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The red line in the P-T diagram is the coexistence curve of two phases: I and II. For instance, II is the vapor and I is the liquid phase of the same compound.

The Clausius–Clapeyron relation, is a differential equation for a phase transition between two phases of a single compound. In a pressure–temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of this curve.

${\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {Q}{T\,(V^{II}-V^{I})}}$

where $\mathrm {d} P/\mathrm {d} T\,$ is the slope of the coexistence curve, Q is the molar heat of transition (heat necessary to bring one mole of the compound from phase I into phase II), T is the absolute temperature, V^I is the molar volume of phase I at given pressure and temperature.

The equation is named after Émile Clapeyron, who defined it around 1834, and Rudolf Clausius.

Derivation

The condition of thermodynamical equilibrium at constant pressure P and constant temperature T between two phases I and II is the equality of the molar Gibbs free energies G,

$G^{I}(P,T)=G^{II}(P,T).\,$

The molar Gibbs free energy of phase α (I or II) is equal to the chemical potential μ^α of this phase. Hence the equilibrium condition can be written as,

$\mu ^{I}(P,T)=\mu ^{II}(P,T),\;$

which holds along the red line in the figure.

If we go reversibly along the lower and upper green line in the P-T diagram, the chemical potentials of the phases change by Δμ^I and Δμ^II, for phase I and II, respectively, while the system stays in equilibrium,

$\mu ^{I}(P,T)+\Delta \mu ^{I}(P,T)=\mu ^{II}(P,T)+\Delta \mu ^{II}(P,T)\;\Longrightarrow \;\Delta \mu ^{I}(P,T)=\Delta \mu ^{II}(P,T)$

From classical thermodynamics it is known that

$\Delta \mu ^{\alpha }(P,T)=\Delta G^{\alpha }(P,T)=-S^{\alpha }\Delta T+V^{\alpha }\Delta P,\quad {\hbox{where}}\quad \alpha =I,II.$

Here S^α is the molar entropy (entropy per mole) of phase α and V^α is the molar volume (volume of one mole) of this phase. It follows that

$-S^{I}\Delta T+V^{I}\Delta P=-S^{II}\Delta T+V^{II}\Delta P\;\Longrightarrow \;{\frac {\Delta P}{\Delta T}}={\frac {S^{II}-S^{I}}{V^{II}-V^{I}}}$

From the second law of thermodynamics it is known that for a reversible phase transition it holds that

$S^{II}-S^{I}={\frac {Q}{T}}$

where Q is the amount of heat necessary to convert one mole of compound from phase I into phase II. Clearly, when phase I is liquid and phase II is vapor then Q ≡ H_v is the molar heat of vaporization. Elimination of the entropy and taking the limit of infinitesimally small changes in T and P gives the Clausius-Clapeyron equation,

${\frac {dP}{dT}}={\frac {Q}{T(V^{II}-V^{I})}}$

Approximate solution

The Clausius-Clapeyron equation is exact. When we make the following assumptions we may perform the integration:

• The molar volume of phase I is negligible compared to the molar volume of phase II, V^II >> V^I

• Phase II behaves like an ideal gas

$PV^{II}=RT\,$

• The transition heat Q is constant over the temperature integration interval. The integration runs from the lower temperature T₁ to the upper temperature T₂.

Under these condition the Clausius-Clapeyron equation becomes

${\frac {dP}{dT}}={\frac {Q}{\frac {RT^{2}}{P}}}\;\Longrightarrow \;{\frac {dP}{P}}={\frac {Q}{R}}\;{\frac {dT}{T^{2}}}$

Integration gives

$\int \limits _{P_{1}}^{P_{2}}{\frac {dP}{P}}={\frac {Q}{R}}\;\int \limits _{T_{1}}^{T_{2}}{\frac {dT}{T^{2}}}\;\Longrightarrow \;\ln {\frac {P_{2}}{P_{1}}}=-{\frac {Q}{R}}\;\left({\frac {1}{T_{2}}}-{\frac {1}{T_{1}}}\right)$

where ln is the natural (base e) logarithm.