Problems and Solutions on Thermodynamics and Statistical Mechanics

Thermod pmics 81

we obtain

a2E a2E a2s - a2s

axaT aTaxl axaT aTax

From __ -- -

a2E

aTax

Thus (aE/a+ = t - T(at/aT),.
Substituting the expression for t, we have (aE/az)T = 0. It follows
that (ac,/az), = 0. Integrating, we get

E(z, T) = E(T) = 6 dE + E(To) = [ gdT + E(To)

= Lr KdT + E(To) = K(T - To) + E(To).

From

dS= %dT+T 1 [(a,),-t]dz aE

T

we find after integration

S(z, 2') = K In T - A - + - + const.

(;l: :)

(b) For an adiabatic process dS = 0 so that

After integration we have

= 0.292AZo ,

Hence fi = TO exp(0.292Alo/K).