= Solution
For two phases of the same pure substance at equilibrium, their chemical potentials are equal. Along their coexistence boundary,
$$
\mu_1=\mu_2,\qquad d\mu=-s_m\,dT+v_m\,dp.
$$
Differentiating the equality gives
$$
-s_{m,1}\,dT+v_{m,1}\,dp
=-s_{m,2}\,dT+v_{m,2}\,dp.
$$
Therefore the <Clapeyron equation> is
$$
\boxed{\frac{dp}{dT}=\frac{\Delta s_m}{\Delta v_m}
=\frac{\Delta s}{\Delta v}
=\frac{L}{T\Delta v}.}
$$
Molar or mass-specific quantities give the same slope if a consistent basis is used. $L=T\Delta s$ follows from equal Gibbs free energies at coexistence.
Solid phases are relatively incompressible and have modest <thermal expansion> over many intervals. If their <entropy> and volume differences vary slowly enough that $\Delta s/\Delta v$ is approximately constant, the boundary is approximately straight. \b[Straightness is an approximation, not a thermodynamic requirement.] Heat-capacity differences, pressure dependence of volume, other transitions and proximity to special points can curve the boundary. The ideal-vapor integrated Clausius-Clapeyron approximation is not appropriate for these two solid phases.
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