Clausius-Clapeyron relation

The Clausius–Clapeyron relation, is an 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 $$dP/dT\,$$ of this curve:



\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{Q}{T\,(V^{II} - V^{I})} , $$ where 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 (the abscissa of the point where the slope is computed), VI is the molar volume of phase I at the pressure and temperature of the point where the slope is considered and VII is the same for phase II. The quantity P is the absolute pressure.

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^{I}(P,T) = G^{II}(P,T). \, $$ The molar Gibbs free energy of phase &alpha; (I or II) is equal to the chemical potential &mu;&alpha; of this phase. Hence the equilibrium condition can be written as,
 * $$ \mu^{I}(P,T) = \mu^{II}(P,T), \;

$$ which holds along the red (coexistence) line in the figure.

If we go reversibly along the lower and upper green line in the figure, the chemical potentials of the phases change by &Delta;&mu;I and &Delta;&mu;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&alpha; is the molar entropy (entropy per mole) of phase &alpha; and V&alpha; 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 &equiv; Hv 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, VII >> VI


 * Phase II satisfies the  ideal gas law

PV^{II} = R T \, $$
 * The transition heat Q is constant over the temperature integration interval. The integration runs 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\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.