Ideal Gas LawIdeal Gas Law

Deduced from Combination of Gas Relationships:

V 1/P, Boyle's LawV 1/P, Boyle s Law

V , Charles's Law

V n, Avogadro's Law

Therefore, V nT/P or PV nT

PV = nRTh R i l t twhere R = universal gas constant

The empirical Equation of State for an Ideal Gas

Boyle’s Law (experimental)Boyle s Law (experimental)

PV (h b l )PV = constant (hyperbolae)

T4

Is T > or < T ??

Temp T

Is T4 > or < T1 ??

Temp T1

Ideal Gas Equation of StateIdeal Gas Equation of State

Ideal Gas LawIdeal Gas Law

PV = nRTwhere R = universal gas constant

R = PV/nT

R = 0 0821 atm L mol–1 K–1

R = 8 314 J mol–1 K–1 (SI unit)

R 0.0821 atm L mol KR = 0.0821 atm dm3 mol–1 K–1

R = 8.314 J mol 1 K 1 (SI unit)

Standard molar volume = 22.4 L mol–1 at 0°C and 1 atm

Real gases approach ideal gas behavior at low P & high T

General Principle!!General Principle!!

E i di t ib t d iblEnergy is distributed among accessibleconfigurations in a random process.

The ergodic hypothesis

Consider fixed total energy with multiple particles and various possible energies for p p gthe particles.

Determine the distribution that occupies h l i f h il bl “Ph the largest portion of the available “Phase

Space.” That is the observed distribution.

Energy Randomness is the basis* of an exponential distribution of of an exponential distribution of

occupied energy levelsn(E) A exp[-E/<E>]

Average Energy <E> ~ kBTr g En rgy E B

n(E) A exp[-E/kBT]

This energy distribution is known as the Boltzmann Distribution.

* Shown later when we study statistical mechanics.

Maxwell Speed Distribution LawMaxwell Speed Distribution Law

2B

3 2mu 2k T21 dN m4

Bmu 2k T2

B

4 u eN du 2 k T

1 dN is the fraction of molecules per unit speed intervalN duN du

Maxwell Speed Distribution LawMaxwell Speed Distribution Law

M t b bl d 2dN kTMost probable speed, ump20 for u = u mp mp

dN kTudu m

8kT

Average speed, <u> or ū 0

8kTu u N u dum

Mean squared speed, <u2> 2 2

0

3kTu u N u dum

0

Root mean square speed 2 3rms

kTu u q p rms m

Distinguish betweenDistinguish betweenSystem & Surroundings

Internal EnergyInternal Energy

Internal Energy (U) is the sum of all potential and kinetic energy for all

U is a state function

potential and kinetic energy for all particles in a system

U is a state functionDepends only on current state, not on path

U = Ufinal - Uinitial

Internal EnergyInternal Energy,Heat, and Work

If heat (q) is absorbed by the system,and work (w) is done on the system,the increase in internal energy (U) is

i n b :

U = q (heat absorbed by the system)

given by:

U = q (heat absorbed by the system)+ w (work done on the system)

Reversible and Irreversible WorkW

Internal Energy (2)Internal Energy (2)U is a state functionU is a state function

It depends only on state, not on path to get thereU = Ufi l - Ui iti lU = Ufinal Uinitial

This means mathematically that dU is anf

exact differential: i

U dU

For now consider a system of constant compositionFor now, consider a system of constant composition.U can then be regarded as a function of V, T and P.Because there is an equation of state relating them,q gany two are sufficient to characterize U.So we could have U(P,V), U(P,T) or U(V,T).

Internal Energy (5)Internal Energy (5)Many useful general relationships are derived fromMany useful, general relationships are derived frommanipulations of partial derivatives, but I will(mercifully) spare you more.( f y) p ySuffice it to say that U is best used for processestaking place at constant volume, with only PV work:Then dU = dqV and U= U2 – U1 = qV

The increase in internal energy of a system in a rigidgy y gcontainer is thus equal to the heat qV supplied to it.

We would prefer a different state function forconstant pressure processes: enthalpy.

Enthalpy DefinedEnthalpy DefinedEnthalpy, H U + PV

At Constant P,

H = U + PV

U = q + w

H = U + PV

q= qP = U - w, w = -PV

q = U + PV= HqP = U + PV= H

At constant V, q = U = H

Comparing H and UComparing H and Uat constant PH = U + PV

1. Reactions that do not involve gasesV 0 and H U

2. Reactions in which ngas = 0V 0 d H UV 0 and H U

3 Reactions in which n 0 3. Reactions in which ngas 0 V 0 and H U

Heat Capacity atHeat Capacity atConstant Volume or Pressure

CV = dqV/dT = (U/T)V

P ti l d i ti f i t l Partial derivative of internal energy with respect to T at constant V

CP = dqP/dT = (H/T)P

Partial derivative of enthalpy Partial derivative of enthalpy with respect to T at constant P

Ideal Gas: CP = CV + nR

Heat Capacity, CHeat Capacity, C

Heat Capacity (J K-1 )Heat Capacity (J K )Heat needed to raise T of system by 1 K

q = CTq

Specific Heat Capacity (J K-1 kg-1 ) Heat needed to raise T of 1 kg by 1 K

q = CSmT

Molar Heat Capacity (J K-1 mol-1 )Heat needed to raise T of 1 mole by 1 KHeat needed to raise T of 1 mole by 1 K

q = CmnT

Endothermic & Exothermic Processes

H P

H = Hfinal - Hinitial

H Positive Positive amount of heat absorbed by the system

H N ti

Endothermic Process

H Negative Negative amount of heat absorbed(i e heat released by the system)(i.e. heat released by the system)

Exothermic Process

Th h i l E tiThermochemical Equations

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

(a combustion reaction) cH = – 890 kJ

R ti t b b l d

Phases must be specified

Reaction must be balanced

H is an extensive property

Sign of H changes when reaction is reversed

Standard StateStandard State

The Standard State of an element is defined to The Standard State of an element is defined to be the form in which it is most stable at 25 °C and 1 bar pressureand 1 bar pressureSome Standard States of elements:

Hg (l) O2 (g) Cl2 (g) Ag (s) C (graphite)

The standard enthalpy of formation (fH°)of an element in its standard state is of an element in its standard state is defined to be zero.

Enthalpies of FormationEnthalpies of Formation

The standard enthalpy of formation ( H°)The standard enthalpy of formation ( f H )of a compound is the enthalpy change for the formation of one mole of compound from the f f f p felements in their standard state.

Designated by superscript o: H°Designated by superscript o: H

For example, CO2:

C (graphite) + O2 (g) CO2 (g)

H° = 393 5 kJ/mol Appendix DrxnH = -393.5 kJ/mol Appendix D

f H° CO2 (g) = -393.5 kJ/mol

E h l i f R iEnthalpies of Reaction

The enthalpy of reaction can be calculated f th th l i f f tifrom the enthalpies of formationof the reactants and products

rxnH° = fH°(Products) rxn f ( )

- fH°(Reactants)

Example: Find rxn H°(usin Standard Enthalpies f F rmati n)(using Standard Enthalpies of Formation)

CH (g) + 2 O (g) CO (g) + 2 H O (l)CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

f H° (from Appendix D, text):CH4 (g) -74.6 kJ/molO2 (g) 02 (g)CO2 (g) -393.5H2O (l) -285 8H2O (l) 285.8

rxnH° = -393.5 -2 (285.8) – 0 – (-74.6) kJ/mol

Therefore, rxnH° = -890.5 kJ/mol

S d LSecond Law

Statement of law

Spontaneity (system vs. surroundings)

Entropy definedEntropy defined

Applied to cyclic processesCarnot engine – valid for all reversible engines

Steps of Carnot CycleSteps of Carnot Cycleq=0

qh absorbed (+) from hot reservoir Th

d b ( )

w done on system (+)

w done by system (–)

q=0w done by system (–)

q discarded (–) into coldqc discarded ( ) into cold reservoir Tcw done on system (+)

Summary of 4 Steps in Carnot CycleStep U qrev wrev S

AB 0RThlnVB/VA

+1678 J–RThlnVB/VA

– 1678 JRlnVB/VA4.19 J/K

BCCv(Tc-Th)–1247 J

0Cv(Tc-Th) –1247 J

0

CD 0 RTclnVD/VC

–1259 J–RTclnVD/VC

+1259 JRlnVD/VC–4.19 J/K

DACv(Th-Tc) +1247 J

0Cv(Th-Tc) +1247 J

0

Net 0R(Th-Tc)lnVB/VA

+419 J–R(Th-Tc)lnVB/VA

– 419 J0

N t k Net work =work done by work done by

system –k d work done on systemsystem

In example above, net work = = 1678 J +1247 J -1259 J – 1247 J 419 J= 419 J

What is the efficiency?

Efficiency of Carnot EngineEfficiency of Carnot Engine

Efficiency, , of Engine:

= Work PerformedHeat AbsorbedHeat Absorbed

/ = w/qh

= Th - Tch c

Th

Entropy change in a general ith id l process with an ideal gas

V1, T1 V2, T21, 1 2, 2Make a two component reversible path:Isothermal (expan or compr) from V1, T1 to V2, T1Isochoric heating or cooling to V2 T2Isochoric heating or cooling to V2, T2

2 2ln lnV TS S S nR C

1 21 1

ln lnVS S S nR CV T

P

PPath 1

Path 2

P1

V

Path 2P2

V2V1

Exam 1Exam

~ 5 problems (weights given) – budget your time Closed book Don’t memorize formulas/constants Don t memorize formulas/constants

You will be given things you needExam will not be heavily numeric, but will emphasize concepts

If it seems lengthy, do another problem & come back laterIf t seems lengthy, do another problem & come back later Understanding homework will be useful