Past exam of the materials science course of the University of Cambridge / 2018 / Paper 1 / 7 / b

For two phases of the same pure substance at equilibrium, their chemical potentials are equal. Along their coexistence boundary,Differentiating the equality givesTherefore the Clapeyron equation isMolar 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. 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.