thermo & stat mech - spring 2006 class 19 1 thermodynamics and statistical mechanics partition...

Thermo & Stat Mech - Spring 2006 Class 19

1

Thermodynamics and Statistical Mechanics

Partition Function


2

Free Expansion of a Gas


3

Free Expansion


4

Isothermal Expansion


5

Isothermal Expansion

Reversible route between same states.

f

iTQd

SđQ = đW + dU Since T is constant, dU = 0. Then, đQ = đW.

dVV

nRTPdVWd

VdV

nRVdV

TnRT

TWd

TQd

dS


6

Entropy Change

2ln2

ln2

nRVV

nRVdV

nRSV

V

The entropy of the gas increased.For the isothermal expansion, the entropy of theReservoir decreased by the same amount. So for the system plus reservoir, S = 0

For the free expansion, there was no reservoir.


7

Statistical Approach

2)!2/(

!

)!2/()!2/(

!

1!0!

!

lnlnln

N

N

NN

Nw

N

Nw

w

wkwkwkSSS

f

i

i

fifif


8

Statistical Approach

2ln2ln

2/ln)]2/ln(ln[

}]2/)2/ln()2/{(2ln[

])!2/ln(2![ln

)!2/(

!lnln

2

nRNkS

N

NNkNNNNkS

NNNNNNkS

NNkS

N

Nk

w

wkS

i

f


9

Partition Function

Z

egNN

Zeg

eg

egNN

j

j

j

j

jj

n

jj

n

jj

jj

FunctionPartition 1

1


10

Boltzmann Distribution

Z

Z

Z

N

U

eg

eg

NUN

eg

egNN

n

jj

n

jjjn

jjj

n

jj

jj

j

j

j

j

ln

1

1

1

1


11

Maxwell-Boltzmann Distribution

Correct classical limit of quantum statistics is Maxwell-Boltzmann distribution, not Boltzmann.

What is the difference?


12

Maxwell-Boltzmann Probability

n

j j

Nj

MB

n

j j

Nj

B

n

j jjj

jFD

n

j jj

jjBE

N

gw

N

gNw

NgN

gw

gN

gNw

jj

11

11

!

!!

)!(!

!

)!1(!

)!1(

wB and wMB yield the same distribution.


13

Relation to Thermodynamics

YdX

dN

dXdX

dNdNdU

dXdX

ddX

dNdNdU

NU

j

jj

j

jj

jjj

jjjj

jjj

jjj

jjj

Call

so ,)(


14

Relation to Thermodynamics

YdXdNTdSdNdU

dX

PdVTdSdUYdXTdSdU

YdXdNdU

YdX

dNdX

dX

dNdNdU

jjj

jjjX

jjj

j

jj

j

jj

jjj

and )(

0 If

)(or

like is This

and


15

Chemical Potential

dU = TdS – PdV + dN

In this equation, is the chemical energy per molecule, and dN is the change in the number of molecules.


16

Chemical Potential

VTN

F

dNPdVSdTdF

SdTTdSdNPdVTdSdF

TSUF

dNPdVTdSdU

,


17

Entropy

j j

jj

jj

jjj

jjj

jj

jjj

n

j j

Nj

MBMB

g

NNNkS

NNNgNkS

NgNkS

N

gwwkS

j

ln

lnln

!lnln

! whereln

1


18

Entropy

)1ln(ln

lnln

ln

NZNkT

US

kTNZNNNNkS

eZ

N

g

N

g

NNNkS

j

jj

jj

jj

kT

j

j

j j

jj

j


19

Helmholtz Function

)1ln(ln

)1ln(ln

)1ln(ln

NZNkTF

NZNkTT

UTUTSUF

NZNkT

US


20

Chemical Potential

Z

NkTZNkT

NNkTNZkT

N

F

NZNkTF

VT

ln)ln(ln

1)1ln(ln

)1ln(ln

,


21

Chemical Potential

kT

Z

Ne

Z

N

kTZ

NkT

kT

So,

Then

ln so ln


22

Boltzmann Distribution

n

jj

jj

n

jj

n

jj

n

jj

jj

j

j

j

j

j

eg

egNN

eg

Ne

egeNN

egeN

1

1

11


23

Distributions

Dirac-Fermi 1

1

Einstein-Bose 1

1

Boltzmann 1

jj

j

jj

j

jj

j

feg

N

feg

N

feg

N

j

j

j


24

Distributions

Dirac-Fermi

1

1

Einstein-Bose

1

1

Boltzmann 1

j

kTj

j

j

kTj

j

j

kTj

j

f

eg

N

f

eg

N

f

eg

N

j

j

j


25

Ideal Gas

1ln2

ln2

3ln

)1ln(ln

2

2

2/3

2

Nh

mkTVNkTF

NZNkTF

h

mkTVZ


26

Ideal Gas

nRTNkTPV

VNkT

V

FP

Nh

mkTVNkTF

NT

1

1ln2

ln2

3ln

,

2


27

Ideal Gas

kTZ

N

U

h

mVZ

h

mV

h

mkTVZ

2

31

2

3ln

ln2

32lnln

22

2/3

2

2/3

2

2/3

2


28

Entropy

2/3

2

2/3

2

2/3

2

2ln

2

5

12

ln2

3

2lnln

)1ln(ln

h

mkT

N

VNkS

h

mkT

N

VNk

T

NkTS

h

mkTVZ

NZNkT

US


29

Math Tricks

For a system with levels that have a constant spacing (e.g. harmonic oscillator) the partition function can be evaluated easily. In that case, n = n, so,

.1for 1

1

1

1

0

000

εβ

n

n

n

n

n

n

n

eex

x

eeeZ n


30

Heat Capacity of Solids

Each atom has 6 degrees of freedom, so based on equipartition, each atom should have an average energy of 3kT. The energy per mole would be 3RT. The heat capacity at constant volume would be the derivative of this with respect to T, or 3R. That works at high enough temperatures, but approaches zero at low temperature.


31

Heat Capacity

Einstein found a solution by treating the solid as a collection of harmonic oscillators all of the same frequency. The number of oscillators was equal to three times the number of atoms, and the frequency was chosen to fit experimental data for each solid. Your class assignment is to treat the problem as Einstein did.


32

Heat Capacity

Heat Capacity

-5

0

5

10

15

20

25

30

0 100 200 300 400

T (K)

Cv

(J/K

)

thermo & stat mech - spring 2006 class 19 1 thermodynamics and statistical mechanics partition...

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