Chapter (5)

Maxwell-Boltzmann distribution statistics and

Application to molecular energies in an ideal gas

Introduction

Statistical mechanics provides a bridge between the macroscopic realm

of classical thermodynamics and the microscopic realm of atoms and

molecules. We are able to use computational methods to calculate the

thermodynamic parameters of a system by applying statistical mechanics.

We must remember that the energies of molecules, atoms, or electrons

are quantized. To describe the systems we must know the energies of the

quantum states and the distribution of particles (i.e., molecules, atoms, or

electrons) among the quantum states. The Schrodinger equation that we

discussed in the section on quantum mechanics provides a method for

calculating the allowed energies. The Boltzmann distribution law that is a

fundamental principle in statistical mechanics enables us to determine

how a large number of particles distribute themselves throughout a set

of allowed energy levels.

Consider a system composed of N molecules, and its total energy E is a

constant. These molecules are independent, i.e. no interactions exist

among the molecules. Countless collisions occur. It is hopeless to keep

track positions, moments, and internal energies of all molecules. All

possibilities for the distribution of energy are equally probable provided

the number of molecules and the total energy are kept the same. That is,

we assume that vibrational states of a certain energy, for instance, are as

likely to be populated as rotational states of the same energy. For

instance, four molecules in a three-level system: the following two

conformations have the same probability.

---------l-l-------- 2 ---------l--------- 2

---------l---------- ---------1-1-1----

---------l---------- 0 ------------------- 0

Imagine that there are total N molecules among which n0 molecules with

energy 0, n1 with energy 1, n2 with energy 2, and so on, where 0 < 1 < 2

< .... are the energies of different states. The specific distribution of

molecules is called configuration of the system, denoted as { n0, n1,

n2, ......}

The above configuration is thus, { 4, 6, 4, 3 }

THE DISTRIBUTION OF MOLECULAR STATES

{N, 0, 0, ......} corresponds that every molecule is in the ground state,

there is only one way to achieve this configuration; {N-2, 2, 0, ......}

corresponds that two molecule is in the first excited state, and the rest

in the ground state, and can be achieved in N(N-1)/2 ways.

A configuration { n0, n1, n2, ......} can be achieved in W different ways,

where W is called the weight of the configuration. And W can be

evaluated as follows,

1. N! different ways to arrange N molecules;

2. ni! arrangements of ni molecules with energy i correspond to the

same configuration;

Example 1:

1. Calculate the number of ways of distributing 3 objects a, b and c

into two boxes with the arrangement {1, 2}.

Answer:

| a | b c |, | b | c a |, | c | a b |.

Therefore, there are three ways 3! / 1! 2! or

| a | c b |, | b | a c |, | c | b a |

Example 2:

Calculate the number of ways of distributing 20 objects into six

boxes with the arrangement {1, 0, 3, 5, 10, 1}.

1

2

3

4

5

6

7

89

10

1112

13

14

15

1617

1

2

3

4

STATE

Answer:

20! / 1! 0! 3! 5! 10! 1! = 931170240

It is easy to calculation the number of ways of distributing of one

molecules (A), the probability (w) is A=1. The number of ways of

distributing of two molecules, the probability (w) is AB=2 or BA=2. The

number of ways of distributing of three molecules, the probability (w) is

ABC, ACB, BAC, BCA, CAB, CBA= 3x2=6. The number of ways of

distributing of three molecules, the probability (w) is ABCD, ABDC,

ADBC, DABC,BCDA, BCAD, BACD, ACDB, CDAB, CDAB,CDBA, CBDA,

BDAC, DABC, DABC, DACB, DCAB, CABD= 4x4=16.

The Distribution of Internal Energies in Molecules

What happens if we apply thermal energy (kT energy) to a mol of

molecules confined in a jar at constant temperature? The molecule takes

up the energy and distributes it among the available energy levels. This

distribution was studied by Boltzmann who showed that:

Ni/Nj = exp (-(Ei-Ej))/kT

Ni = No. of molecules in level i (higher)

Nj = “ “ “ “ j (lower)

Ei = energy of molecules in level I (higher)

Ej = “ “ “ “ “ j (lower)

The ratio, Ni/Nj is the relative occupancy. The Boltzmann distribution, as

the equation is known, applies only when the system is in thermal

equilibrium. It gives the relative occupancy for two energy states, Ei and

Ej. What are some of the properties of this equation? Suppose we look

at rotation at very low temperature i.e. when Nj = No and Ev = Eo = 0

Ni/No = exp (-(Ei-Eo))/kT

Ni = Noe -Ei/kT

ie. at constant temperature there is an exponential decrease in the

number of molecules in the higher energy states (Fig. 1)

Fig. 1

e.g. CO vibration at room temperature (RT = 250C; ΔEv = 25 kJ mol-1)

11 molkJ5.2molkJ25

o

st

i e)stateground(N

)stateexcited1(N = e -10 ≈ 5 10-5 or 0.005

%

i.e. only 0.005% of molecules in first excited state, Ni the rest is in the

ground state, No.

Test of the Distribution at Constant, Low and High Temperatures

What happens if temperature is not kept constant?

Ni/Nj = e -(Ei-Ej)/kT

Ni/Nj = e -Eij/kT

Eij 0

i.e. Ni/Nj = e -0 = 1 i.e. we get a uniform distribution of molecules over all

available energy levels (Fig. 2a)

If Eij >> kT (at low temperatures)

Then Ni/Nj = e - = 0 i.e. only the lowest energy level is occupied; others

are empty (Fig. 2b)

J = 0 J = 1 J = 2 J = 3

Ni

Suppose at high temperature, kT >> Eij

Fig. 2

Since we have several degenerate rotational levels, the equation is

modified to account for this:

Ni/Nj = gi/gj e –(Ei-Ej)/kT ……………………..where g = degeneracy and

g = (2J + 1) = No. of degenerate levels e.g.

EJ

J = 2

J = 1 three degenerate energy levels in J = 1

J = 0

Partition Function

If N = total No. of molecules in a sample:

Ni/N = e –Ei/kT/i e –Ei/kT

Define ie –Ei/kT = Z = Molecular Partition Function

q = measure of availability of states.

If more than one state with energy level Ei (degenerate rotation)

NJ = gi e –EJ

/kT/ J gi e –EJ

/kT

Z = J gi e –E

J/kT

or Z = J gi e –E

J ……………………………..where = 1/kT

Boltzmann Distribution Law.

Microstates and Configurations

Consider a simple assembly of three localized (i.e., distinguishable)

particles that share three identical quanta of energy. These quanta can

be distributed among the particles in ten possible distributions as

illustrated in the figure below.

Fig. 2a

EJ

Fig. 2b

Each of the ten distributions is called a microstate; in our example

the ten microstates fall into three groups, or configurations,

denoted A, B, and C. When dealing with only three particles, we can

count the number of microstates and configurations, but for larger

numbers of particles, we need to calculate them. Our simple

example above can be used to understand how to calculate the

number of microstates and configurations.

For configuration B: The two-quanta parcel of energy can be

assigned to any of the three particles, the one-quantum parcel to

either of the two remaining particles, and the zero-quantum parcel

to the remaining particle. Thus, the total number of ways in which

assignments can be made is

For configuration A: We again have three choices in assigning the

first (three-quanta) parcel of energy, two choices when we assign

the second (zero-quantum), and one choice when we assign the

third (zero-quantum). But because the last two parcels are the

same, the final distribution is independent of the order in which

they are assigned. That is,

distinguishable assignments collapse into one microstate because

the two particles end up in the same quantum level. Thus, the total

number of microstates associated with configuration A is

For configuration C: We have triple occupancy of the first quantum

level and the final distribution is independent of the order in which

the particles are assigned. Thus, we have one microstate.

This approach can be generalized to a larger number (N) of

localized particles. We have N choices of the particle to which we

assign the first parcel of energy, (N-1) choices in assigning the

second, etc., to give a total of

distinguishable possibilities if no two energy parcels are the same.

If some number (na) of the parcels are the same, we can have only

N!/na! distinct microstates. Symbolizing by W the total number of

microstates in a given configuration, we can write

where na represents the number of units occupying some quantum

level, nb represents the number of units occupying some other

quantum level, etc.

This equation can be written in the general form

For our simple example above, we can calculate the number of

microstates in configuration A as

For very large N, it is necessary to deal with natural logarithms.

That is,

We can now apply Stirling's approximation in the form

We can then approximate the expression for the number of

microstates as

Because , we can simplify to

Predominant Configuration

Note that in our simple example, one of the configurations (B) had

a greater number of microstates associated with it than the

others. As N increases, a predominant configuration becomes even

more apparent.

We of course want to be able to identify this predominant

configuration. For large N, it is not practical to do this by the

approach we have illustrated above. However, it can be shown that

the distribution of microstates for large N can be described by a

smooth curve, the sharpness of which increases as the number of

particles increases.

In the graph above, the configuration index number represents the

fraction describing the number of times a microstate

corresponding to a given configuration is observed divided by the

total number of microstates. The predominant configuration

corresponds to the very peak of the curve. That is,

where dX denotes a change from the predominant configuration to

another configuration only infinitesimally different from it. Using

this criterion, we can now develop a simple formula that describes

the predominant configuration.

The Boltzmann distribution law

Consider an isolated macroscopic assembly of N particles, identical

but distinguishable by spatial localization, which share a large number

of energy quanta, each of which suffices to promote a particle from

one quantum level to the next level above it. Because the assembly is

isolated, both N and the energy will remain constant. We now want to

determine for which of the enormous number of configurations that

can be assumed by the assembly will the number of associated

microstates realize its maximum value.

Consider any three successive quantum levels l, m, n, with

associated energies el, em,en and numbers of particles nl, nm, nn,

respectively. For any given configuration, the number of microstates

can be calculated from

Now make a change in the initial configuration such that one particle

is shifted from each of levels l and n into level m. This change

maintains a constant number of particles and energy but creates a

new configuration that differs from the first in that

Hence,

Because we want to calculate the properties of the predominant

configuration, let us assume that the original configuration was in

fact the predominant configuration for which the number of

associated microstates reaches its maximum value, where

Because the change we have made in the configuration is extremely

small, we would expect essentially no change in W. Therefore,

from which we can write

Canceling terms that are common to both sides of the equation and

inverting gives

Rearranging gives

Expanding some of the factorial terms gives

Cancellation gives

But numbers such as 1 and 2 are small in comparison to the large

populations being studied. Hence,

or

This relationship also holds for other successive energy levels. Thus,

for a macroscopic assembly of particles with uniform energy spacing

between their quantum states, we can describe the predominant

configuration as the one for which the following geometric series

applies:

A similar approach can be taken when considering a system in which

the energy spacing is not uniform. Without going through the

derivation, the appropriate equation is

The values of p and q are small, positive integers selected so that the

following equation holds:

Thus,

Since

Then,

Note that l, m, and n are any three quantum levels. Thus, for any two

quantum levels, the indicated function must have exactly the same

value. That is, this function is a constant; it is denoted by beta.

For any two quantum states, i and n, we can write

If i is taken to be the ground state, with population n0 and energy e0

= 0, this equation reduces to

or

This equation is the Boltzmann distribution law. It defines the

predominant configuration for an isolated macroscopic assembly of

identical but distinguishable particles, with any kind of energy

spacing between their quantum states. When a system is said to obey

a Boltzmann distribution, it will be consistent with the above

equation. At equilibrium, the configuration of an isolated macroscopic

assembly is typically that described by the Boltzmann distribution

law.

The Meaning of Beta

The constant beta has a significant physical meaning. We will not

show the derivation, but beta is inversely related to temperature.

That is,

where T is in Kelvin, k has unit of energy/K.

Degeneracy

The Boltzmann distribution law, as derived above, considers the

population of each distinct quantum state. Some of these quantum

states may have the same energy, in which case these states are said

to be degenerate. To account for degeneracy, one simply multiplies

the energy level by the number of quantum states that have that

energy. Degeneracy will be included explicitly in the derivation that

follows.

Derivation of Boltzmann Distribution Law

Consider now how an Avogadro's number (N) of molecules will

distribute themselves throughout their allowed energy levels.

where W is the probability of a given distribution. We want to find

the distribution such that W is a maximum. It is more convenient

mathematically to seek a maximum in ln W. Taking the natural

logarithm of the above equation, and applying Stirling's

approximation to the ni terms gives

In solving for a maximum in ln W, we must impose two conditions, a

constant number of molecules and a constant total energy of the

system. That is,

and

These conditions are imposed by applying Lagrange's method of

undetermined multipliers. Two parameters, a and b, are introduced.

Rather than looking for a maximum in ln W, we seek a maximum in

Thus, we can assure constant N and constant E, and the maximum we

find is still a maximum in ln W. To find the maximum with respect to

ni, we solve

Recall our previous expression for ln W. That is,

N is a constant that is independent of the individual ni's, so we can

write

Substituting, we obtain

Considering the relative populations of two energy levels, i and j, we

can write

This again is the Boltzmann distribution law, in this case with the

degeneracy (gi's) explicitly shown.

c. Partition function

The sum over states (Zustandsumme) is called the partition

function. In principle, it‟s a sum over all the particle states of a system,

and therefore contains the statistical information about the system. All

of the thermodynamic properties of the system are derivable from the

partition function.

i

kT

Ei

eZ

Remember, the partition function is written as a sum over all

microstates, rather than over energy macrostates, so that the

multiplicity factor does not appear explicitly. All of the macroscopic

thermodynamic quantities are obtainable from the microscopic statistics

of the single particle energy states. For instance, internal energy:

V

i

ii

i

kT

E

i

V

T

ZkTU

kT

ZU

NEkT

Z

Z

Ze

kT

E

T

Z i

ln2

2

2

2

Entropy:

NkT

U

N

ZNk

kNkT

E

Z

NNk

kNNNkkS

i

ii

i

ii

ln

ln

lnln

Pressure:

T

TT

V

ZNkTP

NkTN

ZNkT

VTSU

VdV

dST

dV

dUP

PdVdUTdS

ln

ln

2. Probability

a. Ensembles

microcanonical

In an isolated system, every microstate has equal probability of

being occupied. The total energy is fixed. The collection of all the

possible microstates of the system is called the microcanonical ensemble

of states.

canonical

If the probability distribution is the Boltzmann distribution, the

collection of energy states is called the canonical ensemble. We‟ve seen

that such a distribution applies to a system at constant temperature and

fixed number of particles, in contact with an energy reservoir. The

internal energy is not fixed, but we expect only small fluctuations from an

equilibrium value.

grand canonical

If the number of particles is allowed to change, then we have to sum

also over all possible numbers of particles, as well as all possible energy

states, giving the grand canonical ensemble.

b. Average values

The probability that a particle will be observed to occupy a particular

energy state is given by the Boltzmann distribution.

Z

eEP

kT

E

)(

The average value of the particle energy would be computed in the usual

way,

Z

eE

N

NEE i

kT

E

i

i

ii

i

.

The internal energy of the system of N particles is U = N<E>. We are

averaging over a collection of particles whose energies are distributed

according to the Boltzmann distribution.

3. A Couple of Applications

a. Equipartition Theorem

A particle‟s kinetic energy is proportional to the square of its

velocity components. In Cartesian coordinates, 222

zyx vvvK . In a

similar vein, the rotational kinetic energy of a rigid body is also

proportional to the square of the angular velocity, thus 2rotationK . In

the case of a harmonic oscillator, the potential energy is also

proportional to a square, namely the displacement components, 222 zyxU potential . Very often, we approximate the real force acting

on a particle with the linear restoring force of the harmonic oscillator.

Let us consider a generalized quadratic degree of freedom, 2cqqE .

Each value that q takes on represents a distinct particle state. The

energy is quantized, so the q-values are discrete, with spacing q .

The partition function for this “system” is a sum over those q-states

q

kT

cq

q

kT

qE

eeZ

2

.

In the classical limit q is small, and the sum goes over to an integral

kTdyekTdqeZ cykT

cq

00

2

2

The average energy is

kTkTkT

kT

Z

ZE

2

11

2

11

1

1 23

.

So, each quadratic degree of freedom, at equilibrium, will have the

same amount of energy. But, this equipartition of energy theorem is

valid only in the classical limit and high temperature limits. That is,

when the energy level spacing is small compared to kT. We saw

earlier that degrees of freedom can be “frozen out” as temperature

declines.

The Kinetic Theory of Gases

1. Sample of gas huge number of atoms (molecules).

2. Atoms are very small compared the space in which they travel.

3. Atoms have translational energy only (ignore rotational, vibrational or

other internal structure).

Molecular View of Pressure

Atoms have translational energy coming motion in three dimensions.

2 2 2 2

tr x y z

1 1m v v v mv

2 2

Assume gas is in a box with lengths, lx, ly, lz.

Assume gas is in thermodynamic equilibrium

- No net energy transfer between walls of box and gas atoms.

Consider the z-component of an atom‟s velocity as it hits the wall of the

box with lengths lx and ly.

1. Collision reverses vz.

2. vx and vy remain unchanged.

3. Overall speed and energy are unchanged. (elastic collisions)

Force is the time derivative of momentum.

zz z z

z z z

d mvdp dv dvdmF v m m ma

dt dt dt dt dt

Find the change in the momentum of the collision with the wall

z f i z z zp p p m v mv 2mv

For a single atom i, the time average of the force is:

z,i z,i z,i 2 1p F dt F t t

Assuming a constant speed, the time change can be related to the

distance travelled by the atom.

z2 1

z,i

2lt t

v

Consider the force on the wall using Newton‟s third law.

2

z,i z,i z,iz,iW

z,i z,i

2 1 2 1 zz z,i

2m v 2m v m vpF F

t t t t l2l v

Find the total force on the wall by summing over all of the atoms.

2

z,iW W W

z z,i z,i

i z

Nm vF F N F

l

Using the average force, calculate the pressure on the wall

perpendicular to the z-direction.

2 2 2W

z z,i z z,i z,i

z x y x y z

F Nm v l Nm v Nm vFp

A A l l l l l V

Consider that choosing z-direction to calculate the pressure is arbitrary.

1. Pressure is the same on any wall.

2. Average molecular velocity is the same in all directions.

x,i y,i z,iv v v 22 2 22

i x,i y,i z,i z,iv v v v 3 v 2

2 i

z,i

vv

3

2

iNm vp

3V

Pressure can be written in terms of the molecular translational energy.

1. For a single molecule, 2

tr i

1m v

2

2. For “N” molecules, 2 tr

tr tr i tr

2E1 2E N Nm v p pV E

2 3V 3

Each type of molecule has its own translational energy and therefore its

own pressure.

tr,a

a

2Ep

3V tr,b

b

2Ep

3V tr,c

c

2Ep

3V

tr,total tr,a tr,b tr,cE E E E

tr,a tr,b tr,c tra b c

2E 2E 2E 2Ep p p p

3V 3V 3V 3V

Molecular View of Temperature

Temperature is a measure of how energy is distributed in a material

sample.

1. Higher temperature means more modes of motion are available for

a molecule to move.

- Types of motion include translational, rotational, vibrational and

electronic.

2. Lowest temperature implies one mode of motion. (3rd Law of

Thermodynamics)

Thus temperature is a function of energy (microscopic or macroscopic).

1. For a gas without internal structure,

tr trT f or T g E

2. Note the temperature function is different for microscopic and

macroscopic energies.

We can define temperature macroscopically in terms of the ideal gas law.

tr tr

2 3pV nRT E E nRT

3 2

tr

E 3E E RT

n 2

trtr

A

E 3 nRT 3 RTE N

N 2 N 2 N

AN Avogadro 's number

The corresponding microscopic definition follows from a definition of

Boltzmann‟s constant, k.

23

23

A

R 8.314J mol Kk 1.381 10 J K

N 6.022 10 mol

1. R is used for macroscopic quantities (moles)

2. k is used for microscopic quantities (a molecule)

tr

A

3 RT 3kT

2 N 2

Velocity Distributions

RMS (root-mean-square) Velocity 1

2N 2i

i

vv

N

1 1

2 22

tr

3 1 3kT 3RTkT m v v

2 2 m

M

Maxwell (Maxwell-Boltzmann) Distribution Function

Introductory assumptions A gas sample has a broad distribution of speeds.

Let the number of molecules with speed between v and v + dv = dNv

The distribution function relates the fraction of total molecules within a

range of velocities between v and v + dv. (This function is what we want

to find!).

vdNG v dv

N

- Probability of finding molecules with velocity between v1 and v2

2

1

v

v

G v dv

- Probability of finding molecules with any speed is G v dv 1

- G v is a probability density.

To find the distribution function for any speed, G v , first find the

distribution function for velocities in z-direction only, zg v .

zv

z z

dNg v dv

N

One-dimensional velocity distribution function Because our sample is in isotropic space (no preferred direction exists),

the distribution functions for the velocities in the other directions are

similar.

Let us examine the product of probability densities in all three

directions.

x y zg v g v g v v

To find g, take derivative w.r.t. vx.

x y z

x

v vg v g v g v

v v

22 2 2 xx y z x x y y z z

x

vvv v v v 2vdv 2v dv 2v dv 2v dv

v v

xx y z

v vg v g v g v

v v

x y z xx x

x x xx x y z x x y z

g v g v g v v g v v vv v

v v g v v v v v vv g v g v g v v g v g v g v

Similar relationships can be found for the “y” and “z” directions.

y

y y

g v v

v vv g v

z

z z

g v v

v g v v v

yx z

x x z zy y

g vg v g vb

v g v v g vv g v

Since “b” does not depend on vx, vy or vz; it must be a constant.

x

x

x

g vbv

g v

2x

1bv

x 2 2x x x x x x

x

g v 1bv dv dv bv ln g v C g v Ce

g v 2

2 2 2 2 2 2 2

x y z x y z

1 1 1 1 1bv bv bv b v v v bv

3 32 2 2 2 2x y zv g v g v g v Ce Ce Ce C e C e

Evaluate “C” by letting the total probability equal one.

2 2x x

1 1bv bv

2 2x x x xg v dv Ce dv C e dv 1

A table of definite integrals would include: 2ax

0

1e dx

2 a

2x

1bv

2x

1 2 bC e dv C 2 1 C

2 b 2

when b

a2

2x

1bv

2x

bg v e

2

To evaluate “b”, recall that tr

3kT

2

2 2 2 2 2

tr x y z x

1 1 1m v m v v v m 3v

2 2 2

2 2

x x

3 3 kTm v kT v

2 2 m

Alternatively, calculate 2

xv using the probability density.

2xbv

2 2 2 2x x x x x x

bv v g v dv v e dv

2

From a table of integrals, 2

1

22n ax

12n 10 n2

2n !x e dx

2 n!a

To apply the integral, n = 1 and b

a2

,

2x

1bv 2

2 2 2x x x 3

23

b b 2! 1v v e dv 2

2 2 bb

2 1!2

Equate the two calculations of 2

xv .

2

x

1 kT mv b

b m kT

Thus the one-dimensional velocity distribution function for an ideal gas is:

2

tr ,xxmv

2kT kTx

m mg v e e

2 kT 2 kT

Three-dimensional velocity distribution function

The one-dimensional distribution function yields the probability of

finding an atom with a specific component of the velocity, vz, within a

differential range, vz and vz + dvz

To find the three-dimensional distribution function yielding a velocity, v,

within a range of v and v + dv, we need to recognize that an atom with

velocity, v, can travel in any direction. The velocity of a particle moving in

an arbitrary direction can be represented in a three-dimensional

“velocity” space.

An atom with velocity, v, could be anywhere on the sphere with radius, v.

The total probability density of an atom with a velocity between v and v +

dv lies between a sphere of radius, v, and a sphere of radius v + dv in

velocity space..

The volume of this shell is the difference between the volume of the two

spheres.

3 3

shell sphere sphere

2 33 2 3

2 32

2 2

4 4V V v dv V v v dv v

3 3

4 4 4 4 4v 3v dv 3v dv dv v

3 3 3 3 3

4 4 43v dv 3v dv dv

3 3 3

43v dv 4 v dv

3

The three-dimensional distribution function is the product of the one-

dimensional distribution functions and the volume of the shell in velocity

space.

2 2 vx y z

dNG v dv v 4 v dv g v g v g v 4 v dv

N

2 2 2x x xmv mv mv

2 22kT 2kT 2kTx y z

m m mg v g v g v 4 v dv e e e 4 v dv

2 kT 2 kT 2 kT

2 2

3 3mv v

2 22 22kT 2RT

mG v dv e 4 v dv e 4 v dv

2 kT 2 RT

MM

vy

vx

vz

v

}dv

The above equation is known as the Maxwell (Maxwell-Boltzmann)

distribution function.

Graphs of Maxwell-Boltzmann distribution function velocity

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

.55

0

f u 2 298( )

f u 2 398( )

f u 2 2000( )

120 u

- The average velocity of the gas molecules increases as the

temperature increases.

- At low velocities, the shape of the curve resembles a quadratic

function; whereas at moderate to high velocities, the shape of the

curve resembles a Gaussian function.

298 K

398 K

2000 K

0 1 2 3 40

1

2

3

4

4.5

0

f u 2 298( )

f u 44 298( )

f u 146 298( )

f u 146 2000( )

4.50 u

- The average velocity of the gas molecules decreases as the molar

mass increases.

Aside: The speed of sound in a gas medium is proportional to the

average velocity of the gas molecules. Thus, breathing He

makes your voice higher and breathing SF6 makes your voice

lower.

Other Gas Velocity Averages

Average speed

2 2

3 3mv mv

2 22 32kT 2kT

0 0 0

m mv vG v dv v e 4 v dv 4 e v dv

2 kT 2 kT

From a table of integrals: 22n 1 ax

n 10

n!x e dx

2a

23 3 1 1

mv2 2 2 2

32kT1 1

0

m m 1! 8kT 8RTv 4 e v dv 4

2 kT 2 kT mm2

2kT

M

Most probable speed

The most probable speed, vmp, is the speed with the highest probability;

that is, the speed where the velocity distribution is a maximum.

We can find vmp by taking the derivative of the Maxwell-Boltzmann

distribution function, setting it equal to zero and solving for v.

2

3mv

222kT

dG v d me 4 v dv 0

dv dv 2 kT

H2, 298 K

CO2, 298 K

SF6, 298

K

SF6, 2000

K

2 23

mv mv2

2 22kT 2kTm d d

4 e v dv 0 e v dv 02 kT dv dv

2 2 2 2 2mv mv mv mv mv32 22kT 2kT 2kT 2kT 2kT

d 2mv mve v dv e v e 2v 0 e 2ve 0

dv 2kT kT

1 12

2 22

mp

mv 2kT 2kT 2RT2 0 v v

kT m m

M

RMS speed (from above) 1 1

2 22

rms

3kT 3RTv v

m

M

Comparison of all three averages

mp rmsv v v

mp rmsv : v : v 2 : 8 : 3 1.414:1.596:1.732 1:1.128:1.225

We shall now work out population ratios for different kinds of

energy mode at room temperature, 25 oC (298.15 K). We calculate

population distributions for one mole. The denominator of exponent takes

the value R=NA k = 8.314 JK-1mol-1, and RT = 298.15·8.314 = 24789 Jmol-1

RT = 2.48 kJmol-1 (rounding).

Electronic level population For an electronic level difference Δε is quite high. If we take a

hydrogen atom in gas phase Δε = 1000 kJmol-1. The population of the

2s excited state n2s relative to ground state, n1s is

175403

1

2 1048.2

1000exp

e

n

n

s

s , so the number 2s H-

atoms

403

12

enn ss

so that we can be very confident that there are no 2s hydrogen atoms around at room temperature.

Vibrational level population

Next we consider vibrational modes. If we take CO molecules, the

vibrational quantum levels have an energy separation of around Δε=25

kJmol-1. Vibrational energy levels (v = 1, 2, 3…) are much more closely

spaced than electronic ones. Hence

510

0v

1v 10548.2

25exp

e

n

n

If we have a system containing 20000 molecules, than

110520000 5

1v

n

It means that one out of every 20000 molecules is found at the first

vibration excited state in a time average.

Rotational levels are much more closely spaced than vibrational

levels. For CO the rotational energy levels are around 0.05 kJmol-1

apart. We can write

94.2348.2

05.0exp3 02.0

0J

1J

en

n

The factor of 3 in the equation above requires some explanation. It

arises because the first excited level J=1 is actually threefold

degenerate (i.e. there are 3 levels all with the same energy). This

degeneracy factor must be taken into account when assessing the

probability of finding a molecule in a given level.

Example 3: Temperature dependence of population. A comparison of

population density between temperatures 25 and 1000 oC.

Calculate the ratio of molecules in excited rotation, vibration and

electronic energy level to that in the lowest energy level.

Data: T1 = 25 oC, T2 = 1000 oC

first excited energy levels are at 30 cm-1

rotation,

1000 cm-1

vibration,

40 000 cm-1

electronic,

above the lowest one.

Assume that, J = 4 for excited rotation level, and such a level is 2J+1

fold degenerate.

Vibration and electronic states have no degeneracies.

h = 6.626 10-34J/s, c = 3 1010 cm/s, k = 1.38 10-23 J/K, T1 = 298 K, T2 =

1273 K

Rotation

~chc

hh

8.7~

exp91

kT

hc

lowern

uppern and at T2

7.8~

exp2

kT

hc

lowern

uppern

In each set of ten molecules at T2 = 1000 oC only one is at the lowest

energy level.

Vibration

008.0~

exp1

kT

hc

lowern

uppern and at T2

323.0~

exp2

kT

hc

lowern

uppern

A 40-fold increment in population inversion is caused by the increase

in temperature.

Electronic

84

1

1024.1~

exp

kT

hc

lowern

uppern and at T2

20

2

1034.2~

exp

kT

hc

lowern

uppern

Practically, all the molecules are in ground electronic state at room

temperature.

2. Ideal Gas

a. One molecule

introttrans

nstatesinternal

kT

E

jstatesrotational

kT

E

istatesnaltranslatio

kT

E

i

kT

E

ZZZeeeeZnjii

,,,

1

b. Two or more molecules

If the molecules are not interacting, then as before, the partition

function for N molecules is just N

N ZZ 1 or !N

ZZ

N

iN depending on

whether the molecules are distinguishable or not. In the ideal gas, the

molecules are not distinguishable one from another. If the molecules

were in a solid, then they would be distinguishable because their positions

would be distinctly different.

[Note that the rotational partition function is lumped in with the internal

partition function.]

c. Internal partition function

The internal partition function sums over the internal vibrations of the

constituent atoms. We would usually approximate the energy levels by

harmonic oscillator energy levels.

i n

kT

nh i

eZ1

int

The index i labels the vibrational modes, while n labels the uniformly

spaced energy levels for each mode. For instance, the water molecule has

three intermolecular modes. A molecule having more atoms has more

modes. A diatomic molecule has just one mode of vibration.

d. Rotational partition function

A molecule is constrained to a particular shape (internal vibrational

motions apart), which we regard as rotating like a rigid body. The angular

momentum, and therefore the rotational kinetic energy, is quantized,

thusly I

jjE j2

12

, where I is the moment of inertia of the molecule

about the rotational axis. Classically, if is the angular velocity and L is

the magnitude of the angular momentum, then the kinetic energy of

rotation is

I

L

I

IIKErot

222

1 22

2

. Quantum mechanically, the angular

momentum is quantized, so that 22 1 jjL with j equaling an integer.

Now, this applies to each distinct axis of rotation. In three dimensions,

we start with three axes, but the symmetry of the molecule may reduce

that number. The water molecule has three axes, but a carbon monoxide

molecule has only one. Basically, we look for axes about which the

molecule has a different moment of inertia, I. But it goes beyond that.

If the symmetry of the molecule is such that we couldn‟t tell, so to speak,

whether the molecule was turning, then that axis does not count. That‟s

why there are no states for an axis that runs through the carbon and

oxygen atoms of carbon monoxide.

Therefore, a rotational partition function will look something like this for

three axes

3

1

1

i j

kT

jj

rot

i

eZ

.

e. Translational partition function

In an ideal gas, the molecules are not interacting with each other. So the

energy associated with the molecular center of mass is just the kinetic

energy, m

pmv

22

1 22 . The molecule is confined to a finite volume, V, so

that kinetic energy is quantized also.

First consider a molecule confined to a finite “box” of length Lx on the x-

axis. The wave function is limited to standing wave patterns of

wavelengths x

xn

n

L2 , where

nx = 1, 2, 3, 4, . . . This means that the x-component of the momentum is

limited to the discrete values x

x

n

nxL

hnhp

2,

. The allowed values of

kinetic energy follow as

2

22

8 x

xn

mL

nhE .

Naturally, the same argument holds for motion along the y- and z-axes.

x y z

z

z

y

y

x

x

n n n

kT

mL

n

mL

n

mL

nh

tr eZ

2

2

2

2

2

22

888

Unless the temperature is very low, or the volume V is very small, then

the spacing between energy levels is small and we can go over to integrals.

23

2

222

1

0

8

0

8

0

8

2

222

2

22

2

22

2

22

h

mkTV

Lh

mkTL

h

mkTL

h

mkT

e dnednedneZ

zyx

kT

z

mL

nh

y

mL

nh

x

mL

nh

trz

z

y

y

x

x

The quantity QvmkT

h

23

2

2 is the quantum volume of a single molecule.

It‟s a box whose side is proportional to the de Broglie wavelength of the

molecule. In terms of that, the Q

trv

VZ .

[Actually, in the classical form of the partition function, we are

integrating over the possible (continuous) values of particle momentum

and position.

mkTLdxdpeZ x

L

xmkT

p

xtr

x x

20 0

2,

2

The classical partition functions differ from the classical limit of the

quantum mechanical partition functions by factors of h. Because h is

constant, this makes no difference in the derivatives of the logarithm of

Z.]

Putting the parts all together, for a collection of N indistinguishable

molecules

Nrottr ZZZN

Z int!

1

3. Thermodynamic Properties of the Ideal Monatomic Gas

a. Helmholz free energy

Consider the derivative of the partition function with respect to

temperature.

2

2

111ln

kT

Ue

kT

E

ZT

e

eT

Z

ZZ

TkT

E

i

kT

E

kT

E

i

i

i

On the other hand, recall the definition of the Helmholtz free energy.

UT

FTF

T

FTUTSUF

V

V

UT

T

F

T

V

2

2T

U

T

T

F

V

Evidently, we can identify the Helmholtz free energy in terms of the

partition function thusly,

ZkTF ln .

For the monatomic ideal gas,

N

Qv

V

NZ

!

1, whence (using the Stirling

approximation)

VvNNkTNNNvNVNkTF QQ lnln1lnlnlnln .

b. Energy & heat capacity

NkTh

mk

h

mkT

h

mkTNkT

h

mkT

Th

mkTNkT

T

v

vNkT

T

vNkT

T

ZkTU

Q

QV

Q

V

2

322

2

32

22

1lnln

2

25

2

23

2

2

23

2

23

2

2

222

NkNkTTT

UCV

2

3

2

3

c. Pressure

Of course, we‟re going to get the ideal gas law.

V

NkT

V

VNkT

V

VvNNkT

V

FP

NTNT

Q

NT

,,,

lnlnln1ln

d. Entropy and chemical potential

NV

Q

NVT

VvNNkT

T

FS

,,

lnln1ln

T

vNkTVvNNkS

Q

Q

lnlnln1ln

2

3

2

23

2 2

2lnln1ln

h

mkT

TmkT

hNkTVvNNkS Q

2

5ln

2

3lnln1ln

2

2

3

2lnln1ln

2

22

2

Q

Q

Q

Nv

VNkS

NkVvNNkS

h

mk

mkT

hNkTVvNNkS

Q

Q

VT

Q

VT

Nv

VkT

N

NNkTVvNkT

N

VvNNkT

N

F

ln

lnlnln1ln

lnln1ln

,,

Rotation and vibration of diatomic and polyatomic molecules

For a diatomic molecule as well as the three translational degrees of

freedom there are also two rotational and one vibrational degree of

freedom:

z

x

y

NOTE - the molecule could equally well have been drawn along the z axis.

there is nothing special about the y axis.

As shown in the diagram there can be rotation of the whole molecule

about the x and z axes. The atomic masses are concentrated in a tiny

nucleus and the moment of inertia about the y axis is so small that for

the present we will simply state that there is no contribution to the

rotational kinetic energy from rotation about the y axis.

There is one vibrational degree of freedom which is along the y axis.

Degree of freedom - examples are motion in the x, y and z directions.

Motion in each is independent of motion in either of the other two. A

diatomic molecule is able to vibrate and this vibrational motion is

independent of translation in the x, y or z directions. It is another degree

of freedom. Electronic motion introduces other degrees of freedom.

Consider an atom. It can move in the x, y and z directions. It's motion in

any direction is independent of it's motion in the other two. It is said to

have three degrees of freedom.

Consider two atoms moving independently of each other. Between them

they have six degrees of freedom. N independent atoms have 3N degrees

of freedom.

What about a diatomic molecule? There are two atoms but they are not

independent of each other. It turns out that there are still 6 degrees of

freedom. Three of them are the translational of the centre of mass of

the molecule and of the other three, one is the vibration of the molecule

and two are the rotation.

A triatomic molecule has 9 degrees of freedom. Three of these are the

translation of the centre of mass. A linear triatomic has a further four

vibrational and two rotational degrees of freedom. A bent triatomic has

three vibrational and three rotational:

vibration rotation

+ -

+ + -

Rotation and vibration of a diatomic molecule.

Normal modes of vibration of bent triatomic molecules.

Normal modes of vibration of linear triatomic molecules.

Equipartition of energy "When a substance is in equilibrium there is an average energy of kT/2

per molecule of RT/2 per mole associated with each degree of freedom."

According to the equipartition principle the average energy in each

degree of freedom of a gas should be the same, namely RT/2 J.mol-1 .

This implies that for each degree of freedom the contribution to Cv - d

dTRT( / )2

- is constant at R/2 or R for vibration.

This is a good approximation at room temperature for translational and

rotational degrees of freedom but not for vibration. We will see why

later.

Experimental results show that the heat capacity varies with

temperature as follows.

Molar heat capacities of ideal gases in J.K-1 - hard spheres

3N - 3 Trans Rot Vib Cv (J.K-1)

Atom 0 3R/2 0 0 1.5R

Diatomic 3 3R/2 2R/2 2R/2 3.5R

Linear

triatomic

6 3R/2 2R/2 8R/2 6.5R

Bent triatomic 6 3R/2 3R/2 6R/2 6.0R

Examples of observed heat capacities for gases are given in the following

table.

Molar heat capacities of real gases in J.K-1

Temp

(C)

-100 0 100 400 600

Gas Ideal

gas

He, Ne, Ar,

Hg

12.5 12.5 12.5 12.5 12.5 12.5

H2 17.5 20.5 20.8 20.8 20.9 29.1

N2 20.7 20.7 20.8 22.2 22.7 29.1

O2 20.8 20.9 21.0 24.5 25.9 29.1

Cl2 --- 24.5 24.6 26.1 26.7 29.1

H2O (bent) --- --- 26.6 28.5 31.8 49.8

CO2 (linear) --- 28.2 32.1 41.2 45.6 54.0

Why do we have this variation? To understand this we have to see why at

low temperature some gases do not seem to take up as much energy as

would be expected. Qualitatively it is easy to see that if energy levels are

quantised the equpartition theory is on dangerous grounds.

Energy level spacings

How can the deviations of the heat capacities of gases from those

predicted by the equipartition theory be explained?

thermal

energy

rotational

energy

vibrational

energy

electronic

energy

room

temperature

translational

energy

The translational energy levels are so close together that they can be

regarded as being continuous.

The above diagram shows energy levels in a hypothetical molecule. The

example does not correspond to any particular molecule and is purely for

the purpose of showing the participation of different degrees of freedom

in determining the heat capacity of a gas.

The average energy at room temperature is marked. The following should

be noted :

translational energy levels are extremely close together and many are

populated.

rotational spacing is quite small and a few levels will be populated.

vibrational spacing is larger and in this example only the ground and

first excited level will be significantly populated.

the first excited electronic state is very high and will not participate.

Perhaps you can see that as the temperature is increased energy can go

into translational levels without restriction ie. as T is increased the

translational energy will increase as given by 3/2RT.

However energy cannot go freely into electronic energy. In fact

essentially no collisions will be energetic enough to give electronic

excitation and this degree of freedom will contribute nothing to the heat

capacity.

For the inert gases there are no vibrational or rotational degrees of

freedom and the first electronic level is much higher than thermal

energy. The heat capacity is exactly what one would expect for a particle

with only translational degrees of freedom

For many diatomic and polyatomic gases at room temperature energy does

not freely go into vibration. For example all collisions in the range below

the first vibrational level will not give any vibrational excitation and only a

fraction of the rest will. Hence the heat capacity is lower than predicted.

Population of energy levels The following equation from statistical mechanics :

n

N

e

e

ikT

kT

i

i

i

ni = no. of particles in energy level i ; N = total

no. of particles;

tells us that if the temperature is low only the lowest energy levels are

significantly populated. At room temperature k.T is about 0.6 x 10-20 J

which is insufficient to give much vibrational excitation of common

diatomic molecules.

The denominator in this equation is a very important quantity called the

partition function (q) which you will learn more about in Statistical

Mechanics.

Example 4:

A hypothetical molecule has energy levels equally spaced by 4.14 x 10-21 J.

What is the population of the energy levels at temperature 100K, 300K,

428.6K, 1,000K?

4.14

8.28

12.4

E x 10-21

T (K) 100 300 428.6 1,000

q 1.05 1.58 1.99 3.86

1/kT 7.25 x

1020

2.42 x

1020

1.69 x 1020 0.725 x

1020

The length of the horizontal bar represents the population of the state.

It can be seen that as the temperature is increased higher energy levels

are populated as might be expected. Note that the lowest level is always

the most populated.

Note that the equation given above for the population of energy states is

the Boltzmann distribution for distinguishable particles. The following

assumptions were made :

particles are distinguishable

any number of particles can be in the same energy level.

The distribution is different where these two assumptions are not

applicable, eg. for Fermi particles (such as an electron) and Bose particles

(such as a photon).

Example 5:

A hypothetical gas has molecules with only one excited energy level at

0.2ev. Calculate the molar heat capacity of the gas at 0K, 1,000K and

5,000K. Make a qualitative sketch of CV as a function of T and explain it's

form.

1eV = 1.6 x 10-19 J

R = 8.314 J.K-1.mol-1

k = 1.38 x 10-23 J.K-1

NA = 6x1023 molecules per mol

ANSWER Molecule with only one excited level

The fraction F of molecules in the excited level is given by :

kT10x6.1x2.0

kT10x6.1x2.0

19

19

e1

eF

Now 0.2x1.6x10-19/k = 2319

Fe

e

Y

Y

T

T

2319

23191

1

The energy of 1 mole of the gas = E = F x 0.2 x 1.6 x 10-19 x 6 x 1023

J.mol-1 = F x 1.92 x 104 J.mol-1

Therefore CV can be found from :

C E TV V ( / )

by differentiation or computation E at appropriate intervals.

Differentiation Using :

dT

dv

v

u

dT

du

v

1

v

u

dT

d2

and dT

dwee

dT

d ww

22

2

Y1

1

dT

dY

Y1

Y

Y1

1

dT

dY

dT

)Y1(d

Y1

Y

dT

dY

Y1

1

Y1

Y

dT

d

dT

dF

Y e

dY

dT Te

T

T

2319

2

23192319.

.e1

e

T

2319

dT

dF2T2319

T2319

2

CdF

dTx x J K molV 192 104 1 1. . .

When T is very small CV 0

When T = 1,000K

R437.0JK631.310x92.1xdT

dFC

10x1891.009837.01

09837.010x319.2

dT

dF

14

V

3

2

3

When T = 5,000K

R051.0K.J4221.010x92.1xdT

dFC

10x98.216289.01

6289.010x76.92

dT

dF

14

V

6

2

6

Example 6:

A system of five particles each having one degree of freedom and energy

levels equally spaced at 10-20 J has a total energy of 4 x 10-20 J. Calculate

the statistical population of the energy levels.

N( )

0 1 2 3 4 5

1

5

N( )

0 1 2 3 4 5

1

5

N( )

0 1 2 3 4 5

1

5

N( )

0 1 2 3 4 5

1

5

N( )

0 1 2 3 4 5

1

5

5 ways 20 ways 10 ways

30 ways 5 ways

5 4 3 2 1 0 Divide by

70

5 4 3 2 1 0

0 5 20 50 100 175 0 0.0

7

0.2

9

0.71 1.43 2.5

0

1

2

3

0 2 4 6

eN

(e)

av

era

ge

From the Boltzmann equation we have:

K294,1T

kT

1

10

56.0

10.175

100lne17510010x1When

e175ngsinU

Cen

20

2010.20

i

i

i

20

i

i

Using the relation Eav = kT:

K159,1T

kTJ10x8.0E 20

av

How much energy? How fast do the molecules/atoms move?

Translation

Thermodynamics, Maxwell-Boltzmann, equipartition theorem, translational

kinetic energy of gas = (3/2) k T per molecule.

At room temperature, T = 300 K, Etrans = 6.21 x10-21 J molecule-1 = m v2 / 2

For He atom, m = 4 amu, v = 1.36 x103 ms-1 (~3000 mph!!!)

For C6H6 m = 78 amu v = 3.05 x102 ms-1

This is the velocity on average, not of any particular molecule.

Rotation

Similarly the equipartition theorem states that the average rotation

energy of a diatomic molecule is (1/2) k T per molecule per rotational

direction.

At T = 300 K, Erot,x = μ Re2 ωx

2 /2 = (1/2) k T = 2.07 x10-21 J

molecule-1

for H – 35Cl: Re = 1.2746x10-10 m (spectroscopic data)

μ = mH mCl / (mH + mCl) ~ (1 x 35)/(1+35) amu ~

1.614x10-27 kg

ωx ~ 6.288x1012 radians s-1 = 1.000x1012 rev s-1

a revolution every 1.0 ps on average.

Vibration

If we assume a vibrational energy of 10-19 J molecule-1 for NO which has a

force constant k = 1.54x103 Nm-1 and Re = 1.1508x10-10 m, can estimate

(classically) the extent to which the bond stretches. At maximum (or

minimum) R all the energy is Potential Energy (the atoms come to rest at

the limits of the vibration, no Kinetic Energy) therefore (k/2)(R – Re)2 =

10-19 J, from which (R – Re) = 1.14x10-11 m about 10% of Re When R = Re all

the energy is kinetic energy, therefore μ vR2/2 = 10-19 J , from which vR =

4016 ms-1 and from the classical equations of motion, estimate that a

vibration period is 2.26x10-13 s or the vibrational frequency is 4.40x1014 s-

1.Quantum Mechanical Molecular Energy of the Nuclei. The motion of

nuclei in molecules does not obey classical mechanics. The energies of

translational, rotational and vibrations motion of nuclei obey Quantum

Mechanics. In all three cases quantum mechanics finds the energy to be

quantized. Only certain values of energy are allowed. The equations for

the energy involve quantum numbers (1 quantum number for every

independent coordinate).

Translation. The „Particle in the box‟ problem. The energy of motion of a

particle of mass m in 1 D (say along x) and restricted to a path length of L

(e.g. a molecule contained in a box; an electron traveling along a copper

wire) is found to be: En = n2 h 2 / (8mL2) Joules n = 1, 2, 3, … all

positive integer values >0, h = 6.626x10-34 Js (Planck‟s constant), m (mass)

, kg L length of „box‟, m, Units (Js)2/(kg m2) J

In 3 D, a box of lengths A, B, C in the x, y, z directions, the energy is

specified by three quantum numbers.

En,p,q = n2 h 2 /(8m A2) + p2 h 2 /(8m B2) + q2 h 2 /(8m C2) Joules

n = 1, 2, 3, …; p = 1, 2, 3, …; q = 1, 2, 3, …

The sum of the three 1 D energies.

The 3D case demonstrates „degeneracy‟ if the „box‟ is a cube, A = B = C:

En,p,q = (n2+ p2 + q2 )h 2 /(8m A2) J

n = 1, 2, 3, …; p = 1, 2, 3, …; q = 1, 2, 3, …

n = 1, p = 2, q = 3 has energy E1,2,3 = (1+ 4 + 9 )h 2 /(8m A2) = 14 h 2 /(8m A2)

J as do: E1,3,2 E2,1,3 E2,3,1 E3,1,2 E3,2,1

6 different energy levels with the same energy – 6 degenerate levels.

Consider a molecule in a (1D) room: O2 (g) , m = 32 amu = 5.314x 10-26 kg, L

= 10 m

En = 1.03283x10-44 n2 Joules

Points to note about the „Particle in the Box‟ problem:

1) n ≠ 0. The particle cannot have zero translational energy. This has

to do with the Heisenberg uncertainty principle.

2) Energy levels ~ n2, get further apart with increasing n.

3) 1D box is non-degenerate (no 2 energy levels have the same

energy), 3D box has degenerate levels under special circumstances.

Quantized Rotational Energy for a diatomic molecule

and AmThe energy of two masses, ‟ problem.Rigid RotatorRotation. The „

mB at a fixed distance Re rotating about their centre of mass, is found to

be: Erot = (ħ 2 / 2 I) J (J + 1) Joules J = 0, 1, 2, 3 …, I = μ Re2

momentum of inertia, kg m2 ħ = h / 2π Js

μ = mA mB / (mA + mB) reduced mass, kg

Each energy level is degenerate. There are (2 J + 1) levels with the same

energy. There is a second quantum number mJ which has integer values

between – J and + J , that is (2 J + 1) values all with the same energy

determined by J. 2 quantum numbers as there are 2 separate rotational

axes/motions.

For H – 35Cl (data given previously)

Erot = (ħ 2 / 2 I) J (J + 1) = 2.12x10-22 J (J + 1) Joules

In spectroscopy of diatomic molecules, it is usual to write

Erot = B J (J + 1)

B = (ħ 2 / 2 I) = (ħ 2 / 2 μ Re2) Joules is called the rotational

constant and in spectroscopy it is usual to give the value in wavenumbers,

cm-1, E (J) = h c E(cm-1) , c velocity of light, cm s-1 !! For B = 2.12x10-22 ≡

10.7 cm-1

Points to note:

1) Energy levels ~(J2 + J),

get further apart with

increasing J

2) J = 0 is allowed.

3) Degeneracy (2 J + 1)

4) Excellent agreement

with spectroscopy in the microwave region.

Specific heat of ideal gases and the equipartition theorem

Specific heats revisited

The specific heat of a material will be different depending on whether

the measurement is made at constant volume or constant pressure.

J = 0, E0 = 0B, (2J+1) = 1 level

J = 2, E2 = 6B, (2J+1) = 5 levels

J = 3, E3 = 12B, (2J+1) = 7 levels

J = 4, E4 = 20B, (2J+1) = 9 levels

J = 1, E1 = 2B, (2J+1) = 3 levels

E r qu. number energy degeneracy

Molar specific heat at constant volume

cV (1/n) dQ/dT|V

But if dV = 0 then dW = 0 and by the first law of thermodynamics,

dEint = dQ ,

cV = (1/n) Eint/T |V true for all materials.

Specializing to ideal gases, we know Eint(T)

cV = (1/n) dEint/dT or

dEint = ncVdT true for all ideal gases.

Molar specific heat at constant pressure

cP (1/n) dQ/dT|P

and using the first law,

cP = (1/n) [dEint+dW]/dT|P = (1/n)dEint/dT|P + (1/n)dW/dT|P

For an ideal gas, Eint(T) and

dW|P = d(PV)|P = P dV|P = P d(nRT/P)|P = nRdT

cP = cV + R true for all ideal gases.

Equipartition Theorem

Each degree of freedom of a system has an energy ½ kBT.

A degree of freedom is an independent mode of motion: translation,

rotation, and vibration. Let the number of degrees of freedom be

denoted f. Neglecting vibration,

for ideal gases:

N=nNA and R=NAkB

1/ndE/dT cV+R

System f Eint(T) cV cP

m.i.g. 3N 3N ½ kBT 3/2 R 5/2 R

(trans.) 3/2 nRT

d.i.g. 5N 5N ½ kBT 5/2 R 7/2 R

dumbbell(trans.+rot.) 5/2 nRT

p.i.g. 6N 6N ½ kBT 6/2 R 4 R

(trans.+rot.) 6/2 nRT

Ideal Gas

Ideal gas law PV = nRT

First Law of Thermo. dEint = dQin - dWout

Processes isothermal, adiabatic, etc.

Work by gas Wout = ∫p dV

Internal energy of gas Eint(T) = f ½ kBT

An application of the Internal Energy expression.

The Equipartition Theorem and Heat Capacities.

The Equipartition Theorem states: That when quantum effects can be

ignored, the average energy of each „squared term‟ of molecular energy is

½ kT .

translational energy:

½ mvx2 + ½ mvy

2 +½ mvz2 3/2 kT

rotational energy:

½ IA ωA2 + ½ IB ωB

2 + ½ IC ωC2 3/2 kT

(kT for a linear as only 2 rotational axes)

vibrational energy:

½ μ v2 + ½ kx2 kT per vibrational mode

This means that at a temperature T there is (on average) as much energy

in translation as there is in rotation (can seldom ignore the quantum

effects in vibrational motion).

average translational energy per molecule, etc.

Now consider each „bit‟ of the molecular energy.

Average translational energy:

Average rotational energy, linear molecules:

For a non-linear molecule:

kT

q

z

BkTz

r

r

r

r

1

2

1

1

11

/

kTr2

3

evrt

e

e

v

v

r

r

t

t

evtrt

evrt

evtrt

N

z

z

Nz

z

N

z

z

Nz

z

N

zzzz

zzzz

NU

zzzzz

z

z

NZ

ZU

)()(

)()(

)(

1

Average vibrational energy:

now let T ∞, so that quantum effects can be ignored:

hυβ 0

kT per vibration, but only in extreme conditions (very high temperature).

At reasonable temperatures:

<ε t > = 3/2 kT

<ε r > = kT (linear), 3/2 kT (non-linear)

<ε v > 3/2 kT per vibration

if T large (or υ small)

Can use these results to discuss the dependence of Heat Capacities on

Temperature.

As:

h

h

hhh

hh

v

v

v

h

hv

e

eh

eehe

ee

z

z

ee

z

1

111

11

1

11

1

2

1

1

kTh

h

hh

h

hh

e

ehh

h

v

11)1(

)1(1

)1(

1

V

VT

UC

can separate components of CV

Electronic Partition Function

In nearly all cases electronic energies are very large and all terms except

the first are ~ 0

Justifying the value of β

Consider a very simple system and calculate its Internal Energy.

System with equally spaced energy levels:

kN

T

kTN

TNC

CCCCC

V

t

V

e

V

v

V

r

V

t

VV

t

2/3

2/3

ee

ej

egegg

egzenergieselectronicj

je

,2,1

,

210

,

0gze

1

1

32

)(~

1

)1(

1

),2,1,0(,

EZ

Eebut

e

eee

eeZ

nnEE

E

E

EEE

n

nE

n

E

n

n

which is a very simple result for a very simple system.

Now calculate the Internal Energy:

another simple result.

U is a state function of the system; (N, T, V ) fixed, U fixed.

To change U have to change (N, T, V ) to new, fixed values.

But: d U = T dS + p dV and dV = 0 as V is fixed (not variable).

Change U of the system by changing T .

Therefore: U ~ T ,

β -1 ~ T ; β ~ T -1

β = ( kT ) -1

Problems of Ch. (5)

1. Consider a mixture of Hydrogen and Helium at T=300 K. Find the

speed at which the Maxwell distributions for these gases have the

same value.

12

1

1

1

)()(

1

)(

)(

11

EE

E

E

Z

ZU

Tk

vmv

Tk

m

Tk

vmv

Tk

m

BBBB 2exp4

22exp4

2

2

22

2/3

2

2

12

2/3

1

Tk

vmm

Tk

vmm

BB 2ln

2

3

2ln

2

3 2

22

2

11

km/s 6.1107.12

2ln3001038.13ln3

2ln

2

327

23

21

2

1

21

2

2

1

mm

m

mTk

vmmTk

v

m

mB

B

Tk

mvv

Tk

mmTvP

BB 2exp4

2,,

22

2/3

2. Consider an ideal gas of atoms with mass m at temperature T.

(a) Using the Maxwell-Boltzmann distribution for the speed v, find the

corresponding distribution for the kinetic energy (don‟t forget to

transform dv into d). (b) Find the most probable value of the kinetic energy.

(c) Does this value of energy correspond to the most probable value of

speed? Explain.

3. Using the Maxwell-Boltzman function, calculate the fraction of Argon gas

molecules with a speed of 305 m/s at 500 K.[ 0.00141].

4. If the system in problem 1 has 0.46 moles of Argon gas, how many molecules have

the spped of 305 m/s?

5. Calculate the values of vmp, vavg, and vrms for Xenon gas at 298 K. .[ vmp = 194.27

m/s, vavg =219.21 m/s, vrms= 237.93 m/s]. ,

6. From the values calculated above, label the Boltzmann distribution Plot with the

approximate locations of vmp, vavg, and vrms.

7. What will have a larger speed distribution, Helium at 500 K or Argon at 300 K?

Helium at 300 K or Argon at 500 K? Argon at 400 K or Argon at 1000 K?.

8. Consider an assembly of N = 105 particles distributed over the five available

energy levels, j = 0, 1, 2, 3, and 4 shown in the diagram. The energy of each level (in

units of Joules) is given by j = j (100 K) kB, and the degeneracy of each level is gj

= 108. What is the population distribution, the total assembly energy E (J), and the

assembly entropy S (J/K) at a temperature of 100 K? .

dvTk

mvv

Tk

mNdvvNPvdN

BB

2exp4

2

22

2/3

d

mTkmTk

mNd

mdv

mvd

d

dvNPdN

BB

2

2

1exp

8

2

2

2

122/3

TkmTk

mP

BB

exp

24

2 2/3

2/12/3

0

1expexp

2

10 2/12/1

TkTkTk

P

BBB

m

Tkv

dv

vdP B

vv

20 max

max

9. For an ideal gas of classical non-interacting atoms in thermal equilibrium, the Cartesian components of the velocity of are statistically independent. In three dimensions, the probability density distribution for the velocity is

2 2 2

3/22

, , 22 exp

2

x y z

x y z

v v vP v v v

Where 2 kT

m . The energy of a given atom is 21

2E m v .

Find and sketch the probability density distribution for the energy of an atom in (i) three dimension, (ii) two dimension and (iii) one dimension.

10. Given a gas of particles with a Maxwellian distribution for velocity:

(a) What is the most probable velocity?

(b) What is the average velocity?

(c ) What is the root-mean-square speed?

(d) What is the most probable speed?

(e) What is the average speed?

(f) What is the average energy of a particle?.

11. Compute the mean energy of the canonical distribution of mean energy

The system in the representatives statistical ensemble are distributed over their accessible

states in according with the canonical distribution

r

E

E

rr

r

e

eP

the mean energy given by

r

E

r

r

E

r

r

e

Ee

E

where

r

E

r

EZee rr

where Z=

r

Ere

the quantity Z defined the sum over state or partition function

Z

ZE

1=

Zln

12. Using the canonical distribution compute the dispersion of the energy

The canonical distribution implies a distribution of systems over possible energies; the

resulting dispersion of the energy is also readily computed

2222

EEEEE

here

r

E

r

r

E

r

r

e

Ee

E

2

2

but 2

r

EEe r =

r

E

r

r

E rr eEe

2

then

2

22 1

Z

ZE

using the following

Z

Z

12

2

1

Z

Z+

2

21

Z

Z

-

E

2E - 2E

EE

2=

2

2 ln

Z

10. The internal energy of the ideal gas is given by E=E(T) show that for the ideal gas its

internal energy does not depend on its volume

Answer

Let E=E(T,V)

Then we can write mathematically

dVV

EdT

T

EdE

Tv

from the first law

dWdEdQTdS

dVV

vRdE

TdS

1using the above equation for dE

dVV

vR

V

E

TdT

T

E

TdS

TV

11

the entropy as a function of T and V

S=S(T,V)

dVV

SdT

T

SdS

TV

Comparing the equation

VV T

E

TT

S

1

V

vR

V

E

TV

S

TT

1 with the second order differential equation

VT

S

TV

S

22

TV

E

TT

E

TVT

S

V VTVT

211

V

vR

V

E

TTV

S

T TVTV

1

=

VT

E

TV

E

T

2

2

11

comparing the two equation

VT

E

TV

E

T

2

2

11=

TV

E

T

21 the right and the left equations are equal when

01

2

V

E

T which implies E is independent of V