= Clausius-Clapeyron relation
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{title2=$dp/dT=L/(T\Delta v)$}
{wiki=Clausius–Clapeyron_relation}
= Clapeyron equation
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{synonym}
= Clausius-Clapeyron equation
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{synonym}
The <Clapeyron equation> connects the slope of a <phase coexistence curve> to the <latent heat> and the change in <volume (thermodynamics)> during a <first-order phase transition>. A steep boundary can indicate a large <latent heat>, a small change in <volume (thermodynamics)>, or both. This lets a measured boundary constrain caloric properties and predicts how changing <pressure> shifts a <melting> <temperature>.
For a pure substance in bulk <phase coexistence>, label the initial and final <thermodynamic phases> $\alpha$ and $\beta$. Both have the same absolute <temperature> $T$, <pressure> $p$, and <chemical potential>. With $s$, $h$, and $v$ denoting <entropy>, <enthalpy>, and <volume (thermodynamics)> on one common basis, define $\Delta s=s_\beta-s_\alpha$, $\Delta h=h_\beta-h_\alpha$, and $\Delta v=v_\beta-v_\alpha$. Then
$$
\boxed{\frac{dp}{dT}=\frac{\Delta s}{\Delta v}=\frac{L}{T\Delta v}},\qquad L=\Delta h=T\Delta s.
$$
The <derivative> follows the <phase coexistence curve>; it is not a <partial derivative> within one <thermodynamic phase>. Use <molar volume> $v_m$ with molar <latent heat> $L_m$ in $\mathrm{J\,mol^{-1}}$, or <specific volume> $v$ with <latent heat> $L$ in $\mathrm{J\,kg^{-1}}$. Both give $dp/dT$ in $\mathrm{Pa\,K^{-1}}$. For <molar mass> $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a <simple compressible system>, fixed composition, and negligible interface contributions; a curved interface can instead require the <Young–Laplace equation>.
For positive <latent heat> in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting <liquid> to <gas> normally increases <volume (thermodynamics)> and gives a positive slope. The <negative melting-point slope of water> illustrates the opposite sign. At a <thermodynamic critical point>, the coexisting <gas> and <liquid> merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the <Clapeyron equation>.<Statistical Physics by David Tong>[§5.1.2, equation (5.185) and the discussion of the critical endpoint]{ref}
\Image[/clapeyron-coexistence-slopes-and-volume-jump.png]
{title=<Phase coexistence curve> tangents: positive vaporization slope and negative <water> <melting> slope}
A short derivation uses the molar <Gibbs free energy> $g_m$, equal to the <chemical potential> for a pure substance. Equality persists along the boundary, so $dg_{m,\alpha}=dg_{m,\beta}$ with $dg_m=-s_m\,dT+v_m\,dp$. Therefore $-\Delta s_m\,dT+\Delta v_m\,dp=0$. Equality of $g_m=h_m-Ts_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the <Clapeyron equation>.
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