= Solution
At coexistence, the two phases of a pure substance have equal <chemical potential>, or equivalently equal molar <Gibbs free energy>:
$$
\mu_1(T,p)=\mu_2(T,p).
$$
For either phase, $d\mu=-s_m\,dT+v_m\,dp$. Move an infinitesimal distance along the coexistence curve, where the equality must remain true. Subtracting the two differentials gives
$$
0=-\left(s_{m,2}-s_{m,1}\right)dT+\left(v_{m,2}-v_{m,1}\right)dp,
\qquad \frac{dp}{dT}=\frac{\Delta s_m}{\Delta v_m}.
$$
A reversible transformation at this common <temperature> absorbs <latent heat> $L=T\Delta s_m$. Thus
$$
\boxed{\frac{dp}{dT}=\frac{\Delta s_m}{\Delta v_m}=\frac{L}{T\Delta v_m}.}
$$
The derivation assumes <thermodynamic equilibrium> between two bulk phases of a single pure substance, with the same <temperature> and <pressure>, and only ordinary pressure-volume work. Surface-curvature shifts, additional external-field work and composition changes are omitted. It applies to a first-order <phase boundary> where the <latent heat> and <volume> discontinuity are defined. No ideal-gas assumption and no constant-$L$ assumption are needed for this differential relation. At the critical endpoint both discontinuities vanish, so their ratio requires a limiting treatment rather than direct substitution of two zeros.
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