Chapter 19 Statistical thermodynamics: the concepts

Stat

isti

cal T

herm

odyn

amic

sK

inet

ics

Dyn

amic

s

T, P, S, H, U, G, A...

{ri},{pi},{Mi},{Ei}…

How to translate mic into

mac?

The job description

Brute force approach does not work!

2

2

2

2

2

2

2

2

2

2

iii

i

zyxi

iiiiiim EV

222222

22 2

2 EVm

333332

32 3

2 EVm

• 1 mol 100000000000000000000000 particles• ~100000000000000000000000 equations needed to be solved!!!

555552

52 5

2 EVm

111112

12 1

2 EVm

444442

42 4

2 EVm

………

Bad news: We cannot afford it!

Good news:We do not need that detailed description!

How?

Thanks to them

Josiah Willard Gibbs

James Clark Maxwell

Ludwig Boltzmann

Cheng-Ning Yang

Tsung-Dao Lee

Van Hove

Landau

Arrhennius Enrico Fermi

Bose Paul Dirac

Langevin

Einstein

Ising

Mott Anderson Bardeen …

THE magic word

Statistical

We, the observers, are macroscopic. We only need averageof microscopic information.

Spatial and temporal average

T, P, S, H, U, G, A...

{ri},{pi},{Mi},{Ei}...

Still lots of challenges (opportunities) herein!

Contents

The distribution of molecular states

19.1 Configuration and weights

19.2 The molecular partition function

The internal energy and the entropy

19.3 The internal energy

19.4 The statistical entropy

The canonical partition function

19.5 The canonical ensemble

19.6 The thermodynamic information in partition function

19.7 Independent molecules

Assignment for Chapter 19

Exercises:

• 19.1(a), 19.2(b), 19.4(a)

Problems:

• 19.6(a), 19.9(b), 19.11(b), 19.15(a)

• 19.3, 19.7, 19.14, 19.18, 19.22

The distribution of molecular states

EEE E

These particles might be distinguishable

Distribution = Population pattern

•

..............

Enormous possibilities!

E E

E

EE

E

EE

E

E

E

EE

E

..............

Distinguishable particles

E EE EE

Principle of equal a priori probabilities

• All possibilities for the distribution of energy are equally probable.

An assumption and a good assumption.

They are equally probable

E E

E

EE

E

EE

E

E

E

EE

E

..............

E

EE

E

E

They are equally probable

Configuration and weights

{5,0,0,...}

The numbers of particles in the states

{3,2,0,...}

{3,2,0,...}

One configuration may have large number of instantaneous configurations

{N-2,2,0,...}

How many instantaneous configurations?

N(N-1)/2

E

18!/3!/4!/5!/6!

{3,4,5,6}

!...!!!

210 nnnNW

Nnnn

nnn

...

,...},,{

210

210

Configuration and weights

W is huge!

20 particles: {1,0,3,5,10,1} W=931000000

How about 10000 particles with {2000,3000,4000,1000}?

i

iii

iii nnNNnnnNNNW lnln)ln()ln(ln

xxxx ln!ln

ii

nnnN

nN

nnnN

nnnNW

!ln!ln

...)!ln!ln!(ln!ln

!...)!!ln(!lnlnln

210

210!...!!!

210

Stirling’s approximation:

xx exx 21

2!

Wmax

{ni}max {ni}

There is an overwhelming configurationW

Among all possible configurations consistent with the macroscopic constraints, only one is significant and needs to be considered.

When the number of particles reaches certain value, only one configuration is so dominant that all others can be neglected. This is the most important revelation of a system consisting of a large number of particles, leading to a tremendous simplification making it possible to build a theory that can deal with moles of particles. An excellent example of ‘big brings simple’.Of course, fluctuation is intrinsic in a big system.

10500

10200

10100 1050

Equilibrium configuration

The dominating configuration is what we actually observe. All other configurations are regarded as fluctuation.

Eni

ii

Nni

i

The dominating configuration is the configuration with largest weight.

Constant total number of molecules:

Constant total energy :

ji

jij

i

i nn

nN

if 0

if 1,1

Find the distribution with largest lnWi

i i

dnn

WWd

lnln

iiidn 0

iidn 0

ii

ii

i iiiii

i i

dnn

W

dndndnn

W

ln

ln

0ln

i

in

W

j i

jj

ii n

nn

n

NN

n

W lnlnln 1lnln

ln

N

n

NN

n

N

n

NN

iii

Nni

i 1ln1lnln

lnln

i

jj

i

j

j i

jjj

i

j

j i

jj nnn

n

n

nnn

n

n

n

nn

N

nNn

n

W ii

i

ln1ln1lnln

0ln i

i

N

n ieN

ni

Lagrange’s method of undetermined multipliers

z=f(x,y) with g(x,y)=c1, h(x,y)=c2

To find the maximum of z with constraints g and h, we May use

0** dhbdgadz

Boltzmann distribution

i

i

i

i

e

e

N

n

kT

1

Boltzmann constant

ieN

ni jj

jjeNenN

j

iee

1

The molecular partition function(nondegenerate case)

q

ep

i

i

Nn

i

j

ieq

E

1

54

3

2

G

11 g

15 g

14 g13 g

12 g

The molecular partition function(degenerate case)

j levels

jiegq

E

,...,, 3,52,51,5

,...,, 3,32,31,3

,...,, 3,42,41,4

,...,, 3,22,21,2

,...,, 3,12,11,1

G

1g

5g

4g

3g

2g

-1

-1

.

ˆ is invariant under symmetry operation:

ˆ ˆ

ˆ

ˆ

ALL ={ } are the

ˆeigenstates of with the same energy .

i i i

i i i

i i j

i

H

RHR H

RHR R E R

HRR E R

R

H E

The degeneracy—symmetry (recall quantum chemistry course)

Angular momentum

...3,2,1,0 2

)1( 2

22

lI

llI

E

...3,2,1,0 2

)1( 2

22

lI

llI

JE

...3,2,1,0 ,)1( || lllJ

llllmmJ llz ,1,...1, ,

2zJ

1zJ

00 zJ

zJ2zJ

6)12(2 || J

The Rotational Energy Levels (Ch 16)

2aaa I

2

1E Around a fixed-axis

222

2

1

2

1

2

1ccbbaa IIIE Around a fixed-point

aaa IJ c

c

b

b

a

aI

J

I

J

I

JE

222

222

I2

J

I2

JJJE

22c

2b

2a

(Spherical Rotors)

22 1JJJ ,...2,1,0J

2

j I21JJE

I2hcB

2

cI4B

1JhcBJE j

BJJFJF 21

Example: Linear Molecules (rigid rotor)

0

)1(

j levels

)12(j

JhcBJj eJegq i

E

9,53,52,51,5 ,...,,

5,33,32,31,3 ,..,, 7,43,42,41,4 ,...,,

3,22,21,2 ,,

1

G

11 g

95 g74 g53 g32 g

E1

E2

E3

E4

E5

Exercises

E

,...,, 3,52,51,5

,...,, 3,32,31,3

,...,, 3,42,41,4

,...,, 3,22,21,2

,...,, 3,12,11,1

G

11 g

25 g34 g13 g22 g

E1

E2

E3

E4

E5

q=?

E=0,g=1

E=ε,g=2q=?

eq 21

The physical interpretation of the molecular partition function

j levels

jiegq

00

gqlimT

qlimT

q is an indication of the average number of states that are thermally accessible to a molecule at the temperatureof the system.

E, TE, T=0 E, ∞

狀態數加權和

eq

1

1

ieepi 1

Example: Uniform ladder of energy levels (e.g., harmonic vibrator)

...1 32

0

eeeegqj

jj

xxxx 1132 ...1

ieepi 1

eq

1

1

eq 1

E=0,g=1

E=ε,g=1

1,,0

1

qT

eq

E=0,g=1

E=ε,g=1

2,0,

1

qT

eq

E=0,g=1

E=ε,g=1

ep

1

10

e

ep

11

E=0,g=1

E=ε,g=1

E=0,g=1

E=ε,g=1

11

1 00

T

ep

01

01

T

e

ep

E=0,g=1

E=ε,g=1

2/11

10

T

ep

2/111

T

e

ep

E=0,g=1

E=ε,g=1

E=0,g=1

E=ε,g=1

Approximations and factorizations

• Exact, analytical partition functions are rare.

• Various kinds of approximations are employed:

dense energy levels

independent states (factorization of q)

…

Dense energy levels

Xh

mqX

21

2

2

2

22

8mX

hnEn One dimensional box: ,...,n 21

12 nEn 2

2

8mX

h

1

12

n

nX eq dneq n

X

1

12

Xh

mdxeq x

X

21

2

21

21

0

21

2

2

11 2

Independent states (factorization of q)

Zn

Yn

Xnnnn 321321

ZYX qqq

q

Zn

Yn

Xn

Zn

Yn

Xn

Zn

Yn

Xn

3

-

2

-

1

-

--

n all

-

n all

---

321

321321

eee

eeee

XYZh

mq

23

2

2

3V

q

21

21

22 mkT

h

mh

Three-dimensional box:

Thermal (de Broglie) wavelength

(Translational partition function)

2 1

2k B2mE = k T 2thermal

p

thermalp mkT

Why q, the molecular partition function, so important?

• It contains all information needed to calculate the thermodynamic properties of a system of independent particles (e.g., U, S, H, G, A, p, Cp, Cv …)

• It is a kind of “thermal wavefunction”. (Remember the wavefunction in quantum mechanics which contains all information about a system we can possibly acquire)

Find the internal energy U from q

i

iinE

i

iie

q

NE

ii ed

dei

d

dq

q

Ne

d

d

q

Ne

d

d

q

NE

ii

ii

EUU )0(

V

q

q

NUU

)0(

V

qNUU

ln

)0(

Total energy of the system:

q

ep

i

i

Nn

i

At T=0, U=U(0)

A two-level system

e

N

e

eNe

d

d

e

NE

111

1

E=0,g=1

E=ε,g=1 eq 1

U=?

eN

eeN

q

Vq

V

UU

eNU

q

qNU

qNUU

11

1

)0()0(

)()0(

))(ln

()0(

)ln

()0(

2

2

)1(

1

)()(

)0(

e

eNkTv

UvT

Uv

eN

C

UU

A two-level system eq 1

)1ln(ln

]lnlnln[ln

]lnlnln[lnln

lnlnln

lnln

lnlnln

11

1

1

1

ln

eqe

qeeqNqN

qeeNeqNN

N

N

NNW

eq

q

q

qNe

qe

qN

q

qNe

iq

eNW

iq

Neq

Ne

ii

ii

E=0,g=1

E=ε,g=1

W=?

i

ii nnNNW lnlnln

Exercise

A two-level system

e

NE

1

E=0,g=1

E=ε,g=1

E=0,g=1

E=ε,g=1

01

1/ 0/

kT

eNE

E=0,g=1

E=ε,g=1

5.01

1/ 2

1/

kT

eNE

The value of βV

qNUU

ln

)0(

nRTUU2

3)0(

2

3)0(

NUU kTkTnN

nN

nRT

N

A

A 1

For monatomic perfect gas,

d

dV

d

dV

Vq

VV433

31

222

1

2 21

21

21

21

m

h

m

h

d

d

d

d32

3

Vq

V

2

3)0(

2

3)0( 3

3 NU

V

VNUU

3V

q Translational partition function:2

1

2

mh

The statistical entropy

0ln

i

in

W

i

iinUU )0( ii

ii

ii dndndUdU )0(

ii

ii dndUd 0

TdSdqdU rev When no work is done,

ii

idnkT

dUdS

For the most probable configuration:

ii n

Wln

i i

iii

dnkdnn

WkdS ln

i

ii

Wdkdnn

WkdS ln

ln

WkS ln

dN=0

Heat does not change energy levels

Work changes energy levels

Boltzmann formula

WkS lnT0, W=1, S0 (Third law of thermodynamics )

)]1ln([ln

)1ln(

1

1ln

ekNWkS

e

e

eNW

E=0,g=1

E=ε,g=1

The two-level system

)]1ln([1

eNkSe

Negative temperature!

q

ep

i

i

Nn

i

E=0,g=1

E=ε,g=1

eNN

E=0,g=1

E=ε,g=1

T>0, N+<N-

T<0, N+>N-Examples: laser, maser, NMR etc.

E=0,g=1

E=ε,g=1

eq 1

eNUU

1

)0(

)(0/

)(0/ 1

kT

kT

q

q

2)0(/ kTq

E=0,g=1

E=ε,g=1

eq 1

0/ 0/ kTNkS

2ln/ / kTkNS

2

2

)1(

e

eNkvC

0)(0/ kTvC

0)0(/ kTvC

)]1ln([1

eNkSe

E=0,g=1

E=ε,g=1

eq 1

eNUU

1

)0(

)(0/

)(0/ 1

kT

kT

q

q

E=0,g=1

E=ε,g=1

2ln/ 0 kNS

0/ NkS

2

2

)1(

e

eNkvC

0)0(/ kTvC

0)(0/ kTvC

)]1ln([1

eNkSe

E=0,g=1

E=ε,g=1

VT U

S)(1

eNUU

1

)0(

Positive temperature

Negative temperature

)]1ln([1

eNkSe

U and S

qNk

T

UUS ln

0

ii i

ii

ii

iii ppNkN

nnknnNnkS lnlnlnln

qp ii lnln

qNkUUk

qNknk

qpNkNpk

qppNkS

iii

ii

iii

i iiii

ln0

ln

ln

ln

i

ii nnNNW lnlnln WkS ln Nni

i

=1

q

ep

i

i

eq

1

1

Example: Simple harmonic oscillator

V

q

q

NUU

)0(

eN

eeN

Vee

11

])1

1[)(1(

Example: Simple harmonic oscillator

qNk

T

UUS ln

0

)]1ln([

1

1ln/)(

1

1

eNk

eNkT

e

eN

E=0,g=1

E=ε,g=1Exercise: The two-level system

2ln/ )0(/ kTkNS0/ )(0/ kTNkS

)]1ln([1

eNkSe

U=?, S=?

eN

V

ee

N

q

q

NUU

1

)(1

)0(

)]1ln([

)1ln(/)(

ln0

1

1

eNk

eNkT

qNkT

UUS

e

eN

0/ )(0/ kTNkS

2ln/ )0(/ kTkNS

)]1ln([

)1ln(/)(

ln0

1

1

eNk

eNkT

qNkT

UUS

e

eN

The canonical partition function QIndependent system vs interacting system

Molecular partition function q

Ensemble: an imaginary collection of replicationof the actual system with a common macroscopicparameter.

Canonical ensemble: an imaginary collection of replications of the actual system with a common temperature. (N, V,T)

Microcanonical ensemble: an imaginary collection of replications of the actual system with a common energy. (N, V,E)Grand canonical ensemble: an imaginary collection of replicationsof the actual system with a common chemical potential. (μ, V,T)

(closed system)

(isolated system)

(open system)

!...~!~!

~~

10 nn

NW

ionsconfigurat

ousinstantane possible ofnumber

the,...},~,~{on distributi

with system, theof copies~

10 nn

N

,...}~,~{ 10 nn

in~ The number of states with energybetween Ei and Ei+1

Q

e

N

n iEi

~~

i

EieQ

The canonical partition function Q

The most probable configuration:

Find internal energy U from Q

NE~

/~

NEUEUU~

/~

00 N~

Q

ep

iE

i

~ iE

ii

iii eE

QUEpUU

10~0

VV

QU

Q

QUU

ln

01

0

The total energy of the ensemble: E~

The average energy:

The fraction of members of the ensemble in a state with energy Ei:

(classroom exercise)

WN

kWkWkS N ~

ln~~

lnln~1

Qk

T

UUS ln

0

Find entropy S from Q

NWW~~ The total weight :

(5 points bonus!!!)

i

NiiieQ 21

NE iiii 21

NqQ

Independent molecules

N

i

N

ii

qeeeQ iii ...21

(classroom exercise)

(a)For distinguishable independent molecules:

NqQ

(b) For indistinguishable independent molecules:

!NqQ

N

E

E

Monatomic gas(Sackur-Tetrode equation)

3V

q

3

25

lnAnN

VenRS

21

2 mkT

h

3

25

lnp

kTenRS

!lnln

0NkqNk

T

UUS

nRNnRqnR

T

UUS

lnln

0

)ln(

lnlnlnln1lnln2

3

3

25

32

3

3

A

AA

nN

VenR

enNV

enRnNV

nRnRS

AnNN

xxxx ln!ln

Using the Sackur-Tetrode equation

• Calculate the standard molar entropy of gaseous argon at 25 oC

30

25

3

025

3

025

0

ln

lnln0

p

kTenR

N

VeR

nN

VeRS

A

m

An

Sm

11

)106.1(101012.40

molJK 1666.18

}ln{ 31125

212/5

R

RSmNm

Jem

Experimental value:1 1158.84 JK mol

The size of the Ar atoms is not considered in the theory, leading to a bigger value. The discrepancy can be used to estimate the Ar size.

Sackur-Tetrode equation

i

f

V

V

if

nR

aVnRaVnRS

ln

lnln

2

22

8mX

hnEn

As the container expands, X increeasesmore states accessiblefor the systemS increases.

aVnRnN

VenRS

A

lnln3

25