Clausius-Clapeyron relation

PD Image
The red line in the P-T diagram is the coexistence curve of two phases, I and II, of a single system. Phase II may be the vapor and I the liquid phase of the same compound. The lower green line gives the slope in phase I and the upper green line in phase II.

The Clausius–Clapeyron relation is an equation for a system consisting of two phases of a single compound in thermodynamic equilibrium at constant absolute temperature T and constant pressure P. A curve in a two-dimensional thermodynamical diagram that separates two phases in equilibrium is known as a coexistence curve. The Clausius–Clapeyron relation gives the slope of the coexistence curve in the P-T diagram:

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

where Q is the molar heat of transition. It is the heat necessary to bring one mole of the compound from phase I into phase II; it is also known as latent heat. For instance, when phase I is a liquid and phase II is a vapor, then Q ≡ H_v is the molar heat of vaporization. Further, V^I is the molar volume of phase I at the pressure P and temperature T of the point where the slope is considered and V^II is the same for phase II.

The equation is named after Émile Clapeyron, who derived 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,

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

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

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

which holds everywhere along the coexistence (red) curve in the figure.

If we go reversibly along the lower and upper green line in the figure [the tangents to the curve in the point (T,P)], the chemical potentials of the phases, being functions of T and P, change by Δμ^I and Δμ^II, for phase I and II, respectively, while the system stays in equilibrium,

$\mu ^{\mathrm {I} }+\Delta \mu ^{\mathrm {I} }=\mu ^{\mathrm {II} }+\Delta \mu ^{\mathrm {II} }\;\Longrightarrow \;\Delta \mu ^{\mathrm {I} }=\Delta \mu ^{\mathrm {II} }.$

From classical thermodynamics it is known that

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

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

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

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

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

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

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

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. In general, far from the critical point, this inequality holds well for liquid-gas equilibria.

• Phase II satisfies the ideal gas law, where R is the molar gas constant,

$PV^{\mathrm {II} }=RT.\,$

If phase II is a gas and the pressure is fairly low, this assumption is reasonable.

• The transition (latent) heat Q is constant over the temperature integration interval. The integrations run from the lower temperature T₁ to the upper temperature T₂ and from P₁ to P₂.

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

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

where ln(P₂/P₁) is the natural (base e) logarithm of P₂/P₁. We reiterate that for gas-liquid transitions Q = H_v, the heat of vaporization.