= Solution
The <Clapeyron equation> gives the slope of a two-phase <phase coexistence curve>. Let $s_l,s_v$ and $v_l,v_v$ be <entropy> and <volume> per molecule, or consistently per mole. Equality of the two <chemical potentials> and $d\mu=-s\,dT+v\,dP$ imply
$$
0=d(\mu_v-\mu_l)=-(s_v-s_l)dT+(v_v-v_l)dP.
$$
At reversible coexistence the latent <heat> in the same amount convention is $\mathcal L=T(s_v-s_l)$. Therefore
$$
\boxed{\frac{dP}{dT}=\frac{\mathcal L}{T\Delta v},\qquad\Delta v=v_v-v_l.}
$$
The relation assumes <thermodynamic equilibrium> between two bulk phases of a pure substance, neglecting curvature and other work terms. For vaporization both latent <heat> and <volume> change are positive, giving a positive vapor-pressure slope. If the vapor is approximately ideal and its <volume> dominates, the <ideal-vapor Clausius-Clapeyron approximation> gives $d\ln P/dT\simeq\mathcal L/(k_BT^2)$ with latent <heat> per molecule.
Experimentally, a sealed two-phase sample in a temperature-controlled bath supplies the coexistence <pressure> as <temperature> changes. Independent calorimetry and phase-density measurements supply $\mathcal L$ and $\Delta v$, allowing the <pressure> slope to test the equation. The formula connects those distinct measurements and is a practical route to latent heats.
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