Problems and Solutions on Thermodynamics and Statistical Mechanics

Thermodynamics 139

Fig. 1.39.

Solution:
From the Clausius-Clapeyron equation, we have

dp AS and so AS = AV- dP.

dT - AV’ dT

When He3 melts, the volume increases, i.e., AV > 0.

When 0.02 K< T < 0.32 K, because - < 0,AS < 0.

When 0.32 K < T < 1.2 K, because - > 0,AS > 0.

When T = 0.32 K, AS = 0. The results are shown in Fig. 1.39(b).

dP

dP

dT

dT

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The phase transition between the aromatic (a) and fragrant (f) phases

of the liquid mythological-mercaptan is second order in the Ehrenfest scheme,

that is, AV and AS are zero at all points along the transition line p,-f(T).

Use the fact that AV = V,(T,p)-Vf(T,p) = 0, where V, and Vf are the

molar volumes in phase a and phase f respectively, to derive the slope of the

transition line, dp,-t(T)/dT, in terms of changes in the thermal expansion

coefficient, a, and the isothermal compressibility, kT at the transition.

(MIT)

Solution:

Along the transition line, one has

Va(T1 P) = &(TIP)

Thus dV,(T,p) = dVf(T,p).

Since