Answer to Question #168675 in Physics for Desmond

Question

Derive and explain clausius-clapeyron equation

Answer 1

The derivation will be given for a liquid-vapor equilibrium interface but it equally well applies to the interface between any two phases.

Let sv and sl be the specific entropy of the vapor and liquid phases, respectively. The pressure and temperature of the two phases are equal. The chemical potential μ is equal on either side of the phase boundary curve. Therefore the changes dμ in the chemical potential for movements along the phase boundary curve are also equal. This means that

$$dμ_v = V_vdp - s_vdT = V_ldp - s_ldT = dμ_l$$

Solving for dp gives

$$dp =\frac{(s_v - s_l)dT}{V_v - V_l}$$

But the change in entropy Δs for the phase change is just $\frac{L}{T}$ so

$$\frac{dp}{dT} =\frac{L}{T(V_v - V_l)}$$

This is the Clausius-Clapeyron equation.