The Distribution of Energy

Consider 6 particles – Each particle has three energy levels

E3 = 2a

E2 = a

E1 = 0

a

a

ELet N = Total number of particles = 6

N1 = Number of particles in state E1

N2 = Number of particles in state E2

N2 = Number of particles in state E2

Total energy of the system = 4a

N = N1 + N2 + N3 = 6

N1E1 + N2E2 + N3E3 = 4a

The Boltzmann Distribution – Effect of Temperature

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

Energy

Pj

= N

j/N

30K

300K

3000K 30000K

300000K

T,V,NQ

e

A

aT,V,NP

Tk/V,NEj

j

Bj

i

NV,NE

!N

,vqe,V,NQ i

Tk/EE

1

2 B12eN

N Tk/E1Be1

1NN

Tk/E2

Be1

1NN

The system partition function is

for indistinguishable particles.For the two-level system:

The Distribution of Molecules on Two Energy Levels

Box at rest Box weakly shaken Box vigorously shaken

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.00 20000.00 40000.00 60000.00 80000.00 100000.00 120000.00

Tempertaure (K)

N2/

N1

N2/N1 = exp[-E/kBT]

N2/N1 approaches 1 as T

A Two-Level System

A Two-Level System

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000

Temperature (K)

Ni/

N

As T approaches infinity,N1 and N2 approaches 50%

N1/N = 1/(1+exp[-E/kBT])

N2/N = 1/(1+exp[E/kBT])

Energy can be evaluated form the partition function:

For the system: V,N

2B

V,N T

QlnTk

QlnE

For the atom/molecule: V,N

2B

V,N T

qlnTk

qln

NE , and for 1 mol: UNE A

and so we see that V

BA T

qlnTkNU

2

Heat capacity and pressure can also be evaluated from the partition function:

V,Nv T

UC

and ,N

B V

QlnTkP

You should be able to calculate any of these properties for any partition function I give you. Here follows some results for a diatomic molecule. You should understand (and derive completely) how these results were obtained.

The translational partition function (3D): Vh

TMk2T,Vq

23

2B

trans

The rotational partition function (large T): 2

B2

roth

TIk8Tq

The vibrational partition function: T/

T2/

vibvib

vib

e1

eTq

U VC

Translational RT

2

3 R

2

3

Rotational RT R

Vibrational

1e2R

T/vibvib

vib

2T/

T/2vib

vib

vib

e1

e

TR

Total internal energy is the sum of the individual internal energies and total heat capacity is the sum of the individual heat capacities. Note, we have ignored the electronic contribution. Note elecvibrottrans qqqqT,Vq and elecvibrottranstotal

The First Law: Some Terminology

System: Well defined part of the universe

Surrounding: Universe outside the boundary of the system

Heat (q) may flow between system and surroundings

Closed system: No exchange of matter with surroundings

Isolated System: No exchange of q, w, or matter with surroundings

Isothermal process: Temperature of the system stays the same

Adiabatic: No heat (q) exchanged between system and surroundings

THE CONCEPT OF REVERSABILTY

Irreversible processes:

Hot

Cold

Warm

WarmTemperature equilibration

Mixing of two gases

Expansion into a vacuumP = 0

Evaporation into a vacuumP = 0

THE CONCEPT OF REVERSABILTY

REMOVE PINS

Irreversible Expansion

pins

P = 2 atm

Pext = 1 atm

(1)

(2)

P = 2 atm P = 1.999 atm

Step 1

P = 1.998 atm

Step 2Infinite number of steps

Reversible ExpansionP = 1 atm

P = 1 atm

Pext = 1 atm

THE CONCEPT OF REVERSABILTY

Reversible processes:

Po

Po + P

Po

Tiny weight

Condensation(pressure minimally increases by adding tiny weight)

Evaporation(pressure minimally decreases by removing tiny weight)

P

VV1 V2

Pext

w = -Pext(V2 – V1)

P, V1 P, V2

IRREVERSIBLE EXPANSION

Pext

Pext

V1

V2

REVERSIBLE EXPANSION

Pext = Pressure of gas. If the gas is ideal, then Pext = nRT/V

How does the pressure of an ideal gas vary with volume?

P

V

This is the reversible path. The pressure at each point along curve is equal to the external pressure.

1

2ln12

1

2

1

2

1 V

VnRTdV

VnRTdV

V

nRTPdVdVPw

V

V

V

V

V

Vext

REVERSIBLE EXPANSION

P

VVi Vf

Pi

Pf

A

B

The reversible path

i

f

V

VnRTw ln

The shaded area is

IRREVERSIBLE EXPANSION

P

VVi Vf

Pi

Pf

A

B

fif VVPw

The shaded area is

Pext = Pf

Reversible expansion gives the maximum work

REVERSIBLE COMPRESSION

P

VVf Vi

Pf

Pi

A

BThe reversible path

i

f

V

VnRTw ln

The shaded area is

P

Reversible compression gives the minimum work

IRREVERSIBLE COMPRESSION

VVf Vi

Pf

Pi

A

B

fif VVPw

The shaded area is

Pext = Pf

The Carnot Engine

P1

V1

T

Isothermal expansion

q

P2

V2

T

w = -nRT ln(V2/V1)

How can we take it back?

Hot reservoirTh

qh

Heat engine

w

Cold reservoirTc

qc

The Carnot Engine- what is the efficiency?

A

B

Isothermal reversible expansion U = 0, w = -nRTh ln(V3/V1)q = +nRTh ln(V3/V1)

(Th)

(Th)

C

Adiabatic reversible expansion

dU = CpdT = δwq = 0

(Tc)

D

Isothermal reversible compression U = 0, w = -nRTc ln(V2/V4)q = +nRTc ln(V2/V4)

P1

P2

P3

P4

Pres

sure

V1 V2 V3 V4

(Tc)

Adiabatic reversible compression

dU = CpdT = δwq = 0

total work

Statistical Interpretation of Entropy

Consider the following expansion:

V/2 V/2When stopcock is opened:At any given momentProbability that a given molecule is in the LHS bulb = 50 % (or ½)Probability that a given molecule is in the RHS bulb = 50 % (or ½)

Probability that two molecules stay in LHS bulb = ½ ½ = ¼ (25%)

Probability that N molecules stay in LHS bulb = (½)N

Microstates: A particular way of arranging molecules among the positions accessible to them while keeping the total energy fixed.

Consider four molecules in two compartments:

1 way 4 ways 6 ways 4 ways 1 way

Total number of microstates = 16

The most probable(the “even split”)

If N the “even split” becomes overwhelmingly probable

Boltzmann S = kB lnW

Consider spin (or dipole restricted to two orientations)

or W = 2, and S = kB ln21 particle

2 particles

3 particles

1 mole

W = 4, and S = kB ln4, , ,

W = 8, and S = kB ln8

W =2NA, and S = kB ln(2NA) = NA kB ln2 = Rln

Some entropy problemsThe spectroscopic entropy of a diatomic molecule is given by the following equation.

1//

2/523

2ln

1

/1lnln

2ln eT

vibT

rotA

B ge

Te

Te

N

eV

h

TMk

R

Svib

vib

What is meant by residual entropy? (a)Calculate the spectroscopic entropy of CO at 298.15 K and 1 atm.(b)Write down the expression for calculating the (calorimetric) absolute molar entropy from 0 K to 298.15 K. Note that CO undergoes a solid-solid phase transition at 61.6 K. The experimentally determined calorimetric value is ~170 JK -1mol-1..(c) Calculate the residual entropy of CO. Make the correction for residual to the calorimetric value. Compare the two entropies (spectroscopic and calorimetric).

The molar heat capacity of C2H4(g) can be expressed by the following equation over the temperature range 300 K < T < 1000 K.

2

2822826929.60854105.16/

T

K

T

KRTCV

Calculate S if one mole of ethene is heated from 300 K to 600 K at constant volume. There are no phase transitions.

Unit 3 – Real & Ideal Gases

Non-ideality of gases - Fugacity

o

o

T P

PRTTGPTGV

P

Gln,

Standard molar Gibbs energy

1 bar

Can generalize for a real gas ....1 232 PTBPTB

RT

VPPP

o

o

f

TPfRTTGPTG

,ln,

The thermodynamic function fugacity.Note: f (P,T) P as P 0

......exp, 2

32 PTBPTBP

P

f

TPfPPoo

Gibbs energy must be taken relative to some standard stateStandard state of real gas is taken to be the corresponding ideal gas at 1 bar i.e. must “adjust” the real gas to ideal behavior

Real gas (T,P) Ideal gas (T,P)

Real gas (T, P 0) = Ideal gas (T, P 0)

G1

G2G3 oo

P

P f

fRT

P

PRTdPV

P

RTG lnln'

0'1

Fugacity coefficient

P

f Ideal gas, = 1. Fugacity coefficient measures extent of non-ideality

'

0'1

ln dPP

ZP

(Z-1) / P

P (bar)0 P

1

0 P (bar)

ideal

Real

Unit 2 Summary You need to know the Boltzmann equation in terms of probability and fractional occupancy in a two level system. In general, the Boltzmann equation gives the probability, Pj, that a randomly chosen system (from an ensemble of systems in thermal contact) will be in state j with energy Ej(N,V):

T,V,NQ

e

A

aT,V,NP

Tk/V,NEj

j

Bj

The system partition function is

i

NV,NE

!N

,vqe,V,NQ i

for

indistinguishable particles. For the two-level system:

Tk/EE

1

2 B12eN

N , Tk/E1

Be1

1NN

, Tk/E2

Be1

1NN

You should be able to obtain partition functions for simple energy level systems. The Boltzmann equation seeks to find the maximum number of configurations. For a system with large N, there is a configuration with so great a weight that is overwhelms the rest. The system will almost always be found in it, and it will determine the properties of the system. The Boltzmann equation gives the probability of realizing this configuration as a function of energy and temperature. Energy can be evaluated form the partition function:

For the system: V,N

2B

V,N T

QlnTk

QlnE

For the atom/molecule: V,N

2B

V,N T

qlnTk

qln

NE , and for 1 mol: UNE A

and so we see that V

BA T

qlnTkNU

2

Heat capacity and pressure can also be evaluated from the partition function:

V,Nv T

UC

and ,N

B V

QlnTkP

You should be able to calculate any of these properties for any partition function I give you. Here follows some results for a diatomic molecule. You should understand (and derive completely) how these results were obtained.

The translational partition function (3D): Vh

TMk2T,Vq

23

2B

trans

The rotational partition function (large T): 2

B2

roth

TIk8Tq

The vibrational partition function: T/

T2/

vibvib

vib

e1

eTq

The vibrational partition function depends on how we define the zero of energy. If we define it as the bottom of the internuclear potential, we obtain the above expression. However if we define zero as the v = 0 state, then we obtain the following expression:

1/2h E = 0

E = 0 for v = 0 v = 1

v = 2

e1 = -De

e2

Tvib

vibeTq

/1

1

The expression for the rotational partition function above works for high temperatures. Room temperature measurements have to calculated numerically using the following equation.

TkBTkBrot

BB eeTq /6/2 531

From these partition functions the following can be obtained.

U VC

Translational RT

2

3 R

2

3

Rotational RT R

Vibrational

1e2R

T/vibvib

vib

2T/

T/2vib

vib

vib

e1

e

TR

Total internal energy is the sum of the individual internal energies and total heat capacity is the sum of the individual heat capacities. Note, we have ignored the electronic contribution. Note elecvibrottrans qqqqT,Vq and elecvibrottranstotal

WORK, ENERGY & HEAT

Work is the transfer of energy due to unbalanced forces: VPw ext

If Pext is not constant during the process: dVPw f

i

V

V ext

For a reversible process (compression or expansion of an ideal gas):

1

2V

VlnnRTw

The first law of thermodynamics: wqU

wqdU

Path & state functions: know the

difference. Some important processes: ISOLATED SYSTEM: 0U,0w,0q

ISOTHERMAL (no change in T): 0U 0T

1

2revrev V

VlnnRTqw

ADIABATIC (no change in q): 0q

2

1

T

T Vrev dTTCwU

PRESSURE-VOLUME WORK: VqU (constant volume)

Hq

VPUH

PVUH

P

constant pressure

The temperature of a gas decreases in a reversible adiabatic expansion:

2

12

3

1

2

V

V

T

T

(Know the derivation)

ENTHAPLY

At constant pressure: VPUH , and HqP

To calculate the enthalpy from T = 0K to T = T, must take into account any phase transitions.

fus

fus

T

0

T

T

LPfus

SP dTTCHdTTC0HTH , etc. In other words dTCH P

Note also that T

HCP

, and nRCC VP

}

Make sure you know how to calculate enthalpy changes for chemical reactions. You can use Hess’s law and enthalpies of formation to calculate enthalpies of reaction. Know how to calculate enthalpy changes at different temperatures using the following equation:

2

1

T

T PP1r2 s'reactCproductsCTHTH

ENTROPY

Entropy (state function): T

qdS rev

0dS Equilibrium (reversible process) 0dS Spontaneous process in isolated system

Know the Second Law.

Isothermal expansion: 1

2

V

VlnRS , For mixing of two gases:

i

N

1jimix ylnyRS

(Learn the derivation)

The Third Law: Every substance has finite positive entropy, but at 0 K the entropy may become 0, and does so in the case of a perfectly crystalline substance. Determining absolute entropies from calorimetric data:

T

T

gP

vap

vapT

T

LP

fus

fusT

0

SP

vap

vap

fus

fus

T

dTTC

T

H

T

dTTC

T

H

T

dTTCTS

These entropies are usually called “calorimetric entropies.” Note we have assumed the Third Law holds true for calorimetric entropies. Statistical Entropy: WlnkS B

From partition functions:

VBBB

V,NBB T

qlnTNk

N

T,VqlnNkNk

T

QlnTkQlnkS

Note that the above expression relates entropy to the system partition function (Q) and the molecular partition function (q). Make sure you know how to go from the expression for system Q to molecular q, using Stirling’s approximation (lnN! = NlnN – N).

The molar entropy of an ideal monoatomic gas:

A

1e23

2B

N

gV

h

Tmk2lnRR

2

5S

The molar entropy of a diatomic gas is given by:

1eT/vibT/

rotA

2/523

2B gln

1e

T/e1ln

2

Teln

N

eV

h

TMk2ln

R

Svib

vib

You should be able to derive entropy expressions using simple partition functions. You should be able to determine entropies of monoatomic and diatomic molecules from spectroscopic parameters. The entropies are generally called “statistical entropies.” We have assumed the Third Law holds perfectly when calculating calorimetric entropies, i.e. S = 0 at T = 0 K. Spectroscopic entropies are therefore more accurate. Realize that at 0 K some molecules (usually those possessing small dipoles) may get “locked” into two or more possible degenerate states, e.g. CO and linear NNO, W = 2; CH3D, W = 4. You should be able to calculate residual entropies, and add the values to the calorimetric values to get entropies in closer agreement to statistical ones. Helmholtz Energy: 0STUA For constant T and V 0A Spontaneous (irreversible) 0A Not spontaneous 0A Equilibrium Gibbs Energy: STHG For constant T and P 0G Spontaneous (irreversible) 0G Not spontaneous 0G Equilibrium Maxwell Relations Maxwell relations allow us to derive some important thermodynamic equations. You need to be able to derive all of them. Below I have reproduced Table 22.1 from the textbook. Please learn the differential expressions and the derivations for the corresponding Maxwell relations.

Thermodynamic Energy Differential expression Corresponding Maxwell

relations

U

dU = TdS – PdV VS S

P

V

T

H

dH = TdS + VdP

PS S

V

P

T

A

dA = -SdT – PdV

VT T

P

V

S

G

dG = -SdT + VdP

PT T

V

P

S

Two important ones are the temperature and the pressure dependency of the Gibbs energy.

ST

G

P

This gives: 2

P T

H

T

T/G

Gibbs-Helmholtz equation (learn the

proof)

VP

G

T

This gives: 1

2

P

PlnRTG (ideal gas)

Other very important one are: PT T

V

P

S

, which gives 1

2

P

PlnnRS (ideal

gas)

PV

A

T

and ST

A

V

so that using cross derivatives, we find that

TV V

S

T

P

. We end up with 1

2

V

VlnnRS for an ideal gas.

Beware, I may ask to solve the equations using other equations of state. You need to know how internal energy varies with volume.

VT T

PTP

V

U

(learn the proof)

You need to know how enthalpy varies with pressure.

PT T

VTV

P

H

(learn the proof)

For the above two, you may be asked to solve the partial derivative using an equation of state other than the ideal gas equation (see your homework for examples). The Non-Ideality of Gases

At P1 = 1 bar PlnRTTGP,TG o , where TG o = standard molar Gibbs energy.

For non-ideal gases o

o

f

T,PflnRTTGP,TG , where f(P,T) = fugacity of gas.

Using the virial equation for non-ideality:

.....

PTBPTBexp

P

P

f

T,Pf PPoo 2

23

2

Note, the fugacity coefficient P

f , and

P

0

dPP

1Zln , where

RT

VPZ

(Z = compressibility factor)

4. The relationship is VT T

PTP

V

U

For a van der Waals gas bV

R

T

P

V

, and so 2V

a

V

U

T

Integrating VdV

adUV

V

U

u idealideal

2

1 which leads to

idealidealideal

UV

aU

VVaV,TU

11

For ethane a = 5.5818 dm6 bar mol-2, and

idealU = 14.55 kJ mol-1

Using the van der Waals equation

2V

a

bV

RTP

, we can find P from

various molar volumes. We can then determine U as a function of P and plot the two. I have used the molar volume range suggested in problem 22-4.

(a) For an ideal gas V

R

T

P

V

and so 0

V

RTP

V

U

T

(b) For a van der Waals gas, see above. 2V

a

V

U

T

(c) For Redlich-Kwong gas, where BVVT

A

BV

RTP

/

21

BVVT

A

BV

R

T

P/

V

232

14.48

14.49

14.5

14.51

14.52

14.53

14.54

70 270 470 670

P /bar

Mo

lar

En

erg

y /

kJ/

mo

l

BVVT

A

BVVT

AA

BVVT

A

BVVT

A

BVVT

A

BVVT

A

BVVT

AT

BVVT

A

BVVT

AT

BV

RT

BVVT

A

BV

RT

BVVT

A

BV

RP

V

U

////

////

///T

21212121

21212321

232123

2

3

2

2

22

2

22

22

(d) For the virial equation of state 23

121 VTBVTBZ PP , up to the third virial

coefficient. Solving for pressure gives:

3322

V

RTTB

V

RTTB

V

RTP PP

3

3332

222

V

RT

dT

dB

V

RTB

V

RT

dT

dB

V

RTB

V

R

T

P PP

PP

V

3

23

2

22

3

23

332

22

223322

33

3322

22

V

RT

dT

dB

V

RT

dT

dB

V

RT

dT

dB

V

RTTB

V

RT

dT

dB

V

RTTB

V

RT

V

RTTB

V

RTTB

V

RT

V

RT

dT

dB

V

RTB

V

RT

dT

dB

V

RTB

V

RTP

V

U

PP

PP

PPPP

PP

PP

T

And so

dT

dB

VdT

dB

V

RT

dT

dB

V

RT

dT

dB

V

RT

V

U PPPP

T

322

23

3

22

2

2 1

5. We can determine how enthalpy varies with P, since H = U + PV. You simply add the corresponding PV values to the U values determined In the last problem. Remember to convert the PV units into joules. Basically 1 L = 1000 cm3 = 10-3 m3, and 1 bar = 105 Pa. Therefore the conversion factor is 102. But we need kJ not J, and so divide further by 1000. The final conversion factor is 0.1.

We need PT T

VTV

P

H

for the remaining parts of the problem.

(a) For an ideal gas 0

P

RT

P

RT

T

VTV

P

H

PT

(b) For the virial equation of state 2321 PTBPTBZ PP , up to the third virial coefficient.

Solving for volume gives:

RTPTBRTTBP

RTRTPTBRTTB

P

RTV PPPP 3232

RTPdT

dBRPTBRT

dT

dBRTB

P

R

T

V PP

PP

P

33

22

2322

33

22

32

RPTdT

dBRT

dT

dB

P

H

RTPdT

dBRPTBRT

dT

dBRTB

P

RT

RTPTBRTTBP

RT

T

VTV

P

H

PP

T

PP

PP

PPPT

3.