Clausius–Clapeyron relation

Remarks

On the curve the chemical potentials are equal, $\mu_\alpha(T, p(T)) = \mu_\beta(T, p(T))$. Differentiating along the curve and using $d\mu = -s\, dT + v\, dp$ gives $(s_\beta - s_\alpha)\, dT = (v_\beta - v_\alpha)\, dp$, which is part 1. Part 2 follows by putting $v_\beta - v_\alpha \approx RT/p$. If $L$ is also constant, integration gives $\ln(p_2/p_1) = -\frac{L}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$. For melting of ice, $L > 0$ but $v_{\text{liquid}} < v_{\text{ice}}$, so the melting curve has negative slope: higher pressure lowers the melting point. The hypothesis $v_\beta \ne v_\alpha$ is needed; at a critical point both $L$ and $v_\beta - v_\alpha$ tend to $0$.

Going deeper

For a student

Part 1 is exact and holds for any first-order transition of a pure substance, including solid–solid ones. Part 2 is the form used for vapour pressure curves: plotting $\ln p$ against $1/T$ gives nearly a straight line of slope $-L/R$. This is a standard way to measure latent heats. For continuous transitions, where $L = 0$ and $\Delta v = 0$, the relation becomes $0/0$, and it is replaced by the Ehrenfest equations.