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Problem Set #2
Due after class on Wed April 14
th
.
Answer key posted afterwards on website.
Office Hours
Cancelled for Wed April 14
th
.
Held on Tue April 13
th
, 10:3011:30 am.
Practice Midterm 1
(see website)
Midterm 1
Wed April 21
st
.
Material covered through end of Fri April 16
th
.
Chapters 19, 20, 21, and 22.
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View Full Document Boltzmann Equation of Entropy (S = k
B
lnW).
W = number of microstates accessible to system (W
∝
V
N
, ideal gas)
k
B
= Boltzmann constant = 1.38 x 10
23
J K
1
= R/N
A
S
total
= S
1
+ S
2
= k
B
(ln
W
1
+ ln
W
2
)
where W
total
= W
1
*W
2
Boltzmann Entropy for System with 2fold degeneracy.
W = 2
N
2
ln
2
ln
B
N
B
system
Nk
k
S
=
=
N particles distributed between two states with equal probability
(e.g. N spin1/2 nuclei or CO in crystal lattice)
Entropy of Mixing.
{}
{ }
2
2
1
1
ln
ln
y
R
y
y
R
y
S
mix
−
−
=
∆
2
1
1
1
n
n
n
y
+
=
2
1
2
2
n
n
n
y
+
=
Statistical Entropy (pp 829838)
Entropy of a Large Collection of Molecules
A collection of
N
=10
23
molecules distributed among energy levels as
...
,
,
2
1
0
n
n
n
,
(i.e. n
0
= number of molecules with energy
ε
0
etc.)
!...
!
!
!
2
1
0
n
n
n
N
W
=
!...
ln
!
ln
!
ln
!
ln
ln
2
1
0
n
n
n
N
W
−
−
−
=
We can simplify ln
N!
using Stirling’s approximation (
n
i
assumed to be large numbers):
N
N
N
N
−
≅
ln
!
ln
...
)
ln
(
)
ln
(
)
ln
(
)
ln
(
ln
2
2
2
1
1
1
0
0
0
−
−
−
−
−
−
−
−
≅
n
n
n
n
n
n
n
n
n
N
N
N
W
Since
∑
=
i
n
N
, all nonln terms (N and n
i
) cancel to give
∑
−
≅
i
i
n
n
N
N
W
ln
ln
ln
So that
)
ln
ln
(
ln
∑
−
=
=
i
i
B
B
n
n
N
N
k
W
k
S
.
N =
Σ
n
i
)
ln
ln
(
i
i
B
n
n
N
N
k
S
Σ
−
=
See Stirling’s approximation
(Math Chapter J pp 809815)
(Stirling’s approximation)
( )
N
N
N
N
−
=
ln
!
ln
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View Full Document Deriving
T
dq
dS
rev
=
from Boltzmann Eq (S = k
B
lnW)
For reversible process at constant volume:
S = k
B
ln
W
≅
k
B
(
N
ln
N

∑
n
i
ln n
i
)
()
( )
∑
−
=
∑
−
≅
=
i
i
B
i
i
B
B
n
n
d
k
n
n
N
N
d
k
W
d
k
dS
ln
0
ln
ln
ln
(Note,
0
)
(ln
ln
ln
=
=
=
N
N
N
N
N
dN
N
d
and
N
N
N
)
∑
∑
−
−
=
∑
∑
−
−
=
i
i
i
B
i
i
B
i
i
B
i
i
B
n
dn
n
k
n
dn
k
n
d
n
k
n
dn
k
dS
ln
ln
ln
∑
−
=
∑
−
−
=
∑
∑
−
−
=
i
i
B
i
i
B
i
B
i
i
B
n
dn
k
n
dn
k
dn
k
n
dn
k
dS
ln
0
ln
ln
∑
−
=
i
i
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This note was uploaded on 06/22/2010 for the course CHEM 21360 taught by Professor Ame during the Spring '09 term at East Los Angeles College.
 Spring '09
 Ame

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