Question

Derive Clausius-Clapeyron equation for the
first-order phase transition.

Answer 1

Two phases are in equilibrium when

$$\mu_1(T,P)=\mu_2(T,P)$$

So

$$d\mu_1(T,P)=d\mu_2(T,P)$$

$$\left(\frac{\partial \mu_1}{\partial T}\right)_P dT+\left(\frac{\partial \mu_1}{\partial P}\right)_T dP=\left(\frac{\partial \mu_2}{\partial T}\right)_P dT+\left(\frac{\partial \mu_2}{\partial P}\right)_T dP$$

Using Maxwell's relations

$$\left(\frac{\partial \mu}{\partial T}\right)_P=-S$$

$$\left(\frac{\partial \mu}{\partial P}\right)_T =V$$

we obtain

$$-S_1dT+V_1dP=-S_2 dT+V_2 dP$$

Finally

$$\frac{dP}{dT}=\frac{S_2-S_1}{V_2-V_1}$$

Since

$$T\Delta S=Q$$

we also have

$$\frac{dP}{dT}=\frac{Q}{T(V_2-V_1)}$$