= Solution
At equilibrium between two <thermodynamic phases> of a pure substance, their molar <Gibbs free energies> are equal. Along a coexistence line,
$$
g_1(T,P)=g_2(T,P),\qquad
\mathrm dg=-s\,\mathrm dT+v\,\mathrm dP.
$$
Subtracting the two differentials gives
$$
-(s_2-s_1)\,\mathrm dT+(v_2-v_1)\,\mathrm dP=0.
$$
For a nonzero volume change,
$$
\boxed{\frac{\mathrm dP}{\mathrm dT}
=\frac{\Delta s}{\Delta v}
=\frac{\Delta h}{T\Delta v}.}
$$
The second equality uses $\Delta g=\Delta h-T\Delta s=0$ on coexistence. Total <entropy> and volume changes may be used instead if they refer to the same amount of material. The ratio must use one consistent transformation direction, so reversing the labels changes both signs and leaves the slope unchanged. This is the general <Clapeyron equation>; the ideal-vapour approximation often called Clausius-Clapeyron makes further assumptions not needed for two solids.
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