= Solution
<Phase equilibrium> minimizes <Gibbs free energy> at specified <temperature>, <pressure> and composition. Coexisting phases have equal <temperature>, <pressure> and <chemical potential>. For a pure substance, a first-order transition has continuous equilibrium <Gibbs free energy> but a jump in its first derivatives, such as <entropy> or volume; the <latent heat> is $T\Delta S$. Continuous transitions lack that <latent heat> and can instead have singular or discontinuous higher responses.
The van der Waals molar equation
$$
p=\frac{RT}{v-b}-\frac a{v^2}
$$
combines an excluded-volume correction with attractive interactions. Its inflection conditions give
$$
\boxed{v_c=3b,\quad T_c=\frac{8a}{27Rb},\quad p_c=\frac{a}{27b^2}.}
$$
Below $T_c$, the oscillatory isotherm contains a mechanically unstable segment with $(\partial p/\partial v)_T>0$. Equilibrium instead replaces the loop by a constant-pressure coexistence line. Equality of chemical potentials gives the <Maxwell construction>,
$$
\int_{v_\ell}^{v_g}[p(v,T)-p_{\rm sat}]\,dv=0.
$$
On isothermal compression the gas reaches saturation, then converts progressively into liquid at fixed <pressure>. The mean <molar volume> obeys $v=(1-x)v_\ell+xv_g$, where $x$ is the gas fraction. Metastable superheated or supercooled states can persist when <nucleation> is delayed; they are not the global equilibrium selected by this construction.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-2-b2-phase-coexistence.png]
{title=Illustrative reduced van der Waals coexistence and a generic pure-substance phase diagram}
A <pressure>–<temperature> <phase diagram> contains coexistence curves separating solid, liquid and gas. Three phases meet at a <triple point>; the liquid–gas coexistence curve ends at a <thermodynamic critical point>, above which liquid and gas can be connected without crossing a first-order phase boundary. A generic solid–liquid line continues beyond the liquid–gas critical <temperature>. For one component the <Gibbs phase rule> is $F=3-P$, so a three-phase equilibrium fixes both <temperature> and <pressure>.
Along a two-phase coexistence curve, $d\mu=-s\,dT+v\,dp$ for each phase gives
$$
\boxed{\frac{dp}{dT}=\frac{\Delta s}{\Delta v}=\frac{L}{T\Delta v}.}
$$
This is the Clapeyron equation, valid with consistent molar or specific quantities. For vaporization, neglecting the liquid volume and treating the vapour as ideal gives $\Delta v\simeq RT/p$, hence $d\ln p/dT=L/(RT^2)$. With approximately constant molar <latent heat>,
$$
\boxed{\ln\frac{p_2}{p_1}=-\frac LR\left(\frac1{T_2}-\frac1{T_1}\right).}
$$
This Clausius–Clapeyron approximation explains the rapid rise of vapour <pressure> with <temperature>. It fails close to the <thermodynamic critical point>, where phase volumes converge, and is not a universal expression for melting. The sign of a melting-line slope depends on its volume change; water's denser liquid gives a familiar negative slope.
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