Thermo & Stat Mech - Spring 2006 Class 19 1 Thermodynamics and Statistical Mechanics Partition...
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Thermo & Stat Mech - Spring 2006 Class 19
1
Thermodynamics and Statistical Mechanics
Partition Function
Thermo & Stat Mech - Spring 2006 Class 19
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
Thermo & Stat Mech - Spring 2006 Class 19
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.
Thermo & Stat Mech - Spring 2006 Class 19
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Statistical Approach
2)!2/(
!
)!2/()!2/(
!
1!0!
!
lnlnln
N
N
NN
Nw
N
Nw
w
wkwkwkSSS
f
i
i
fifif
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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Partition Function
Z
egNN
Zeg
eg
egNN
j
j
j
j
jj
n
jj
n
jj
jj
FunctionPartition 1
1
Thermo & Stat Mech - Spring 2006 Class 19
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
Thermo & Stat Mech - Spring 2006 Class 19
11
Maxwell-Boltzmann Distribution
Correct classical limit of quantum statistics is Maxwell-Boltzmann distribution, not Boltzmann.
What is the difference?
Thermo & Stat Mech - Spring 2006 Class 19
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.
Thermo & Stat Mech - Spring 2006 Class 19
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Relation to Thermodynamics
YdX
dN
dXdX
dNdNdU
dXdX
ddX
dNdNdU
NU
j
jj
j
jj
jjj
jjjj
jjj
jjj
jjj
Call
so ,)(
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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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.
Thermo & Stat Mech - Spring 2006 Class 19
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Chemical Potential
VTN
F
dNPdVSdTdF
SdTTdSdNPdVTdSdF
TSUF
dNPdVTdSdU
,
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
19
Helmholtz Function
)1ln(ln
)1ln(ln
)1ln(ln
NZNkTF
NZNkTT
UTUTSUF
NZNkT
US
Thermo & Stat Mech - Spring 2006 Class 19
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Chemical Potential
Z
NkTZNkT
NNkTNZkT
N
F
NZNkTF
VT
ln)ln(ln
1)1ln(ln
)1ln(ln
,
Thermo & Stat Mech - Spring 2006 Class 19
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Chemical Potential
kT
Z
Ne
Z
N
kTZ
NkT
kT
So,
Then
ln so ln
Thermo & Stat Mech - Spring 2006 Class 19
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
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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Ideal Gas
1ln2
ln2
3ln
)1ln(ln
2
2
2/3
2
Nh
mkTVNkTF
NZNkTF
h
mkTVZ
Thermo & Stat Mech - Spring 2006 Class 19
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Ideal Gas
nRTNkTPV
VNkT
V
FP
Nh
mkTVNkTF
NT
1
1ln2
ln2
3ln
,
2
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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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
Thermo & Stat Mech - Spring 2006 Class 19
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
Thermo & Stat Mech - Spring 2006 Class 19
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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.
Thermo & Stat Mech - Spring 2006 Class 19
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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.