The Clapeyron Equation
The Clapeyron attempts to answer the question of what the shape of a two-phase coexistence line is. In the \(P-T\) plane, we see the a function \(P(T)\), which gives us the dependence of \(P\) on \(T\) along a coexistence curve.
Consider two phases, denoted \(\alpha\) and \(\beta\), in equilibrium with each other. These could be solid and liquid, liquid and gas, solid and gas, two solid phases, et. Let \(\mu_\alpha (P, T)\) and \(\mu_\beta (P, T)\) be the chemical potentials of the two phases. We have just seen that
\[\mu_\alpha (P, T) = \mu_\beta (P, T) \label{14.1} \]
Next, suppose that the pressure and temperature are changed by \(dP\) and \(dT\). The changes in the chemical potentials of each phase are
\[ d \mu_{\alpha} (P, T) = d \mu_{\beta} (P, T) \label{14.2a} \]
\[\left( \dfrac{\partial \mu_{\alpha}}{\partial P} \right)_T dP + \left( \dfrac{\partial \mu_{\alpha}}{\partial T} \right)_P dT = \left( \dfrac{\partial \mu_{\beta}}{\partial P} \right)_T dP + \left( \dfrac{\partial \mu_{\beta}}{\partial T} \right)_P dT \label{14.2b} \]
However, since \(G(n, P, T) = n \mu (P, T)\), the molar free energy \(\bar{G} (P, T)\), which is \(G(n, P, T)/n\), is also just equal to the chemical potential
\[\bar{G} (P, T) = \dfrac{G(n, P, T)}{n} = \mu (P, T) \label{14.3} \]
Moreover, the derivatives of \(\bar{G}\) are
\[\left( \dfrac{\partial \bar{G}}{\partial P} \right)_T = \bar{V}, \: \: \: \: \: \: \: \left( \dfrac{\partial \bar{G}}{\partial T} \right)_P = -\bar{S} \label{14.4} \]
Applying these results to the chemical potential condition in Equation \(\ref{14.2b}\), we obtain
\[\begin{align} \left( \dfrac{\partial \bar{G}_\alpha}{\partial P} \right)_T dP + \left( \dfrac{\partial \bar{G}_\alpha}{\partial T} \right)_P dT &= \left( \dfrac{\partial \bar{G}_\beta}{\partial P} \right)_T dP + \left( \dfrac{\partial \bar{G}_\beta}{\partial T} \right)_P dT \\[5pt] \bar{V}_\alpha dP - \bar{S}_\alpha dT &= \bar{V}_\beta dP - \bar{S}_\beta dT \end{align} \label{14.5} \]
Dividing through by \(dT\), we obtain
\[\begin{align} \bar{V}_\alpha \dfrac{\partial P}{\partial T} - \bar{S}_\alpha &= \bar{V}_\beta \dfrac{\partial P}{\partial T} - \bar{S}_\beta \\[5pt] (\bar{V}_\alpha - \bar{V}_\beta) \dfrac{\partial P}{\partial T} &= \bar{S}_\alpha - \bar{S}_\beta \\[5pt] \dfrac{dP}{dT} &= \dfrac{\bar{S}_\alpha - \bar{S}_\beta}{\bar{V}_\alpha - \bar{V}_\beta} \end{align} \label{14.6} \]
The importance of the quantity \(dP/dT\) is that is represents the slope of the coexistence curve on the phase diagram between the two phases. Now, in equilibrium \(dG = 0\), and since \(G = H - TS\), it follows that \(dH = T \: dS\) at fixed \(T\). In the narrow temperature range in which the two phases are in equilibrium, we can assume that \(H\) is independent of \(T\), hence, we can write \(S = H/T\). Consequently, we can write the molar entropy difference as
\[\bar{S}_\alpha - \bar{S}_\beta = \dfrac{\bar{H}_\alpha - \bar{H}_\beta}{T} \label{14.7} \]
and the pressure derivative \(dP/dT\) becomes
\[\dfrac{dP}{dT} = \dfrac{\bar{H}_\alpha - \bar{H}_\beta}{T (\bar{V}_\alpha - \bar{V}_\beta)} = \dfrac{\Delta_{\alpha \beta} \bar{H}}{T \Delta_{\alpha \beta} \bar{V}} \label{14.8} \]
a result known as the Clapeyron equation, which tells us that the slope of the coexistence curve is related to the ratio of the molar enthalpy between the phases to the change in the molar volume between the phases. If the phase equilibrium is between the solid and liquid phases, then \(\Delta_{\alpha \beta} \bar{H}\) and \(\Delta_{\alpha \beta} \bar{V}\) are \(\Delta \bar{H}_\text{fus}\) and \(\Delta \bar{V}_\text{fus}\), respectively. If the phase equilibrium is between the liquid and gas phases, then \(\Delta_{\alpha \beta} \bar{H}\) and \(\Delta_{\alpha \beta} \bar{V}\) are \(\Delta \bar{H}_\text{vap}\) and \(\Delta \bar{V}_\text{vap}\), respectively.
For the liquid-gas equilibrium, some interesting approximations can be made in the use of the Clapeyron equation. For this equilibrium, Equation \(\ref{14.8}\) becomes
\[\dfrac{dP}{dT} = \dfrac{\Delta \bar{H}_\text{vap}}{T (\bar{V}_g - \bar{V}_l)} \label{14.9} \]
In this case, \(\bar{V}_g \gg \bar{V}_l\), and we can approximate Equation \(\ref{14.9}\) as
\[\dfrac{dP}{dT} \approx \dfrac{\Delta \bar{H}_\text{vap}}{T \bar{V}_g} \label{14.10} \]
Suppose that we can treat the vapor phase as an ideal gas. Certainly, this is not a good approximation so close to the vaporization point, but it leads to an example we can integrate. Since \(PV_g = nRT\), \(P \bar{V}_g = RT\), Equation \(\ref{14.10}\) becomes
\[\begin{align} \dfrac{dP}{dT} &= \dfrac{\Delta \bar{H}_\text{vap} P}{RT^2} \\[5pt] \dfrac{1}{P} \dfrac{dP}{dT} &= \dfrac{\Delta \bar{H}_\text{vap}}{RT^2} \\[5pt] \dfrac{d \: \text{ln} \: P}{dT} &= \dfrac{\Delta \bar{H}_\text{vap}}{RT^2} \end{align} \label{14.11} \]
which is called the Clausius-Clapeyron equation. We now integrate both sides, which yields
\[\text{ln} \: P = -\dfrac{\Delta \bar{H}_\text{vap}}{RT} + C \nonumber \]
where \(C\) is a constant of integration. Exponentiating both sides, we find
\[P(T) = C' e^{-\Delta \bar{H}_\text{vap}/RT} \nonumber \]
which actually has the wrong curvature for large \(T\), but since the liquid-vapor coexistence line terminates in a critical point, as long as \(T\) is not too large, the approximation leading to the above expression is not that bad.
If we, instead, integrate both sides, the left from \(P_1\) to \(P_2\), and the right from \(T_1\) to \(T_2\), we find
\[\begin{align} \int_{P_1}^{P_2} d \: \text{ln} \: P &= \int_{T_1}^{T_2} \dfrac{\Delta \bar{H}_\text{vap}}{RT^2} dT \\[5pt] \text{ln} \: \left( \dfrac{P_2}{P_1} \right) &= -\dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \\[5pt] &= \dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{T_1 - T_1}{T_1 T_2} \right) \end{align} \label{14.12} \]
assuming that \(\Delta \bar{H}_\text{vap}\) is independent of \(T\). Here \(P_1\) is the pressure of the liquid phase, and \(P_2\) is the pressure of the vapor phase. Suppose we know \(P_2\) at a temperature \(T_2\), and we want to know \(P_3\) at another temperature \(T_3\). The above result can be written as
\[\text{ln} \: \left( \dfrac{P_3}{P_1} \right) = -\dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{1}{T_3} - \dfrac{1}{T_1} \right) \label{14.13} \]
Subtracting the two results, we obtain
\[\text{ln} \: \left( \dfrac{P_2}{P_3} \right) = -\dfrac{\Delta \bar{H}_\text{vap}}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_3} \right) \label{14.14} \]
so that we can determine the vapor pressure at any temperature if it is known as one temperature.
In order to illustrate the use of this result, consider the following example:
At \(1 \: \text{bar}\), the boiling point of water is \(373 \: \text{K}\). At what pressure does water boil at \(473 \: \text{K}\)? Take the heat of vaporization of water to be \(40.65 \: \text{kJ/mol}\).
Solution
Let \(P_1 = 1 \: \text{bar}\) and \(T_1 = 373 \: \text{K}\). Take \(T_2 = 473 \: \text{K}\), and we need to calculate \(P_2\). Substituting in the numbers, we find
\[\begin{align} \text{ln} \: P_2(\text{bar}) &= -\dfrac{(40.65 \: \text{kJ/mol})(1000 \: \text{J/kJ})}{8.3145 \: \text{J/mol} \cdot \text{K}} \left( \dfrac{1}{473 \: \text{K}} - \dfrac{1}{373 \: \text{K}} \right) = 2.77 \\[5pt] P_2(\text{bar}) &= (1 \: \text{bar}) \: e^{2.77} = 16 \: \text{bar} \end{align} \nonumber \]