Solution of statistical thermodynamics

1. Gewinner Sin
a
g
a
, 15 Mei 2022
SolusiQuiz
K
a
nonik
a
l Ensemble

2. Quiz1
C
a
nonic
a
l Ensemble with p
a
rtition function 1
Z =
(2πmKT)3N/2
N!h3N
VN
Canonical ensemble with partition function is shown as below
Calculate the Entropy and prove that if Gibbs paradox exist or not
Answer
We have already known that entropy is the negative derivation of Helmholtz Free Energy written as :
Therefore, we should
fi
nd the Helmholtz using :
So, we will get
F = − kTln
(
(2πmkT)3N/2
N!h3N
VN
)
By extracting
variable T, we get
F = − kTln
(
(2πmk)3N/2
N!h3N
VN
T3N/2
)

3. By applying ln to the equation :
F = − kTln
(
(2πmk)3N/2
N!h3N
VN
T3N/2
)
F = − kTln
(
(2πmk)3N/2
N!h3N
VN
)
−
3NkT
2
lnT
Remember
use chain rule u’v+v’u
d(xlnx)
dx
= 1 + lnx
−S = − kln
(
(2πmk)3N/2
N!h3N
VN
)
−
3Nk
2
(1 + lnT)
S = kln
(
(2πmk)3N/2
N!h3N
VN
)
+
3Nk
2
+
3Nk
2
lnT
S = kln
(
(2πmk)3N/2
N!h3N
VN
)
+
3Nk
2
+ lnT
3Nk
2
After we get F, then, we can calculate S
−S =
∂(−kTln
(
(2πmk)3N/2
N!h3N
VN
)
−
3NkT
2
lnT)
∂T
V
S = kln
(
(2πmk)3N/2
N!h3N
VN
T
3Nk
2
)
+
3Nk
2
Remember
1.
2.
3.
ln(a * b) = lna + lnb
ln
(
a
b )
= lna − lnb
lnax
= xlna

4. Suppose there is a gas of the same type occupying two di
ff
erent spaces with same volume V, same temperature T, same pressure p, number of the
same N particles, so they will have the same entropy S and energy the same total U. When the two spaces are combined, the previously mentioned
quantities will become
Gibbs paradox
V,T,P,N, and U V,T,P,N, and U 2V,T,P,2N, and 2U
From the
fi
rst, we expect
that the Entropy will be 2
times, however is it true?,
let’s calculate
S =
3N
2
kln(2πmk) − klnN! − 3Nklnh + NklnV + 3Nk2
lnT +
3Nk
2
N and V become 2 times
Stot =
3(2N)
2
kln(2πmk) − kln(2N)! − 3(2N)klnh + (2N)kln(2V) +
3(2N)k
2
lnT +
3(2N)k
2
= Smix
Stot = S + S =
3(2N)
2
kln(2πmk) − 2klnN! − 3(2N)klnh + (2N)klnV +
3(2N)k
2
lnT +
3(2N)k
2
Smix ≠ Stot
Gibbs paradox exist
S = kln
(
(2πmk)3N/2
N!h3N
VN
)
+
3Nk
2
+ lnT
3Nk
2
S = 2kln
(
(2πmk)3N/2
N!h3N
VN
)
+ 2
3Nk
2
+ 2lnT
3Nk
2
S = kln
(
(2πmk)3N/2
N!h3N
VN
)
+
3Nk
2
+ lnT
3Nk
2

5. Quiz2
C
a
nonic
a
l Ensemble with p
a
rtition function 2
Canonical ensemble with partition function is shown as below
Calculate the Free energy and equation of states
Answer
We should
fi
nd the Helmholtz using :
So, we will get F = − kTln
(
(2πmkT)3N/2
N!h3N {
VN
+
N(N − 1)
2
VN−1
a
})
By simplifying the equation, we get
F = − kTln
(
(2πmk)3N/2
N!h3N
T3N/2
)
− kTln
(
VN
+
N(N − 1)
2
VN−1
a
)
Z =
(2πmkT)3N/2
N!h3N {
VN
+
N(N − 1)
2
VN−1
a
}
Remember
1.
2.
3.
ln(a * b) = lna + lnb
ln
(
a
b )
= lna − lnb
lnax
= xlna

6. Equation of states
−p =
∂(kTln
(
(2πmk)3N/2
N!h3N
T3N/2
)
− kTln (VN
+
N(N − 1)
2
VN−1
a))
∂V
V
−p =
∂
(
−kTln (VN
+
N(N − 1)
2
VN−1
a))
∂V
V
No variable V
Remember
dlnx
dx
=
1
x
−p = − kT
1
((VN +
N(N − 1)
2
VN−1a))
(
NVN−1
+
N(N − 1)(N − 1)
2
VN−2
a
)
p = kT
(
NVN−1
+
N(N − 1)2
2
VN−2
a
)
((VN +
N(N − 1)
2
VN−1a))
Furthermore, we can extract variable N and V
p = kT
N
V
(
VN−1
+
(N − 1)2
2
VN−2
a
)
((VN−1 +
N(N − 1)
2
VN−2a))
Again, extract variable VN−1
p = kT
N
V
(
VN−1
(1 +
(N − 1)2
2
1
V
a
)
((VN−1(1 +
N(N − 1)
2
1
V
a))
p =
NkT
V
(
1 +
(N − 1)2
2
1
V
a
)
((1 +
N(N − 1)
2
1
V
a))

7. p =
NkT
V (
1 −
N − 1
2
a
V )
p ≃
NkT
V (
1 −
N − 1
2
Na′

pV2 )
(
p +
a′

′

V2 )
V ≃ NkT
a′

′

=
N − 1
2
Na′


8. Quiz3
Tr
a
nsl
a
tion
a
l
a
nd Vibr
a
tion
a
l P
a
rtition
Gas has translation and vibrational state, de
fi
ne the partition function and heat capacity
Answer
Partition function of vibrational and translational
Z = ZtransZrot
J is quantum number of angular momentum

9. Calor Capacity is expressed by
Find the energy
E = NkT2 ∂
∂T
ln
V
h3
(2πmkT)3/2
∑
j
(2j + 1)e−j(j+1)h2
/8π2
kT
E = NkT2 ∂
∂T
3
2
lnT +
∂
∂T
ln
∑
j
(2j + 1)e−j(j+1)h2
/8π2
kT If T is ~ then
becomes 0
E =
3
2
NkT2 ∂
∂T
lnT E =
3
2
NkT2 1
T
E =
3
2
NkT
Cv =
dE
dT
Cv =
d(
3NkT
2
)
dT
Cv =
3Nk
2
For more information, please read
page 181-189 from Diktat

10. Quiz4
Semi Cl
a
ssic
a
l
Find Z of semi classical gas and de
fi
ne the energy and pressure
Answer
Partition function of vibrational, translational, rotation, electronic and spin
Z = ZtransZrotZvibZelecZspin
Example, we calculate only translation
Semi classical gas can not be di
ff
erentiate from others
Ztrans =
V
h3
(2πmkT)3/2
)N
N!
E = kT2 ∂
∂T
ln(
V
h3
(2πmkT)3/2
)N
N!
)
E =
3
2
NkT

11. E0 = 1 state non generate (0 eV)
E1 = 3 states
Quiz5
Prob
a
bility
δE = 0.66eV
E1 = 0.66eV
E0 = 0eV
E1 − E0 = 0.66eV
E1 − 0eV = 0.66eV
E1 = 0.66eV
Remember
p = e
−E
kT
Probability
At 44 K
P generate
pnongenerate = e
0eV
k * 44K = 1
pdegenerate = e
0.66eV
k * 44K
Pnon−generate =
1 * Pnon−generate
1 * Pnon−generate + 3 * Pdegenerate
Please, calculate by yourself as a
training
At 77 K
P generate
pgenerate = e
0eV
k * 44K = 1
pdegenerate = e
0.66eV
k * 44K
Pnon−generate =
1 * Pnon−generate
1 * Pnon−generate + 3 * Pdegenerate
Please, calculate by yourself as a
training