# Clausius-Clapeyron relation

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The red line in the P-T diagram is the coexistence curve of two phases, I and II, of a single-component system. Phase I may be the vapor and II the liquid phase of the component. 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 single-component system[1] consisting of two phases 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:

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

where Q is the molar heat of transition. It is the heat necessary to bring one mole of the compound constituting the system from phase II into phase I; it is also known as latent heat. For phase transitions with constant P, it is the molar transition enthalpy. For instance, when phase II is a liquid and phase I is a vapor, then QHv is the molar heat of vaporization (also known as molar enthalpy of evaporation). 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 published it in 1834, and Rudolf Clausius, who put it on firm thermodynamic basis in 1850.

## 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,

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

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

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

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 ΔμI and ΔμII, for phase I and II, respectively, while the system stays in equilibrium,

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

From classical thermodynamics it is known that

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

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

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

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

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

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

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

## Approximate solution

The Clausius-Clapeyron equation is exact. When the following assumptions are made, it may be integrated:

• The molar volume of phase II is negligible compared to the molar volume of phase I: V I >> V II. In general, far from the critical point, this inequality holds well for liquid-gas equilibriums.
${\displaystyle PV^{\mathrm {I} }=RT.\,}$
If phase I 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

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

Integration gives

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

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

For example, at the top of Mount Everest, atmospheric pressure is about a third of its value at sea level. Using the values R = 8.3145 J/K and Q = 40.65 kJ/mol, the equation gives T2 = 1/[1/373.16 + 8.3145 ln(3) / (40.65 103) ] = 344 K (71 °C) for the boiling temperature of water. It takes many hours to boil an egg on top of the Mount Everest in water of 71 °C.

## Application

PD Image
P-T diagram of water. Schematic and qualitative.

In the figure on the right an impression of part of the phase diagram of water (H2O) is shown. The three phases are: I (vapor), II (liquid) and III (solid). The Clausius-Clapeyron equation

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

gives the slope of the three coexistence (equilibrium, red) curves. The boiling line T-C has a positive slope, because the heat of vaporization Q = Hv is positive and the molar volume of the vapor V I is (much) larger than that of the liquid V II. For a similar reason the sublimation (transition from solid to vapor) line T-S has a positive slope. The sublimation line T-S is steeper than the boiling line T-C because the heat of sublimation is larger than the heat of vaporization, while in the neighborhood of crossing point T (the triple point) the difference in molar volumes is nearly equal. The melting line T-M has a negative slope, which occurs only for a few compounds (among them water). Consider

${\displaystyle {\frac {dP}{dT}}={\frac {Q_{\mathrm {melt} }}{T(V^{\mathrm {II} }-V^{\mathrm {III} })}}={\frac {S^{\mathrm {II} }-S^{\mathrm {III} }}{V^{\mathrm {II} }-V^{\mathrm {III} }}}}$

The molar volume VIII of the solid is larger than the molar volume VII of the liquid, while the enthalpy of melting Qmelt = SIISIII > 0 (a liquid has a larger entropy than the corresponding solid, it costs energy to melt a solid). Compounds that have melting lines with negative slope, have a solid phase of lower density than the liquid phase. For example, it is very well-known that solid water (ice) floats on liquid water.

The point C is the critical point. Beyond this point the vapor and the liquid are one indistinguishable fluid. There is no phase transition or a phase interface to the right of the melting line and above the line of constant P = Pc. The gas/liquid fluid also appears to the right of the line of constant T = Tc.

For water the critical point C is at (Tc, Pc) = (647.1 K, 220.6 bar) (a bar is almost equal to an atmosphere and is 100 kpascal) and the triple point T is at (Tt, Pt) = (273.16 K, 6.1173 mbar).

The Clausius-Clapeyron relation also holds for transitions between different crystalline phases.

## Reference

1. See J.C.M. Li, Clapeyron Equation for Multicomponent Systems, Journal of Chemical Physics, vol. 25, pp. 572–574 (1956) for a generalization to systems of more than one component.