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4 Thermodynamics and Statistical Physics
From the law of equipartition of energy we have
dE t
dT
=
3
2
k;
dE
dT
=
β
2
k
(2)
Hence,
K
η
=
5
2
.
3
2
+
β
2
k
m
(3)
We can express the result in terms of C ν and γ . From the law of equipartition of energy
C ν =
(3 + β)
2
.
k
m
; C p =
(5 + β)
2
.
k
m
whence γ =
C p
C ν
= 1 +
2
3 + β
or β =
5 − 3γ
γ − 1
(4)
Furthermore
C ν =
k
m(γ − 1)
(5)
Combining (3), (4) and (5)
K
ηC ν
=
1
4
(9γ − 5)
4.3.2 Maxwell’s Thermodynamic Relations
4.21 Let f (x, y) = 0
( 1 )
d f =
∂ f
∂ x
y
dx +
∂ f
∂ y
x
dy = 0
( 2 )
Equation of state can be written as f (P, V, T ) = 0. By first law of thermodynamics
dQ = dU + dW
(3)
By second law of thermodynamics
dQ = T ds
(4)
for infinitesimal reversible process
dW = pdV
(5)