗
• The number of collisions equals the number of particles with their centersin V :
number of collisions = n∗πσ2v4t (38.6)
• The collision frequency = n∗πσ2v
For the above model we need to find the average collision frequency. Since the
molecules are moving relative to one another we must find the average relative
velocity, v12 = h|v1 − v2|i
It can be shown that
v12 =
r16RT
πM=√2v =⇒ collision
frequency=√2n∗πσ2v. (38.7)
From the above expression one defines the mean free path λ to be
λ =v/√
2n∗πσ2v/n∗=LP
RT=RT√2PLπσ2
(38.8)
Example: Ar at SATP (T = 298 K, P = 1 bar):
v = 380.48ms,
collisionfrequency
= 5.25× 109 s−1,
λ = 72.5 nm
253
39. The Rate Laws of Chemical
Kinetics
Thermodynamics described chemical systems in equilibrium. For the study of
chemical reactions it is important understand systems that can be very far from
equilibrium. For this we turn to the field of chemical kinetics.
We can, from thermodynamics, address the question; Will the reaction occur?
We need kinetics, however, answer the question: How fast will the reaction occur?
39.1. Rate Laws
Consider a general four component reaction
aA + bB = cC + dD (39.1)
The time dependence of this reaction can be observed by following the disap-
pearance of either of the reactants or appearance of either of the products. That
is,
−d[A]dt
or − d[B]dtor
d[C]dtor
d[D]dt
(39.2)
BUT this is ambiguous because a moles of A reacts with b moles of B and a does
not, in general, equal b. We must account for the stoichiometry.
254
254
We define the reaction velocity as
v =1
vi
d[I]dt
(39.3)
where vi = −a,−b, c or d and I = A, B, C, or D.
This definition is useful but must be used with caution since for complicated
reactions all the v’s may not be equal. An example of this is
aA +½
bB → cC + dDb0B0 → c0C0 + d0D0
(39.4)
A rate law is the mathematical statement of how the reaction velocity depends
on concentration.
v = f(conc.) (39.5)
For the most part, rate laws are empirical.
Many, but certainly not all, rate laws are of the form
v = k[A1]xA1 [A2]xA2 · · · [An]xAn . (39.6)
The reaction is said to be of order xAi in species Ai and it is of overall orderPi xAi .
In general an overall reaction is made up of so called elementary reactions
Reactant = Product overall rxn (39.7)
Reactant → Intermediates → Product
Note that we shall use an equal sign when talking about the overall reaction and
arrows when talking about the elementary reactions
Example
255
Let
2A + 2B = C + D (39.8)
be the overall reaction. One possible set of elementary steps could be
elementary rxn molecularity
A + A → A0 Bimolecular
A0 → A00 Unimolecular
A00 + 2B→ C + D Trimolecular
.
The rate laws for elementary reactions can be determined from the stoichiometry
molecularity elementary rxn rate law
Unimolecular A→ Product v = k[A]
Bimolecular A + A → Product v = k[A]2
Bimolecular A + B → Product v = k[A][B]
Trimolecular A + A + A → Product v = k[A]3
Trimolecular A + A + B → Product v = k[A]2[B]
Trimolecular A + B + C → Product v = k[A][B][C]
.
Conversely, rate laws for overall reactions can not be determined by stoichiometry.
Connection to thermodynamicsConsider the overall or elementary reaction
aA + bBkfkr
cC + dD (39.9)
where kf is the rate constant for the reaction to proceed in the forward direction
and kr is the rate constant for the reaction to proceed in the reverse direction.
Now, at equilibrium vf = vb which implies
kf [A]a[B]b = kr[C]c[D]d (39.10)
256
bringing kr to the LHS and [A][B] to the RHS we get
kfkr=[C]c[D]d
[A]a[B]b= K 0
c (39.11)
where K 0c is the thermodynamic equilibrium “constant.”
So, we have succeeded in connecting thermodynamics to kinetics BUT we have
done so through the ratio of rate constants. The velocity of a reaction is lost
in this ratio and hence we still can not determine the speed of a reaction from
thermodynamics.
Examples of rate lawsConsider the (overall) reaction between molecular hydrogen and molecular iodine,
H2 + I2 = 2HI. (39.12)
The observed rate laws are vf = kf [H2][I2] and vr = kr[HI]2. This suggests that the
reaction is elementary. In fact, the reaction is not elementary. Moral: Kineticsis very much an empirical science.
Next consider the reaction between molecular hydrogen and molecular bromine,
H2 + Br2 = 2HBr. (39.13)
The observed rate law for this reaction is very complicated,
v =k[H2][Br2]1/2
1 + k0[HBr][Br2]
,
this does not obey any common form.
The above two example are seemingly very similar but they have very different
observed rate laws. Moral: Kinetics is very much an empirical science.
Objectives of chemical kinetics
257
• To establish empirical rate laws
• To determine mechanisms of overall reactions
• To empirically study elementary reactions
• To establish theoretical links to statistical mechanics and quantum mechan-ics
— This involve nonequilibrium thermodynamics–more difficult
• To study chemical reaction dynamics
— the dynamics of molecular collisions that result in reactions
39.2. Determination of Rate Laws
Concentrations c(t) are measured not rates. To obtain the rate from the concen-
tration we must take its time derivative dc(t)dt
. That is we must measure c(t) as a
function of time and find the rate of change of this concentration curve.
The rates of chemical reactions vary enormously from sub-seconds to years. Con-
sequently no one experimental technique can be used.
• For slow reactions (hrs/days) almost any technique for measuring the con-centration can be used.
• For medium reactions (min) either a continuous monitoring technique or a
stopping technique can be used
— A stopping technique used rapid cooling or destruction of the catalyststo stop a reaction at a given point.
• Very fast (sec/subsec) reactions cause problems because the reaction goesfaster than one can mix the reactants.
258
39.2.1. Differential methods based on the rate law
Methods based directly on the rate law rely on the determination of the time
derivative of the concentration.
The main problem with such a method is that randomness in the concentration
measurements gets amplified when taking the derivative.
1. Method of initial velocities
• for v = k[A]x[B]y rate laws.
• initially v0 = kaxby where a and b are the initial concentrations of A
and B respectively
• taking the log of both sides gives lnv0 = ln[kaxby] = ln k+x ln a+y ln b
• a and b can be varied independently so both x and y can be determined.
• problems
1. if the concentration drops very sharply
2. if there is an induction period
2. Method of isolation
• for v = k[A]x[B]y rate laws
• start with initial concentrations a and b equal to the stoichometry; thisgives the overall order of x+ y
• flood with, say, A so v ≈ kax[B]y
39.2.2. Integrated rate laws
The above differential methods look directly at the rate law which is a differential
equation. The differential equation is not solved.
259
We now solve the differential equations to yield what are called the integrated
rate law.
The differential equations (rate law) and their solutions (integrated rate law) are
simply listed here for a few rate laws.type rate lawa) integrated rate lawa)
1st order 1vi
d[I]dt= k[I] [I] = [I0]e
vikt
2nd order 1vi
d[I]dt= k[I]2 1
[I]= 1
[I0]− vikt
nth orderb) 1vi
d[I]dt= k[I]n 1
[I]n−1 =1
[I0]n−1− (n− 1)vikt
enyzme 1vi
d[I]dt= k[I]
km+[I]km ln
[I0][I]+ ([I0]− [I]) = −vikt
a)[I] is the concentration of one of the reactants in an elementary reaction and
vi is the stoichiometric factor for [I] (n.b., vi is a negative number).b)The order need not be an integer. For example n = 3/2 is a three-halves
order rate law.
260
40. Temperature and Chemical
Kinetics
40.1. Temperature Effects on Rate Constants
An empirical rate constant was proposed by Arrhenious:
d ln k
dT=
Ea
RT 2or (40.1)
d ln k
d(1/T )=
Ea
R, (40.2)
where Ea is the Arrhenious activation energy.
Integration of the above yields
ln k = lnA− Ea
RT=⇒ k = Ae−
EaRT (40.3)
(A is the constant of integration). This is the Arrhenious equation
Recall the equilibrium constant can also be obtained from kinetics
K 0c =
kfkr' Ka. (40.4)
Now, take the log of this:
lnKa = ln
∙kfkr
¸= ln kf − ln kr. (40.5)
261
261
Substituting the Arrhenious equation for the rate constants gives
lnKa = ln
∙Afe
−EafRT
¸− ln
hAre
−EarRT
i(40.6)
= ln
∙Af
Ar
¸+
Ear −Eaf
RT
40.1.1. Temperature corrections to the Arrhenious parameters
The Arrhenious parameters A and Ea are constants.
Theoretical approaches to reaction rates predict rate constants of the form
k = aT je−E0/RT . (40.7)
Forcing this to coincide with the Arrhenious implies
Ea = E0 + jRT (40.8)
and
A = aT jej (40.9)
We can verify this by starting with the Arrhenious equation and substituting the
above expressions,
k = Ae−EaRT = aT jeje−
E0+jRTRT = aT jej/ e−j/ e
−E0RT = aT je
−E0RT√
(40.10)
40.2. Theory of Reaction Rates
Simple collision theory (SCT)
• Bimolecular reactions (A,B)
• Reaction rate determined by molecular collisions
262
— Collision frequency for A–B collisions
zAB = πσABL2
s8RT
πLμ[A][B] (40.11)
where μ ≡ mAmBmA+mB
is the reduced mass and σAB is the collision diameter.
• The maximum reaction velocity is vmax = zABL, but intuitively the actual
reaction velocity will be less because
— the ability to react depends on orientation =⇒ a steric factor p
— a minimum amount of collisional energy is required=⇒ e−Emin/RT
• The actual reaction velocity is
v =pzABe
−Em inRT
L(40.12)
• The rate constant for a bimolecular reaction is
k =v
[A][B](40.13)
so SCT predicts
k =pzAB e
−Em inRT
L
[A][B]= pπσABL
s8RT
πLμe−
Em inRT (40.14)
263
• Comparison to the (temperature corrected) Arrhenious equation suggests
A = pπσABL
s8RT
πLμe12 (40.15)
and
Ea = Emin +1
2RT (40.16)
Activated complex theory (ACT)
• An intermediate active complex is formed during the reaction, e.g.,
A + B → (AB)‡ → products. (40.17)
ACT is not limited to bimolecular reactions.
• The active complex is a state in the thermodynamic sense, thus we can applythermodynamics to it.
• For the above example, the equilibrium constant is defined as
K‡a =
a‡
aAaB
low'conc.
[‡][A][B]
(40.18)
264
• Definition: transmission factor, f
— accounts for the fraction of activated complex that becomes product.
— From statistical mechanics, it can be shown that f = kbT/h where kbis Boltzmann’s constant and h is Planck’s constant.
• The reaction rate constant for reactants going to products for ACT is
k = fK‡a =
kbT
hK‡
a (40.19)
• Thermodynamics tells us that
4G‡ = −RT lnK‡a (40.20)
which can be written as
K‡a = e−
4G‡RT = e−
4H‡RT e
4S‡R (40.21)
where 4G‡ = 4H‡ − T4S‡.
• The ACT reaction rate constant now becomes
k =kbT
he−
4H‡RT e
4S‡R . (40.22)
This is Eyring’s equation
40.3. Multistep Reactions
Up to now, the reactions we have studied have been single step reactions.
In general, there is many steps from initial reactants to final products.
Reactions may occur in series or in parallel or both, in what is called a reaction
network.
Parallel reactions:
265
• Parallel reactions are of the form, for example,
A + B1k1→ C (40.23)
A + B2k2→ D
• The rate constant for the disappearance of [A] is simply the sum of the tworate constants: k = k1 + k2
Series reactions:
• Series reactions necessarily include and intermediate product. They are ofthe form
A k1→ B k2→ C (40.24)
• The concentrations of A, B and C are determined by the system of differen-tial equations:
−d[A]dt
= k1[A]
d[B]dt
= k1[A]− k2[B]
d[C]dt
= k2[B],
which, when solved yields
[A] = [A0]e−k1t
[B] =k1[A0]k2 − k1
¡e−k1t − ek2t
¢[C] = [A0]− [A]− [B] = [A0]
µ1− k2e
−k1t − k1ek2t
k2 − k1
¶• See in class animation
266
40.4. Chain Reactions
Chain reactions are reactions which have at least one step that is repeated indef-
initely. The simplest chain reactions have three distinct steps (discussed below)
Chain reactions are extremely important in polymer chemistry
Steps of a chain reaction
1. Initiation: Typically a molecule M reacts to form some highly reactive rad-
ical
M→ R·.
267
2. Propagation: The radical formed in the initiation step reacts with some so
molecule M0 to form another molecule M00 and another radical R0·. This steprepeats an indefinite number of times.
R·+M0 →M00 +R0·.
3. Termination: The radicals interact with each other or with the walls of the
container to forma stable molecule
R0·+R0·→M000
or
R0·+wall→ removed
268
41. Gases and the Virial Series
Unlike liquids and solids, a particular particle has much less significant interactions
with the other particles.
This simplifies the theoretical treatment of gases.
We will now look in detail at the gases.
41.1. Equations of State
Recall from last semester several of the equations of states for gases.
• The ideal gas equation of state
PV = nRT. (41.1)
The equation of state can also be expressed in term of density ρ = mV
ρ =mP
nRT. (41.2)
• The van der Waals gas equation of state
P =nRT
V − nb− n2a
V 2(41.3)
or
P =RT
Vm − b− a
V 2m
, (41.4)
where the parameter a accounts for the attractive forces among the particles
and parameter b accounts for the repulsive forces among the particles
269
269
• BerthelotP =
nRT
V − nb− n2a
TV 2=
RT
Vm − b− a
TV 2m
(41.5)
• DietericiP =
nRTe−anRTV
V − nb=
RTe−a
RTVm
Vm − b(41.6)
• Redlich-Kwang
P =nRT
V − nb− n2a√
TV (V − nb)=
RT
Vm − b− a√
TVm (Vm − b)(41.7)
41.2. The Virial Series
Definition: Compressibility Factor: z = PVnRT
= PVmRT
.
• z is unity for an ideal gas because for such a gas PV = nRT.
• For a real gas z must approach unity upon dilution ( nV→ 0).
• z can be expended in a power series called the virial series.
The virial series in powers of nVis
z = 1 +B(T )³ nV
´+ C(T )
³ nV
´2+D(T )
³ nV
´3+ · · · , (41.8)
or
z = 1 +B(T )
µ1
Vm
¶+ C(T )
µ1
Vm
¶2+D(T )
µ1
Vm
¶3+ · · · . (41.9)
B(T ), C(T ), etc. are called the virial coefficients.
Conceptually B(T ) represents pair-wise interaction of the particles, C(T ) repre-
sents triplet interactions, etc.
270
41.2.1. Relation to the van der Waals Equation of State
Recall the van der Waals equation
P =RT
Vm − b− a
V 2m
(41.10)
multiply both sides by VmRTto get
PVmRT
=Vm
R/ T/R/ T/Vm − b
−V/
m
RT
a
V2/m
=Vm
Vm − b− a
RTVm
=1
1− bVm
− a
RTVm(41.11)
but PVmRT
= z so
z =1
1− bVm
− a
RTVm. (41.12)
The first term is of the form 11−x which has the power series expansion
1
1− x= 1 + x+ x2 + · · · . (41.13)
Therefore
z = − a
RTVm+ 1 +
b
Vm+
µb
Vm
¶2+ · · · . (41.14)
the first term is proportional to 1Vmand so it can be combined with the 1
Vmterm
in the series expansion, hence
z = 1 +³b− a
RT
´ 1
Vm+
µb
Vm
¶2+ · · · . (41.15)
This series can now be compared term by term to the virial series to give expression
for the virial coefficients:
B(T ) =³b− a
RT
´, C(t) = b2, D(T ) = b3, etc. (41.16)
271
41.2.2. The Boyle Temperature
The temperature at which B(T ) = 0 is called the Boyle temperature, Tb.
The virial series at Tb becomes
z(T = Tb) = 1 + 0
µ1
Vm
¶+ C(T )
µ1
Vm
¶2+D(T )
µ1
Vm
¶3+ · · ·
= 1 + C(T )
µ1
Vm
¶2+D(T )
µ1
Vm
¶3+ · · · . (41.17)
The lowest order correction are now³
1Vm
´2. The gas behaves more like an ideal
gas at Tb then for other temperatures.
41.2.3. The Virial Series in Pressure
One can also expand the compressibility factor in pressure
z = 1 +B0(T )P + C 0(T )P 2 +D0(T )P 3 + · · · . (41.18)
The relation of this expansion to the one in 1Vmcan be obtained. One finds (see
homework)
B0(T ) =B(T )
RT, (41.19)
272
C 0(T ) =C(T )−B(T )2
(RT )2(41.20)
and
D0(T ) =D(T )− 3B(T )C(T )− 2B(T )3
(RT )3(41.21)
41.2.4. Estimation of Virial Coefficients
The virial coefficients can be estimated using empirical equations and tabulated
parameters.
• Estimates based on Beattie-Bridgeman constants:
B(T ) = B0 −A0RT− c
T 3, (41.22)
C(T ) =A0a
RT−B0b−
B0c
T 3, (41.23)
D(T ) =B0bc
T 3. (41.24)
where A0, B0, a, b, c are tabulated constants
• Estimates based on critical values (we will discuss critical values shortly, fornow treat them as empirical parameters):
B(T ) =9RTc128Pc
µ1− 6T
2c
T 2
¶. (41.25)
273
42. Behavior of Gases
42.1. P, V and T behavior
We shall briefly consider the P, V and T behavior of dense fluids (e.g., liquids).
Taking volume as a function of P and T, we consider the total derivative
dV (T, P ) =
µ∂V
∂T
¶P
dT +
µ∂V
∂P
¶T
dP. (42.1)
We can change this from a extensive property equation to an intensive property
equation by dividing by V :
dV
V=1
V
µ∂V
∂T
¶P| z
α
dT +1
V
µ∂V
∂P
¶T| z
−κT
dP.
α is the coefficient of thermal expansion.
• At a given pressure, α describes the change in volume with temperature.
• Positive α means the volume of the fluid increases with increasing temper-ature.
κT is the isothermal compressibility
• At a given temperature, κT describes the change in volume with pressure.
• Positive κT means the volume of the fluid decreases with increasing pressure.
• κT is different from z, the compressibility factor.
274
274
42.1.1. α and κT for an ideal gas
As an exercise we shall calculate α and κT using the ideal gas equation of state
(n.b., it is, of course, absurd to treat a liquid as an ideal gas). Starting with the
ideal gas law: V = nRTP
.
κT =−1V
µ∂V
∂P
¶T
=−1V
Ã∂¡nRTP
¢∂P
!T
=−1V
µ−nRT
P 2
¶(42.2)
=1
(PV )| z =nRT
nRT
P=
1
n/ R/ T/n/ R/ T/P
=1
P
and
α =1
V
µ∂V
∂T
¶P
=1
V
Ã∂¡nRTP
¢∂T
!P
=1
V P|z=n/R/T
n/ R/ =1
T(42.3)
42.1.2. α and κT for liquids and solids
In general, the compressibility and expansion of liquids (and solids) are very small.
So one can expand the volume in a Taylor series about a known pressure, P0.
At constant T
V (P ) = V0 +
µ∂V
∂P
¶T| z
−V0κT
(P − P0) +
µ∂V
∂P
¶2T
(P − P0)2 + · · · (42.4)
so,
V (P ) ≈ V0 [1− κT (P − P0)] . (42.5)
This approximation is quite good even over a rather large pressure range (P−P0 =100 atm or so).
275
Likewise at constant P
V (T ) = V0 +
µ∂V
∂T
¶P| z
V0α
(T − T0) +
µ∂V
∂T
¶2T
(T − T0)2 + · · · (42.6)
so,
V (T ) ≈ V0 [1 + α(T − T0)] . (42.7)
As one final point, we can apply the cyclic rule for partial derivatives to determine
the ratio ακT:
α
κT=
¡∂V∂T
¢P
−¡∂V∂P
¢T
cyclic=rule
µ∂P
∂T
¶V
(42.8)
42.2. Heat Capacity of Gases Revisited
This section is a review from the first semester with an additional example beyond
the ideal gas.
42.2.1. The Relationship Between CP and CV
To find how CP and CV are related we begin with
CP =
µ∂H
∂T
¶P
,H = U + PV (42.9)
so
CP =
µ∂ (U + PV )
∂T
¶P
=
µ∂U
∂T
¶P
+ P
µ∂V
∂T
¶P
(42.10)
note¡∂U∂T
¢Pis not CV we need
¡∂U∂T
¢V. Use an identity of partial derivativesµ
∂U
∂T
¶P
=
µ∂U
∂T
¶V
+
µ∂U
∂V
¶T
µ∂V
∂T
¶P
(42.11)
276
thus
CP =
µ∂U
∂T
¶V
+
µ∂U
∂V
¶T
µ∂V
∂T
¶P
+ P
µ∂V
∂T
¶P
(42.12)
= CV +
µ∂V
∂T
¶P
∙µ∂U
∂V
¶T
+ P
¸.
Recall the expression for internal pressure¡∂U∂V
¢T= T
¡∂P∂T
¢V− P . Then
CP = CV +
µ∂V
∂T
¶P
∙T
µ∂P
∂T
¶V
− P/ + P/¸
(42.13)
Finally
CP = CV + T
µ∂V
∂T
¶P
µ∂P
∂T
¶V
(42.14)
For solids and liquids: µ∂V
∂T
¶P
= V α,
µ∂P
∂T
¶V
=α
κT(42.15)
so
CP = CV +α2TV
κT(42.16)
For gases we need the equation of state which often is conveniently explicit in P
or V but not both
1. Explicit in P : Replace µ∂V
∂T
¶P
with −¡∂P∂T
¢V¡
∂P∂V
¢T
(42.17)
2. Explicit in V : Replace µ∂P
∂T
¶V
with −¡∂V∂T
¢P¡
∂V∂P
¢T
(42.18)
277
Examples
1. Ideal gas (equation of state: PV = nRT ): This equation is easily made
explicit in either P or V so we don’t need any of the above replacements
CP = CV + T
µ∂V
∂T
¶P
µ∂P
∂T
¶V
(42.19)
= CV + TnR
P
nR
V=
nRT
PV= nR
Thus CP = CV + nR or CPm = CVm +R
2. One term viral equation (equation of state: V = nRTP+nB). This is explicit
in V so use case 2 above
CP = CV + T
µ∂V
∂T
¶P
µ∂P
∂T
¶V
= CV − T
µ∂V
∂T
¶P
¡∂V∂T
¢P¡
∂V∂P
¢T
(42.20)
The partial derivatives areµ∂V
∂T
¶P
=nR
P+ nB0,
µ∂V
∂P
¶T
= −nRTP 2
, (42.21)
so
−¡∂V∂T
¢P¡
∂V∂P
¢T
= −nRP+ nB0
−nRTP 2
=n/ P (R+ PB0)
n/ RT. (42.22)
Thus
CP = CV + T/µnR
P+ nB0
¶ÃP (R+ PB0)
RT/
!(42.23)
= CV + nR
µ1 +
PB0
R
¶2or
CPm = CVm +R
µ1 +
PB0
R
¶2(42.24)
278
42.3. Expansion of Gases
Expanding gases do work:
−w =Z V2
V1
PexdV (42.25)
As we learned last semester the value of w depends on Pex during the expansion.
Recall that if the expansion is reversible, there is always an intermediate equi-librium throughout the expansion. Namely Pgas = Pex. So,
−wrev =Z V2
V1
PgasdV (42.26)
For an ideal gas (P = nRTV) this becomes
−wrev =Z V2
V1
nRT
VdV = nRT ln
µV2V1
¶(42.27)
Also recall that −wrev is the maximum possible work that can be done in an
expansion. −wrev = −wmax.
42.3.1. Isothermal and Adiabatic expansions
We shall consider two limits for the expansion of gases
1. Isothermal expansion T is constant
2. Adiabatic expansion q = 0.
Isothermal expansion
• For the case of a ideal gas, U(T, V ) = U(T ) (independent of V ). So for
isothermal expansion 4U = 0 = q + w =⇒ q = −w.
279
Adiabatic expansion
• Since q = 0, dU = dw = −PexdV = −PdV (reversible).
• For an ideal gasdU = −PdV = −nRT
VdV (42.28)
42.3.2. Heat capacity CV for adiabatic expansions
Considering an ideal gas going adiabatically from (T1, V1) to (T2, V2).
Recall
CV =
µ∂U
∂T
¶V
=⇒ dU = CV dT (42.29)
So from above
CV dT =−nRTV
dV =⇒ CV dT
T=−nRdV
V(42.30)
Going from (T1, V1) to (T2, V2):Z T2
T1
CV
TdT =
Z V2
V1
−nRV
dV. (42.31)
If CV (T ) is reasonably constant over the internal T1 to T2 then this is approxi-
mately
CV ln
µT2T1
¶= −nR ln
µV2V1
¶(42.32)
where CV =12(CV (T1) + CV (T2)) . Or, in terms of molar heat capacity
CVm ln
µT2T1
¶= −R ln
µV2V1
¶(42.33)
280
42.3.3. When P is the more convenient variable
What if P is the more convenient variable? Then use H instead of U
Let us still consider an adiabatic expansion
H = U + PV, dH = dU + PdV + V dP (because both P and V can, in general,
change)
dH = dq + dw + P/ dV/ + V dP (42.34)
dH = V dP.
Now,
CP =
µ∂H
∂T
¶P
=⇒ dH = CpdT = V dP (42.35)
For an ideal gas this becomes
CpdT =nRT
PdP (42.36)
Going from (T1, P1) to (T2, P2):Z T2
T1
CP
TdT =
Z P2
P1
nR
PdP. (42.37)
If CP (T ) is reasonably constant over the internal T1 to T2 then this is approxi-
mately
CP ln
µT2T1
¶= nR ln
µP2P1
¶(42.38)
where CP =12(CP (T1) + CP (T2)) . Or, in terms of molar heat capacity
CPm ln
µT2T1
¶= R ln
µP2P1
¶(42.39)
From the above two cases
ln
µT2T1
¶=
R
CPm
ln
µP2P1
¶=−RCVm
ln
µV2V1
¶(42.40)
281
So
ln
µP2P1
¶= −CPm
CVm| z ≡γ
ln
µV2V1
¶(42.41)
hence
ln
µP2P1
¶= −γ ln
µV2V1
¶= γ ln
µV1V2
¶= ln
µV1V2
¶γ
(42.42)
Thus µP2P1
¶=
µV1V2
¶γ
⇒ P2Vγ2 = P1V
γ1 , (42.43)
but PiVγi are arbitrary so this implies PV γ = constant (** NOTE: The axes
should be reversed **)
42.3.4. Joule expansion
Consider a gas expanding adiabatically against a vacuum (Pex = 0). In this case
q = 0 (adiabatic) and w = 0 (since −dw = PexdV ).
282
This implies
4U = q + w = 0. (42.44)
Internal energy is constant.
We want to find¡∂T∂V
¢U.
Identity: µ∂T
∂V
¶U
= −µ∂T
∂U
¶V| z
1/CV
µ∂U
∂V
¶T
=1
CV
µ∂U
∂V
¶T
(42.45)
For an ideal gas¡∂U∂V
¢T= 0 (since U(T, V ) = U(T )). Thus in as much as the
gas can be considered ideal¡∂T∂V
¢U= 0. That is, for Joule type expansion the
temperature of the gas does not change. For real gases this is not strictly equal
to zero.
42.3.5. Joule-Thomson expansion
Consider the adiabatic expansion as illustrated by the figure below
283
The work done on the left is
wL = −P14V = −P1(0− V1) = P1V1. (42.46)
The work done on the right is
wR = −P24V = −P2(V2 − 0) = −P2V2. (42.47)
Now,
4U = U2 − U1 = wL + wR = P1V1 − P2V2 (42.48)
Thus
U2 + P2V2 = U1 + P1V1 ⇒ H2 = H1 (42.49)
For Joule-Thomson expansion the enthalpy is constant.
We want to find¡∂T∂V
¢H≡ μ. (the Joule-Thomson coefficient).
Identity: µ∂T
∂P
¶H
= −µ∂T
∂H
¶P| z
1/CP
µ∂H
∂P
¶T
=1
CP
µ∂H
∂P
¶T
= μ (42.50)
284
Recall the useful identity µ∂H
∂P
¶T
= V − T
µ∂V
∂T
¶P
(42.51)
Thus
μ =−V + T
¡∂V∂T
¢P
CP(42.52)
Example: The one term virial equation: (equation of state PV = nRT + nB)
μ =1
CP
µ−nRTP
− nB +nRT
P+ nTB0
¶(42.53)
μ =−B + TB0
CPm.
Limts:
• Low T : B0 is positive and B is negative, so μ is positive–the gas cools upon
expansion
• High T : B0 is nearly zero and B is positive, so μ is negative–the gas warms
upon expansion
• The Joule-Thomson inversion temperature is the temperature where μ = 0.
285
43. Entropy of Gases
43.1. Calculation of Entropy
Entropy must be calculated along reversible paths. This is not a problem though
since entropy is a state function.
Entropy change for changes in temperature.
• At constant V :
— dU = dq + dwdq=CV dT=⇒ dU = CV dT, but also dU = TdS. So
dS =CV
TdT =⇒4S =
Z T2
T1
CV
TdT. (43.1)
At constant P : (use H = U + PV instead of U)
— dH = dU+PdV +V dP = dq−PdV +PdV +V dP . So dH = dqdq=CP dT=⇒
dH = CPdT, but also dHdq=TdS= TdS. So
dS =CP
TdT =⇒4S =
Z T2
T1
CP
TdT. (43.2)
Isothermal expansion of an ideal gas (PV = nRT ):
• Recall that for isothermal expansion of an ideal gas dU = 0 = TdS − PdV
⇒ dS = PdVT
.
286
286
• Using the equation of state
dS =nRdV
V=⇒ 4S =
Z V2
V1
nR
VdV = nR ln
V2V1
. (43.3)
• Using the equation of state to express V1 and V2 in terms of P1 and P2.
dS = nR lnV2V1= nR ln
n/ R/ T/P2
n/ R/ T/P1
= −nR ln P2P1
. (43.4)
If two variables change in going from the initial to final states break the path into
two paths in which only one variable changes at a time.
Entropy of Mixing of an ideal gas
• Since the gas is ideal, there are simply two separate equations:
4SA = nAR lnVA + VB
VA, 4SB = nBR ln
VB + VAVB
(43.5)
and
4Smix = 4SA +4SB (43.6)
287
• Recall Avogadro’s principle: n ∝ V for an ideal gas. So.
4Smix = R
⎛⎜⎜⎜⎝nA lnnA + nB
nA| z 1/XA
+ nB lnnB + nA
nB| z 1/XB
⎞⎟⎟⎟⎠ = −R (nA lnXA + nB lnXB)
(43.7)
43.1.1. Entropy of Real Gases
Consider the question: How does S → Sideal as P → 0 ?
Use Maxwell relation¡∂S∂P
¢T= −
¡∂V∂T
¢Pand single term viral equation, V =
nRTP+ nB.
So µ∂S
∂P
¶T
= −µ∂V
∂T
¶P
= −nRP− nB0 (43.8)
Hence
dS =
µ−nR
P− nB0
¶dP
→→=⇒ S2 − S1 = −nR ln
P2P1− nB0(P2 − P1) (43.9)
For an ideal gas B0 = 0, so
Sideal2 − Sideal1 = −nR ln P2P1
(43.10)
Thus
S2 − S1 = Sideal2 − Sideal1 − nB0(P2 − P1) (43.11)
Letting P1 → 0 and P2 → P θ (Standard pressure 1 bar), this becomes
S2 − S/ ideal
1= Sideal2 − S/ ideal
1− nB0(P2 − P1) (43.12)
Defining Sideal2 , P2 → P θ as Sθ. So,
S(P θ) = Sθ − nB0P θ (43.13)
288
The entropy at any P and T can be obtained expresses as
S(T, P ) = Sideal(T, P )− nB0P (43.14)
Thus
S(T, P ) = Sθ(T )− nR lnP
P θ− nB0P (43.15)
289
Key Equations for Exam 3
Listed here are some of the key equations for Exam 3. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• The Maxwell’s distribution of speeds is
F (v) = 4π
µm
2πkbT
¶ 32
e−mv2
2kbT v2. (43.16)
• The average speed of a particle is
hvi =r8RT
πM(43.17)
• The mean free path isλ =
RT√2PLπσ2
(43.18)
290
290
• The reaction velocity isv =
1
vi
d[I]dt
(43.19)
• The relation between the rate constant and the thermodynamic equilibriumconstant is
Kc =kfkr
(43.20)
• The Arrhenious equationk = Ae−
EaRT (43.21)
• Important thermodynamic relation:
4G = 4H − T4S (43.22)
• Eyring’s equation is
k =kbT
he−
4G‡RT =
kbT
he−
4H‡RT e
4S‡R (43.23)
• The van der Waals gas equation of state:
P =RT
Vm − b− a
V 2m
. (43.24)
• Compressibility Factor:z =
PV
nRT=
PVmRT
. (43.25)
• The virial series is
z = 1 +B(T )
µ1
Vm
¶+ C(T )
µ1
Vm
¶2+D(T )
µ1
Vm
¶3+ · · · . (43.26)
• Relation between heat capacities for an ideal gas:
CPm = CVm +R (43.27)
291
Part VIII
More Thermodyanmics
292
292
44. Critical Phenomena
44.1. Critical Behavior of fluids
The point on the top of the coexistence curve is called the critical point. It is
characterized by a critical temperature, Tc, and a critical density ρc.
Law of rectilinear diameters: The average density [ρave =12(ρliq + ρvap)] is
linear in temperature.
293
293
44.1.1. Gas Laws in the Critical Region
The vapor pressure of a substance is taken from the gas laws as the pressures
where A1 = A2 in the above figure.
Simple gas laws do not work well near critical points.
294
44.1.2. Gas Constants from Critical Data
Consider the van der Waals equation at the critical point (Pc, Tc, Vmc)
Pc =RTc
Vmc − b− a
V 2mc
. (44.1)
There is an inflection point ( dPdVm
= 0, d2P
dV 2m= 0) at the critical point. So, setting
the first and second derivatives at the critical point equal to zero we get
dP
dVm
¯c
=−RTc
(Vmc − b)2+2a
V 3mc
= 0 (44.2)
andd2P
dV 2m
¯c
=2RTc
(Vmc − b)3− 6a
V 4mc
= 0 (44.3)
solving these three equations for Pc, Tc and Vmc gives
Vmc = 3b, (44.4)
Tc =8a
27bR, (44.5)
Pc =a
27b2. (44.6)
These values can be used to find the compressibility factor, z, at the critical point
zc =PcVmc
RTc=3
8= 0.375. (44.7)
Notice that both a and b whose values depend on the particular gas have dropped
out. That is (for the van der Waals Equation) zc = 0.375 for all gases.
The other equations of state give similar resultsvan der Waals Berthelot Dieterici Redlich-Kwong
zc 3/8 = 0.375 3/8 = 0.375 2/e2 ' 0.27 0.33
295
44.2. The Law of Corresponding States
We have found that zc is predicted by the equations of state to be independent of
the particular gas. This is actually not too far from the truth experimentally.
One can define unitless “reduced” variables Tr = T/Tc, Pr = P/Pc, and Vr = V/Vc.
Then zr =PrVrRTr
.
zr is a “universal” function–it is nearly the same for all gasses.
∗ ∗ See Fig. 1.18 Laidler&Meiser ∗ ∗
44.3. Phase Equilibrium
Consider a homogeneous substance consisting of two phases α and β at a constant
T and V.
Suppose some amount of material, dn, goes from α→ β
• (dAα)T = −PdVα − μαdn
• (dAβ)T = −PdVβ + μβdn
• (dA)T,V = −P= 0 since V is constantz | (dVα + dVβ) +
¡μβ − μα
¢dn
For a spontaneous process A deceases (dA < 0)
At equilibrium dA = 0. This implies μβ = μα is the condition for equilibrium.
When α, β denote liquid (or solid) and vapor phases, then for a given T , the
pressure of the system when μβ = μα is the called the vapor pressure of the
material at temperature T.
296
For phase changes at constant T and P then (dG)T,P =¡μβ − μα
¢dn. So again
μβ = μα is the condition for equilibrium.
44.3.1. The chemical potential and T and P
How does μ vary with T and P?
Generally for homogeneous substances,
dG = −SdT + V dP + μdn (44.8)
Now,
S = −µ∂G
∂T
¶P,n
(44.9)
So, µ∂S
∂n
¶P,T
= − ∂
∂n
∂G
∂T= − ∂
∂T
∂G
∂n= −
µ∂μ
∂T
¶P
. (44.10)
But S = nSm(T, P ) so, µ∂μ
∂T
¶P
= −Sm. (44.11)
Similarly, µ∂μ
∂P
¶T
= Vm. (44.12)
Now the total differential of μ is
dμ(T, P ) =
−Smz | µ∂μ
∂T
¶P
dT +
Vmz | µ∂μ
∂P
¶T
dP (44.13)
dμ(T, P ) = −SmdT + VmdP
297
44.3.2. The Clapeyron Equation
At equilibrium μβ = μα so,
−SmαdT + VmαdP = −SmβdT + VmβdP (44.14)
NowdP
dT=
Smα − Smβ
Vmα − Vmβ=−4φSm−4φVm
4S=4HT=4φHm
T4φVm(44.15)
This is the Clapeyron Equation
dP
dT=4φHm
T4φVm(44.16)
44.3.3. Vapor Equilibrium and the Clausius-Clapeyron Equation
The above Clapeyron equation applies to any phase transition; consider the liquid-
vapor phase transition.
Now
4vV = Vm,vap − Vm,liq ' Vm,vap (44.17)
Assuming the vapor phase obeys the ideal gas equation of state,
4vV =RT
P(44.18)
Substituting this into the Clapeyron equation gives
dP
dT=4vHm
T RTP
=4vHmP
RT 2(44.19)
Collecting the T ’s on one side of the equation and the P ’s on the other we get
dP
P=4vHm
R
dT
T 2(44.20)
Now we identify dPP= d(lnP ) and dT
T 2= −d(1/T ) so this becomes
d(lnP ) = −4vHm
Rd(1/T ) (44.21)
298
Rearranging again leads to
d(lnP )
d(1/T )= −4vHm
R(44.22)
This is the Clausius-Clapeyron equation.
44.4. Equilibria of condensed phases
Examples
• Solid—liquid
— ice—water, most other common liquids
• Solid—solid
— rhombic sulfur—monoclinic sulfur
— grey tin—white tin
— graphite—diamond
For example a diamond at STP is metastable with respect to graphite.
“A diamond is not forever!”
At equilibrium μα = μβ this implies (for incompressible liquids and solids)
μªα + Vmα(P − Pª) = μªβ + Vmβ(P − Pª) (44.23)
This can be rearranged so that terms independent of pressure (the standard chem-
ical potentials) are one side and the terms that depend of pressure are on the other
side
μªα − μªβ = (Vmβ − Vmα) (P − Pª) (44.24)
299
Thus for any given T only one P allows for equilibrium.
Recall the Clapeyron equation
dP
dT=4fHm
T4fVm=
Hmβ −Hmα
T (Vmβ − Vmα)(44.25)
We make the good approximation that 4fHm is independent of T and solve the
Clapeyron equationZ →
→
dT
T=4fVmdP
4fHm⇒ ln
TfTªf
=4fVm(P − Pª)
4fHm(44.26)
where Tªf is the freezing temperature at standard pressure (1 bar).
44.5. Triple Point and Phase Diagrams
Definitions
• Phase Diagram: A graph of P vs. T for a system which shows the lines
of equal chemical potential
• Critical Point: The terminal point of the liquid-vapor line. At temper-atures above the critical point there is no distinction between vapor and
liquid.
• Triple Point: The point where all three phases coexist in equilibrium:
μsolid = μliq = μvap (44.27)
300
45. Transport Properties of Fluids
Transport properties of matter deal with the flow (or flux) of some property along
a gradient of some other property.
Flux: movement of something through a unit area.
We now consider three transport properties of fluids:
1. Diffusion: The flux of material down a concentration gradient
2. Viscosity: The flux of momentum down a velocity gradient
3. Thermal Conductivity: The flux of energy down a temperature gradient
∗ ∗ See Transport Phenomena handout ∗ ∗
45.1. Diffusion
At equilibrium concentration on a bulk solution will be uniform.
So if there exists a concentration gradient there will be a net flux, J, of material
from high concentration to low concentration so as to establish an equilibrium.
J =1
A
dn
dt(45.1)
301
301
The flux of material through a plane depends on the concentration difference
J = −DdC
dx=⇒ 1
A
dn
dt= −DdC
dx
where D is the diffusion constant
1
A
dn
dt= −DdC
dx(45.2)
This is Fick’s first law of diffusion (in one dimension).
The change in concentration in a lamina between x and dx with time is given by
the flux in minus the flux out of the lamina:
∂C
∂t=
J(x)− J(x+ dx)
dx= −∂J
∂x(45.3)
Using Fick’s first law for J∂C
∂t=
∂
∂xD∂C
∂x. (45.4)
If D is truly constant we get Fick’s second law of diffusion:
∂C
∂t= D
∂2C
∂x2. (45.5)
302
The solution of this partial differential equation depends on the boundary condi-
tions. Numerous methods of solution exist for this equation but they are beyond
the scope of the course.
The solution for two special boundary conditions are of interest and will simply
be presented here without derivation
1. Point source solution
C(x, t) =C0
2√πDt
e−x2
4Dt (45.6)
2. Step function solution
C(x, t) = C0
"1
2− 1√
π
Z x√4Dt
0
e−y2
dy
#(45.7)
=1
2C0
∙1− erf
µx√4Dt
¶¸=
1
2C0 erfc
µx√4Dt
¶where erf and erfc are tabulated functions respectively called the error func-
tion and complementary error function.
45.2. Viscosity
Viscosity, η, is the resistance to differential fluid flow, i.e., The tendency of a
liquid to flow at the same velocity throughout.
303
The frictional (viscous) force is F = ηAdvdx. (The units of η are mass
lenght·time . 1 poise
= gcm·s .)
Poiseuille’s Formula
• Applies to Laminar (nonturbulent) flow
• For a liquid flowing trough a tube (radius r, length l), the volume of flow
4V in time 4t is4V
4t= −πr
44P
8ηl(45.8)
where 4P is the driving pressure, i.e., the difference in pressure on either
side of the tube.
• For a gas4V
4t=
πr2
16ηl
µP 2i − P 2
f
P0
¶(45.9)
where Pi is the inlet pressure, Pf is the outlet pressure and P0 is the pressure
at which the volume is read.
Stoke’s law: spheres falling through fluids
304
• The frictional force (exerted upwards) is proportional to velocity: Ff = −fv.Stokes showed f = 6πηr
• Gravitational force (exerted downwards): Fg = 4πr3
3(ρ−ρ0)g, where g is the
gravitational acceleration (9.8 m/s2).
• Terminal velocity is reached when Ff + Fg = 0 giving
−fvterm +4πr3
3(ρ− ρ0)g = 0 (45.10)
vterm =4πr3(ρ− ρ0)g
3f
using f = 6πηr
vterm =4π/ r3/(ρ− ρ0)g
3¡6π/ ηr/
¢ =2r2(ρ− ρ0)g
9η(45.11)
• Related to diffusion constant:
D =kT
f
f=6πηr=
kT
6πηr(45.12)
45.3. Thermal conductivity
(This section closely follows parts of chapter 8 in Transport Phenomena by R.B.
Bird, W.E. Stewart and E. N. Lightfoot Wiley New York 1960)
The thermal conductivity, κ, of a material is a measure of the tendency of energy
in the form of heat to flow through the material.
Consider a slab of solid material of area A between two large parallel plates a
distance D apart. The plates are held at constant but different temperatures T1and T2 (T1 > T2) for a sufficiently long time that a steady state exists.
305
Under such conditions, a linear steady state temperature distribution across the
material is established. And a constant rate of heat flow dqdtis needed to maintain
the temperature difference 4T = (T1 − T2)
1
A
dq
dt= −κ4T
D. (45.13)
If we take the limit where D becomes infinitesimally small (D→ dx) we obtain a
differential form of this equation:1
A
dq
dt= Qf = −κ
dT
dx, (45.14)
where Qf is the heat flux. This is called Fourier’s law of heat conduction(one-dimensional version).
Thermal conductivities are positive quantities so Fourier’s law says that heat flow
down a temperature gradient, i.e., from hot to cold.
45.3.1. Thermal Conductivity of Gases and Liquids
∗ ∗ See Reduced thermal conductivity handout ∗ ∗
From this handout we see that typically the thermal conductivity of gases at low
densities increases with increasing temperature, whereas the thermal conductivity
of most liquids decrease with increasing temperature.
306
45.3.2. Thermal Conductivity of Solids
For the most part, the thermal conductivity of solids have to be determined ex-
perimentally because many factors contributing to the thermal conductivity are
difficult to predict.
In general metals are better heat conductors than nonmetals and crystals are
better heat conductors than amorphous materials.
Dry porous materials are poor heat conductors
Rule of Thumb: Thermal conductivity and electrical conductivity go hand inhand.
The Wiedemann, Frantz and Lorenz equation relates the thermal conductivity to
electrical conductivity, κel for pure metals:
κ
κelT= L = const. (45.15)
where L is the Lorenz number (typically 22 to 29 × 10−9 V2/K2).
The Lorenz number is taken as constant because it is only a very weak function
of temperature with a change of 10 to 20% per 1000 degrees being typical.
The Wiedemann, Frantz and Lorenz equation breaks down at low temperature
because metals become superconductive. There is no analog to superconductivity
for thermal conductivity.
307
46. Solutions
Solutions are mixtures of two or more pure substances. So, in addition to the
parameters needed to characterize a pure substance, one also needs to keep track
of the amount of individual species in solution
46.1. Measures of Composition
There are several measures of composition of solutions
• mole ratio r = n1n2
• mole fraction X2 =n2
n1+n2, X1 = 1−X2
• molality m = 1000X2
M1X1, where M1 is the molecular weight of species 1
• Molarity c2 = n2L solution
46.2. Partial Molar Quantities
Thermodynamic properties, in general change upon mixing
4mix = properties of soln −X
properties of pure. (46.1)
For example,
4mixV = Vsoln − Vsolute − Vsolvent (46.2)
Consider a thermodynamic quantity, say, volume.
308
308
In general, it is a function of T, P, n1 and n2: V (T, P, n1, n2). So, the total
derivative is
dV =
µ∂V
∂T
¶P,n1,n2
dT +
µ∂V
∂P
¶T,n1,n2
dP +
µ∂V
∂n1
¶T,P,n2
dn1 +
µ∂V
∂n2
¶T,P,n1
dn2,
(46.3)³∂V∂ni
´T,P,nj
≡ Vi, the partial molar volume.
Similarly
dG =
µ∂G
∂T
¶P,n1,n2
dT +
µ∂G
∂P
¶T,n1,n2
dP +
µ∂G
∂n1
¶T,P,n2
dn1 +
µ∂G
∂n2
¶T,P,n1
dn2,
(46.4)³∂G∂ni
´T,P,nj
≡ μi.
So now for the more general case of mixtures the chemical potential of a species
of the partial molar free energy for that species, rather than simply the molar free
energy as it was earlier.
46.2.1. Notation
The study of solutions brings with it a large number of symbols which we collect
here for future reference.Material
Pure liquid i V •i H•
i S•i G•i
Pure liquid i per mole V •mi H•
mi S•mi μ•i
Whole solution V H S G
Solution/(total moles) Vm Hm Sm Gm
Partial molar of i in solution Vi Hi Si μi
Apparent molar (of solute) φV φH
Reference state V ªi Hª
i Sªi μªi
309
46.2.2. Partial Molar Volumes
Consider the partial molar volume
For constant T and P
dV = V1dn1 + V2dn2 (46.5)
Now, Vi depends on concentration, so change each amount of substance propor-
tional to the amount substance present,
dn1 = n1dλ, dn2 = n2dλ. (46.6)
So,
dV =¡V1n1 + V2n2
¢dλ
dλ=⇒ V = V1n1 + V2n2 (46.7)
That is, the total volume of the solution is equal to the sum of the partial molar
volumes each weighted by their respective number of moles.
The total volume, however, is not necessarily the mole weighted sum of the vol-
umes of each component in its pure (unmixed) state. More specifically
4mixV = V − (V •m1n1 + V •
m2n2) (46.8)
=¡V1n1 + V2n2
¢− (V •
m1n1 + V •m2n2)
=¡V1 − V •
m1
¢n1 +
¡V2 − V •
m2
¢n2
4mixV can be positive, negative or zero.
For example,
1. one unit of baseballs are mixed with one unit of basketballs. 4mixV < 0.
2. one unit of baseballs are mixed with one unit of books. 4mixV > 0.
310
46.3. Reference states for liquids
For liquids there are two more convenient ideal states
1. neat (pure) solvent limit
1. all neighboring molecules are same as the given molecule
2. the ideal state for Raoult’s law
2. infinite dilution limit
1. all neighboring molecules are different than the given molecule
2. the ideal state for Henry’s law
Raoult’s law limit Henry’s law limit
46.3.1. Activity (a brief review)
Recall that activity gives a measure of the deviation of the real state from some
reference state
311
Also recall that the mathematical definition of activity ai of some species i is
implicitly stated as
limζ→ζª
aig(ζ)
= 1 (46.9)
where g(ζ) is any reference function (e.g., pressure, mole fraction, concentration
etc.), and ζª is the value of ζ at the reference state.
This implicit definition is awkward so for convenience one defines the activity
coefficient as the argument of the above limit,
γi ≡aig(ζ)
(46.10)
which we can rearrange as
ai = γig(ζ). (46.11)
The definition of activity implies that γi = 1 at g(ζª) (the reference state)
That is γi → 1 as the real system approaches the reference state.
Connecting with the chemical potential we saw last semester that the deviation
of the chemical potential at the state of interest versus at the reference state is
determined by the activity at the current state (the activity at the reference state
is unity by definition).
μi − μªi = RT ln ai. (46.12)
46.3.2. Raoult’s Law
In discussing both Raoult’s law and Henry’s law, we are describing the behavior
of a liquid solution by measuring the vapor (partial) pressures of the components
312
For simplicity we consider here only a two component solution.
dG = μ1dn1 + μ2dn2. (46.13)
Take differential change along a line of constant concentration, so
dG = (μ1n1 + μ2n2) dλ (46.14)
then
G = μ1n1 + μ2n2. (46.15)
Recall that
4mixG = G(soln)−G(pure components) (46.16)
Hence,
4mixG = μ1n1 + μ2n2 − μ•1n1 − μ•2n2 (46.17)
= (μ1 − μ•1)n1 + (μ2 − μ•2)n2.
Now,
μ1 − μ•1 = RT lnaia•i
low P' RT lnPi
P •i
, (46.18)
where Pi is the vapor pressure of the ith component above the solution.
313
Thus
4mixG = RT
µn1 ln
a1a•1+ n2 ln
a2a•2
¶(46.19)
or at low P
4mixG = RT
µn1 ln
P1P •1
+ n2 lnP2P •2
¶(46.20)
46.3.3. Ideal Solutions (RL)
Raoult’s Law:
Pi = XiP•i (46.21)
That is, the vapor partial pressure of a component of a mixture is equal to the
mole fraction of the component times the vapor pressure that the component
would have if it were pure.
The change in free energy upon mixing for solutions ideally obeying Raoult’s law
is
4id(RL)mix G = RT
Ãn1 ln
X1P•1/
P •1/+ n2 ln
X2P•2/
P •2/
!(46.22)
4id(RL)mix G = RT (n1 lnX1 + n2 lnX2) (46.23)
Again, this is for an ideal solution in the Raoult’s Law sense.
From
S = −µ∂G
∂T
¶P
and H = −µ∂ (G/T )
∂ (1/T )
¶P
, (46.24)
the entropy of mixing for an ideal Raoult solution is
4id(RL)mix S = −R (n1 lnX1 + n2 lnX2) (46.25)
and the enthalpy of mixing is
4id(RL)mix H = 0 (46.26)
314
(since G/T is independent of 1/T ).
The Reference State (RL)
Let us apply the definition of activity for the Raoult’s law reference state.
The reference function is g(ζ) = ζ = Xi. and the reference state is Xi = 1
So,
limXi→1
a(RL)i
Xi= 1 (46.27)
implies
a(RL)i = γ
(RL)i Xi, (46.28)
and γ(RL)i → 1 as Xi → 1
Deviations from Raoult’s Law
Raoult’s law is a purely statistical law. It does not require any kind of interaction
among the constituent particle making up the solution.
Since, in reality, there are specific interactions between particles, real solutions
generally deviate from Raoult’s law.
The physical interpretation of deviation from Raoult’s law is
• positive deviation: the molecules prefer to be around themselves rather thanother types of molecules.
• negative deviation: the molecules prefer to be around other types of mole-cules than themselves.
• no deviation: the molecules have no preference.
315
It is very important to note that this deviation from Raoult’s law is a property of
the solution and NOT any given component.
For example, for a given component, mixing with one substance may lead to
a positive deviation but mixing with another substance may lead to a negative
deviation.
Positive deviation from Raoult’s lawNegative deviation from Raoult’s law
46.3.4. Henry’s Law
Henry’s Law:Pi = kXiXi, (46.29)
where kXi is the Henry’s law constant,
kXi= lim
Xi→0
µPi
Xi
¶(46.30)
Henry’s law applies to the solute not to the solvent and becomes more correct for
real solution as the concentration of solute goes to zero (Xi → 0), i.e., at infinite
dilution.
316
The Reference State (HL)
Referring to the definition of activity again we see that the reference function is
g(ζ) = ζ = Xi. and the reference state is now Xi = 0
So,
limXi→0
a(HL)i
Xi= 1 (46.31)
implies
a(HL)i = γ
(HL)i Xi, (46.32)
and γ(HL)i → 1 as Xi → 0
If instead of mole fraction, molality or molarity is used then
a(HL)i = γ(HL)mi
mi (46.33)
and
a(HL)i = γ
(HL)Mi
Mi (46.34)
respectively.
Comparison of Raoult’s Law and Henry’s Law
Both Raoult’s law and Henry’s law become better approximations for real solu-
tions as the solution becomes pure. But, they apply to opposite species in the
solution. Raoult’s law applies to the dominant species, X1 → 1, whereas Henry’s
law applies to the subdominant species X2 → 0. So, in summary
• Raoult’s law: γ1 → 1 as X1 → 1
• Henry’s law: γ2 → 1 as X2 → 0
317
46.4. Colligative Properties
Colligative properties: Properties of dilute solutions that are independent ofthe chemical nature of the solute
Examples
• Freezing point depression
• Boiling point elevation
• Vapor pressure lowering
• Osmotic pressure
We will consider the examples of freezing point depression and osmotic pressure
46.4.1. Freezing Point Depression
At Tf (freezing point), μ1(solid)| z μs1
= μ1(soln).
318
Using the Raoult’s law reference state (since we are interested in the behavior of
the dominant species), μ1(soln) = μ•1 +RT ln a1:
μs1 = μ•1 +RT ln a1 (46.35)
Rearranging this and taking the derivative with respect to T yields
∂
∂T
→
→ln a1 =
1
RT(μs1 − μ•1) =⇒
∂ ln a1∂T
=−1RT 2
µ∂μs1∂T− ∂μ•1
∂T
¶(46.36)
Now, using ∂μ∂T= H and integrating we getZ →
→d ln a1 =
µ−1RT 2
(Hs1 −H•
1)
¶dT =
4fH
RT 2dT (46.37)
ln a1 =
Z Tf
T•f
4fH
RT 2dT
For small changes in the freezing point we may approximate T by T •f in the
integrand. So,
ln a1 'Z Tf
T•f
4fH
RT •2fdT =
−4fH
RT •2fΘ, (46.38)
where Θ ≡ T •f − Tf . The freezing point depression is
Θ = −RT •2f ln a1
4fH
46.4.2. Osmotic Pressure
We consider the osmotic pressure at a constant temperature, T. (so, dG = V dP ).
319
In the above figure μ1(left) = μ1(right), hence
μ•1 = μ•1 +RT ln a1 + V1Π, (46.39)
where V1 is the partial molar volume of the solvent in solution (difficult to measure)
and Π is the hydrostatic (osmotic) pressure.
From the above equation
ln a1 =V1Π
RT(46.40)
Now we make the approximations V1 = V •m1, a1 = X1 = 1−X2:
ln(1−X2) =V •m1Π
RT(46.41)
For dilute solutions X2 is small so ln(1−X2) may be expanded as
ln(1−X2) = −X2 +X22
2− X3
2
3− · · · ' −X2, (46.42)
but X2 =n2
n1+n2' n2
n1for dilute solutions. Thus
n2n1' V •
m1Π
RT=⇒ n2 '
V •1z |
n1V•m1Π
RT, (46.43)
320
or,
Π =n2V •1|z'c
RT = cRT, (46.44)
where c is the concentration of the solute.
Note the similarity of this equation with the ideal gas equation: P = cRT. Thus
the solute in a very dilute solution behaves as if it were an ideal gas.
321
47. Entropy Production and
Irreverisble Thermodynamics
We have seen that thermodynamics tells us if a process will occur and kinetics
tells us how fast a process will occur.
These two areas of physical chemistry appear to be rather disjoint.
We now we consider thermodynamics of nonequilibrium states and investigate
how (and how fast) these state move towards equilibrium.
This allows us to make a stronger connection between thermodynamics and ki-
netics.
The main concept of this approach is the idea of entropy production and, ulti-
mately, entropy production per unit time–how fast we are producing entropy.
47.1. Fundamentals
We know the difference between reversible and irreversible processes from before.
However, we will state their respective definitions here in a manner best suited
for this chapter.
322
322
Reversible process: dynamical equations are invariant under time inversion(t→−t).
• e.g., the one dimensional wave equation,
1
c
∂2u
∂t2=
∂2u
∂x2t→−t=⇒ 1
c
∂2u
∂(−t)2 =∂2u
∂x2=⇒ 1
c
∂2u
∂t2=
∂2u
∂x2, (47.1)
is invariant under time reversal
Irreversible process: dynamical equations are not invariant under time inver-sion (t→−t).
• e.g., the one dimensional heat equation,
1
κ
∂T
∂t=
∂2T
∂x2t→−t=⇒ 1
κ
∂T
∂(−t) =∂2T
∂x2=⇒ −1
κ
∂T
∂t=
∂2T
∂x2, (47.2)
is not invariant under time reversal.
We will be concerned with the change in entropy, dS, which can be split into two
components dS = deS + diS.
Definitions
• deS is the change in entropy due to interactions with the exterior environ-
ment.
• diS is the change in entropy due to internal changes of the system
The quantity diS is called the entropy production.
323
Splitting up dS into these two parts permits an easy discussion of both open and
isolated systems–the difference between the two appearing only in deS.
General criteria for irreversibility:
• diS = 0 (reversible change)
• diS > 0 (irreversible change)
For isolated systems have diS = dS and the principle of Clausius, diS = dS ≥ 0,holds.
47.2. The Second Law
As you might expect, the second law underlies all the concepts of this chapter.
We need a “local” formulation of the second law:
• Absorption of entropy in one part of the system, compensated by a sufficientproduction in another part is prohibited
— i.e., in every macroscopic region of the system the entropy production
due to irreversible processes is positive.
This is simply another in our long list of alternative statements of the second law.
324
I
II
Considering the above figure of an isolated system, we write the principle of Clau-
sius as
dS = dSI + dSII ≥ 0. (47.3)
The local formulation statement implies
diSI ≥ 0 and diS
II ≥ 0 (47.4)
and the possibility of, for example, diSI < 0 and diSII > 0 such that di¡SI + SII
¢>
0 is excluded.
47.3. Examples
The idea of entropy production can be applied to any of the processes we have
talked about; mixing, phase changes, heat flow, chemical reactions, etc. As exam-
ple we now consider the last two of these: heat flow and chemical reactions.
325
47.3.1. Entropy Production due to Heat Flow
Recall from the lecture on transport phenomena that the heat flux Qf is given by
Qf = −κ4T
D(47.5)
We are now interested in exposing the time dependence, so, using Qf =q4twe get
q
4t= −κA4T
D(47.6)
in differential form this isdq
dt= −κAdT
dx. (47.7)
Example: Find the entropy production in a system consisting of two identical
connected blocks of metal (I and II), one of which is held at temperature T1 and
the other at T2 (take T1 > T > T2) where T is the temperature at the interface.
326
Considering the whole system
dS =dqIT1+
dqIIT2
=
deSz | deqIT1
+deqIIT2
+
diSz | diqIT1
+diqIIT2
. (47.8)
The quantity deqj is the amount of heat supplied by the environment to hold block
j at its fixed temperature.
Furthermore the heat going out of I through the connecting wall is equal to the
heat coming into II through the connecting wall:
diqI = −diqII . (47.9)
Using this we see that the entropy production is
diS = diqI
µ1
T1− 1
T2
¶, (47.10)
which we see is positive because diqI < 0 when T1 > T2.
We have still not made a connection to kinetics.
To do so we must consider the entropy production per unit time diSdt.
327
For this examplediS
dt=
diqIdt
µ1
T1− 1
T2
¶From chapter 24 we know
diqIdt
=−Aκ4T
D. (47.11)
So,diS
dt=−Aκ4T
D
µ1
T1− 1
T2
¶(47.12)
To determine T we use the fact that the heat flow out of I is equal to the heat
flow into II:diqIdt
=−diqIIdt
. (47.13)
Using the above expression for heat flow gives us T since,
−κ/ A/D/
¡T1 − T
¢= −κ/ A/
D/¡T − T2
¢⇒ T =
T1 + T22
; (47.14)
a result we might have guessed.
47.3.2. Entropy Production due to Chemical Reactions
Definitions:
1. Chemical affinity: a ≡ − (4rxnG)T,P = −P
i viμi and a ≡ − (4rxnA)T,V =
−P
i viμi
2. Extent of reaction: ξ is defined by dξ = dnivi, where ni is the number of moles
of the ith component and vi the stoichiometric factor of the ith component.
328
• e.g., for the reaction N2 + 3H2 → 2NH3
dξ =dnN2(−1) =
dnH2(−3) =
dnNH3(2)
(47.15)
and
a = 2μNH3 − μN2 − 3μH2 (47.16)
The connection to kinetics: reaction rate v = dξdt
The connection to thermodynamics:
(dA)T,V =Xi
μidni =Xi
viμi| z −a
µ1
vi
¶dni| z
dξ
= −adξ (47.17)
but
(dA)T,V =
dqz | (dU)T,V − TdS ⇒ dS =
dq
T−
−adξz | (dA)T,V
T(47.18)
so
dS =
deSz|dq
T+
diSz|adξ
T(47.19)
The entropy production per unit time for a chemical reaction is a function of both
the chemical affinity and of the reaction rate
diS
dt=a
T
dξ
dt=a
Tv ≥ 0 (47.20)
We see that for a spontaneous process the entropy production per unit time is
positive. This is because a = − (4rxnA)T,V is positive as is v.
329
Simultaneous Reactions
For N simultaneous chemical reactions, the entropy production per unit time
generalizes todiS
dt=1
T
NXj=1
ajvj ≥ 0. (47.21)
The second law requires that the total entropy production for simultaneous reac-
tions is positive. It says nothing about the entropy production of the individual
component reactions other then the sum of all the component entropy productions
must be positive.
For example in a system of two coupled reactions we could have a1v1 < 0, a2v2 > 0
such that a1v1 + a2v2 > 0.
47.4. Thermodynamic Coupling
Processes may be what is called thermodynamically coupled such that a process
that normally is not thermodynamically favored can be coupled to another process
that is thermodynamically favored so as to allow for the unfavorable process to
proceed spontaneously.
We just saw an example of such a situation with the discussion of simultaneous
reactions.
Thermodynamic coupling need not be confined to coupling between the same
types of processes.
That is, diffusion is the flux of matter down a concentration gradient. The so-
called Soret effect is flux of matter down a temperature gradient. Conversely, the
so-called Dufour effect is heat flux down a concentration gradient
330
The following table lists a number of thermodynamically coupled phenomena
Flux q m material Q (charge)
Gradient
T Thermoconductivity Thermomechanicaleffect Soret effect Seebeck effect
P Mechanocaloriceffect
Hydrodynamicflow
Reverseosmosis Potential of flow
C Dufour effect Osmosis Diffusion Nernst Potential
ε Peltier effect Electrophoresis Migration Electoconductivity
47.5. Echo Phenonmena
Consider an ensemble that is perturbed away from thermal equilibrium by some
means such as by applying a field.
If the perturbation is released the system will begin to evolve in time as it heads
back towards the thermalized equilibrium state.
The ensemble evolves in two ways
• Reversibly
— A second perturbation can “undo” or reverse the evolution.
• Irreversibly
— The evolution towards equilibrium cannot be undone–it is irreversible
Example: The spin echo in pulsed NMR
• A radio frequency pulse prepares an ensemble of nuclear spins such that
they are all spinning coherently.
331
• A strong signal is seen because all the spinning nuclei cooperate.
• Each nucleus is in a slightly different environment so each spin frequency isslightly different.
• The different environment (spin frequencies) cause the ensemble spinningnuclei to dephase
• Dephasing causes a decrease in the observed signal because now not all nucleiare cooperating.
• Now a radio pulse with the opposite phase is applied to make the nuclei spinin the opposite direction
• This undoes or reverses the dephasing process and the signal regains strength
• The full signal is not recovered however since all the while random ther-
malization is taking place to irreversibly destroy the coherence among the
nuclei.
• This cannot be undone with the second radio pulse.
332
Key Equations for Exam 4
Listed here are some of the key equations for Exam 4. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
• The Clapeyron Equation is
dP
dT=4φHm
T4φVm. (47.22)
• The Clausius-Clapeyron equation is
d(lnP )
d(1/T )= −4φHm
R(47.23)
• Fick’s first law of diffusion is
1
A
dn
dt= −DdC
dx(47.24)
333
333
• Fick’s second law of diffusion:∂C
∂t= D
∂2C
∂x2. (47.25)
• Relation between the viscosity and the diffusion constant:
D =kT
f
f=6πηr=
kT
6πηr. (47.26)
• Fourier’s law of heat conduction is1
A
dq
dt= Qf = −κ
dT
dx. (47.27)
• Mixing
4mix = properties of soln −X
properties of pure. (47.28)
• Chemical potentialμ = μª +RT ln a (47.29)
• Raoult’s Law:Pi = XiP
•i (47.30)
• Raoult’s law reference
a(RL)i = γ
(RL)i Xi, γ
(RL)i → 1 as Xi → 1 (47.31)
• Henry’s Law:Pi = kXiXi. (47.32)
where kXiis the Henry’s law constant,
kXi = limXi→0
µPi
Xi
¶. (47.33)
• Henry’s law reference
a(HL)i = γ
(HL)i Xi, γ
(HL)i → 1 as Xi → 0. (47.34)
334
Index
absorption spectroscopy 241activity 146, 311
mathematical definition of 146activity coefficient 146, 312adiabatic expansion 280
and heat capacity 280adiabatic wall 120angular momentum
addition of 202classical 192eigenfunctions for 199, 219jj coupling 202LS coupling 202quantum numbers 199, 219spin 201
angular momentum quantum num-ber 52
antibonding orbital 71Arrhenious activation energy 261Arrhenious equation 261, 291
temperature corrected 262atomic orbitals 49
chemists picture 50physicists picture 50
aufbau principle 58average value theorem 29Berthelot gas 13, 270binominal coefficient 90blue sky 81Bohr model 18
Bohr radius 19Boltzmann distribution 10, 96, 131Boltzmann’s equation 90, 97, 124,
131bond order 77bonding orbital 71Born model 170
corrections to 175enthalpy of solvation 174entropy of solvation 174free energy of solvation 173, 178partition coefficient 174
Born—Oppenheimer approximation62, 99, 235, 240
and the Franck—Condon princi-ple 243
bosons 56Boyle temperature 272chain rule
for partial derivatives 107character table
for the C2v group 225chemical affinity 328chemical potential 144
for a salt 161relation to activity 148relation to Gibbs free energy
145relation to Helmhotz free en-
ergy 145
335
335
Clapeyron equation 298, 300, 333Clausius-Clapeyron equation 299,
333coefficient of thermal expansion 274coexistence curve 293colligative properties 318commutator 30, 189completeness 191complimentary variables 30compressibility factor
at the critical point 295compressibilty factor 270, 291configuration 90confluent hypergeometric functions
65correspondence principle 41critical point 300cyclic rule 14, 108cylindrical symmetry 69Debye—Huckel limiting law 164, 178Debye—Huckel theory 163Debye—Huckel—Guggenheim equation
164Debye’s law 129, 133degeneracy 186
of the ensemble 98diathermic wall 120diatomic molecules
electron-electron potential en-ergy operator for 61
electronic kinetric energy oper-ator for 61
electronic wavefunction for 62Hamiltonian for 61nuclear kinetic energy operator
for 61nuclear-electron potential energy
operator for 61
nuclear-nuclear potential energyoperator for 61
Schrodinger equation for 62Dieterici gas 270diffusion 301diffusion constant 302eigenfunction 5eigenvalue 5eigenvalue equation 190electric dipole approximation 79, 231electrolytes
strong 161electrophoretic effect 167elementary reactions 255
and stoichiometry 256molecularity 256
emission spectroscopy 241enemble 89ensemble average 103, 132enthalpy 136entropy 105
change for changes in temper-ature 286
change for isothermal expansion286
change for mixing 287of real gases 288
entropy production 322, 323due to chemical reactions 328due to heat flow 326
equation of state 116for a Berthelot gas 118for a Dieterici gas 118for a Redlich—Kwang gas 118for a van der Waals gas 117for an ideal gas 116for gases 269
equilibrium constant 135
336
equlibrium constant 153Euler’s identity 4expansion
of gases 111reversible 114
extent of reaction 328Eyring’s equation 265, 291fermions 56Fick’s first law 302, 333Fick’s second law 302, 334first law of thermodynamics 121,
133flipping coins 90fluctuation 92fluorescence 242
stokes shift 242Fourier’s law of heat conduction 306,
334Franck—Condon integral 243Franck—Condon principle 243free energy
Gibbs 138Helmholtz 137
fugacity 147fundamental transistions 66general equlibrium 151generalized displacement 110generalized force 110gerade 69Gibb’s free energy 106Gibbs-Duhem equation 163good theory 16group
mathematical definition of 222multiplication table 223
group theory 221Hamiltonian operator 27Hamitonian
classical 27harmonic oscillator 38
energy levels for 40, 44, 86potential energy 39Schrodinger equation for 39
heat 109sign convention 110
heat capacity 115, 133Heisenberg uncertainty principle 30
and the harmonic oscillator 41helium 55
electron-electron repulsion term55
Hamiltonian 55Helmholtz free energy 106Henry’s law 311, 316, 334Henry’s law constant 316, 334Hermite polynominals 40hot bands 66Hund’s rule 205hydrogen atom
ioniztion energy of 19hydrogen molecule 74hydrogenic systems 46
energy levels for 49, 86Hamiltonian 47normalization constant 49, 85potential energy for 47Schrodinger equation for 47wavefunction (no spin) 49wavefunction (with spin) 52
ideal solutionRaoult’s law 314
immiscible solutions 153infrared spectroscopy 66internal energy 103, 121intramolecular vibrational relaxation
(IVR) 242
337
inversion symmetry 69operator 69
ion mobility 166and current 168
ion transfer 174IR spectroscopy 231
and the character table 232isothermal compressibility 274isothermal expansion 279Joule expansion 282Joule-Thomson expansion 283kinetic theory of gases 250Lagrange multipliers 95Laguerre polynominals 49laminar flow 304law of corresponding states 296law of rectilinear diameters 293Legendra polynomials 200linear combinations of atomic or-
bitals (LCAO) 72Lorenz number 307many electron atom
Hamlitonian for 59maximal work 113Maxwell relations 140Maxwell’s distribution of speeds 252,
290mean free path 253, 290mean ionic activity 162mean ionic activity coefficient 162method of initial velocities 259method of isolation 259microstate 90Mie scattering 84mirror plane symmetry 70molar heat capacity 115molecular collisions
simple model for 252
molecular hydrogen ion 67Hamiltonian for 67
molecular orbital diagram 76molecular orbitals 68molecular rotations 235
asymmetric tops 239centrifugal stretching 236linear tops 238polyatomic molecules 237spherical tops 239symmetric tops 238vibrational state dependence of
236molecular vibrations 228molecule
Scrodinger equation for 78momentum operator 5Morse oscillator 64
energy levels for 65, 86Schrodinger equation for 65wavefunction for 65
Morse potential 64, 86, 240force constant associated with
9Taylor series expansion of 8
normal modes 229operator
Hermitian 189ladder 195linear 189symmetry 222
operator algebra 187orientation quantum number 53orthogonality 191overtone transitions 66parameters
extensive 109intensive 109
338
particle in a box 31, 181energy levels 183energy levels for 34, 44, 218features of the energy levels 35normalization constant for 33potenial energy 31Schrodinger equation for 32three dimensional 183three dimensional energy levels
185three dimensional wavefunction
185wavefunction for 183wavefunctions for 34, 44, 218
particle on a ring 194boundary conditions 194energy levels for 195, 218Hamitonian for 194wavefunctions for 195, 218
partition coefficient 154and drug delivery 155for the Born model 174
partition functioncanonical 96, 131electronic 101grand canonical 97isothermal—isobaric 97microcanonical 96molecular 100rotational 101, 132translational 101, 132vibrational 101, 132
Pauli exclusion principle 56consquences of 58
perturbation theory 207example of the quartic oscilla-
tor 208phase diagram 300
Poiseuille’s formula 304polarizability 79postulate I (of quantum mechan-
ics) 22postulate II (of quantum mechan-
ics) 24postulate III (of quantum mechan-
ics) 25pressure 104principle of Clausius 125, 324principle quantum number 52probability amplitude 22probability distribution 22PV work 111, 133Raman scattering 80Raman spectroscopy 66, 233
and the character table 234Raoult’s law 311, 312, 314, 334
deviations from 315reference state 315
rate law 255rate laws 254
determination of 258integrated 259
Rayleigh scattering 80Rayleigh scattering law 81, 82, 87reaction velocity 255, 291reciprocal rule 108red sunsets 82Redlich-Kwang gas 270reference states 147relationship between CP and CV
139, 276relaxation effects 167rigid rotor 200
degeneracy of 235, 248energy 235, 247
rotational energy levels 200, 219
339
degeneracy of 200rotational Hamiltonian 200rule of mutual exclusion 234Rydberg constant 20SATP 120Schrodinger equation
time dependent 214time independent 27
second law“local” formulation 324
second law of thermodynamics 126,133
statements of 127simple collision theory 262Slater determinant 58
for lithium 59solar system model 17solvation 169solvophobic effect 176specific heat 115spherical harmonic functions 48, 200spin 201
quantum number 51, 53wavefunction 51
spin orientationquantum number 51, 53
spin-orbitcoupling 205Hamiltonian 205interaction energy 205
spontaneous process 142state function 121
table of important ones 136Sterlings approximation 92Stoke’s law 167, 304STP 120superposition 191systems
types of 108temperature 115term symbols 204thermal conductivity 301
of gases 306of liquids 306
thermal equilibrium 120third law of thermodynamics 128,
133tips for solving problems 2total derivative 107transfer matrix 11triple point 300two level system 211
‘left’ and ‘right’ states 213, 219Hamiltonian for 212
Tyndall scattering 84ungerade 69van der Waals equation
340