1000 Solved Problems in Modern Physics

266 4 Thermodynamics and Statistical Physics

From the law of equipartition of energy we have

dEt

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 ofCνandγ. From the law of equiparti-

tion of energy

Cν=

(3+β)

2

.

k

m

; Cp=

(5+β)

2

.

k

m

whenceγ=

Cp

Cν

= 1 +

2

3 +β

orβ=

5 − 3 γ

γ− 1

(4)

Furthermore

Cν=

k

m(γ−1)

(5)

Combining (3), (4) and (5)

K

ηCν

=

1

4

(9γ−5)

4.3.2 Maxwell’s Thermodynamic Relations ..............

4.21Letf(x,y)=0(1)

df=

(

∂f

∂x

)

y

dx+

(

∂f

∂y

)

x

dy=0(2)

Equation of state can be written asf(P,V,T)=0. By first law of thermody-

namics

dQ=dU+dW (3)

By second law of thermodynamics

dQ=Tds (4)

for infinitesimal reversible process

dW=pdV (5)