# 1.0.s: The Clapeyron Equation

The analysis in the two previous sections can be repeated for any phase change of a pure substance. Let $$\alpha$$ and $$\beta$$ denote the two phases that are at equilibrium.

$\alpha \rightleftharpoons \beta$

Let $$\overline{G}_{\alpha }$$, $$\overline{S}_{\alpha }$$, and $${\overline{V}}_{\alpha }$$ represent the Gibbs free energy, the entropy, and the volume of one mole of pure phase $$\alpha$$ at pressure $$P$$ and temperature $$T$$. Let $$\overline{G}_{\beta }$$, $$\overline{S}_{\beta }$$, and $$\overline{V}_{\beta }$$ represent the corresponding properties of one mole of pure phase $$\beta$$. The equations

$d\overline{G}\left(\alpha \right)=\overline{V}_\alpha dP-\overline{S}_{\alpha }dT$

and

$d\overline{G}\left(\beta \right)= \overline{V}_{\beta } dP- \overline{S}_{\beta }dT$

describe the changes in the Gibbs free energy of a mole of $$\alpha$$ and a mole of $$\beta$$ when they go from one $$\alpha-\beta$$-equilibrium state at $$P$$ and $$T$$ to a second $$\alpha-\beta$$-equilibrium state at $$P+dP$$ and $$T+dT$$. Since these Gibbs free energy changes must be equal, we have

\begin{align*} d\overline{G}\left(\beta \right)-d\overline{G}\left(\alpha \right) &=\left({\overline{V}}_{\beta }-{\overline{V}}_{\alpha }\right)dP-\left({\overline{S}}_{\beta }-\overline{S}_{\alpha }\right)dT \\[4pt] &=\Delta \overline{V}dP-\Delta \overline{S}dT \\[4pt] &=0 \end{align*}

and

$\frac{dP}{dT}=\frac{\Delta \overline{S}}{\Delta \overline{V}}$

where $$\Delta \overline{S}$$ and $$\Delta \overline{V}$$ are the entropy and volume changes that occur when one mole of the substance goes from phase $$\alpha$$ to phase $$\beta$$. Since $$\Delta \overline{S}=\Delta \overline{H}/T$$, the condition for equilibrium between phases $$\alpha$$ and $$\beta$$ becomes

$\frac{dP}{dT}=\frac{\Delta \overline{H}}{T\ \Delta \overline{V}} \label{Clap1}$

Equation \ref{Clap1} is known as the Clapeyron equation.