Derivation of Clausius-Clapeyron Equation

The Clausius-Clapeyron equation was initially proposed by German physicist Rudolf Clausius in 1834 and further developed by French physicist Benoît Clapeyron in 1850. This equation is extremely useful in characterizing a discontinuous phase transition between two phases of a single constituent.

We know from fundamental thermodynamics that the total variation of Gibbs free energy ($G$) is expressed as:

$$dG = VdP - SdT \quad \text{--- (Equation 1)}$$

Let us consider a single-constituent equilibrium containing two balancing state phases:

$\text{Phase-1} \rightleftharpoons \text{Phase-2}$

Where Phase-1 may be solid, liquid, or gas, and Phase-2 may be liquid or vapor, depending upon the nature of the phase transformation (melting, vaporization, or sublimation).

For Phase-1, the differential change in free energy is given by:

$$dG_1 = V_1dP - S_1dT \quad \text{--- (Equation 2)}$$

And for Phase-2, the corresponding change in free energy is:

$$dG_2 = V_2dP - S_2dT \quad \text{--- (Equation 3)}$$

At equilibrium, the system exhibits zero net change in free energy ($dG_1 = dG_2$, i.e., $\Delta G = 0$). Equating Equation 2 and Equation 3 yields:

$$V_2dP - S_2dT = V_1dP - S_1dT$$ $$(V_2 - V_1)dP = (S_2 - S_1)dT$$ $$\Delta V \cdot dP = \Delta S \cdot dT$$ $$\frac{dP}{dT} = \frac{\Delta S}{\Delta V} \quad \text{--- (Equation 4)}$$

If $\Delta H$ is the latent heat of phase transformation taking place at transition temperature ($T$), then the entropy change ($\Delta S$) is defined as:

$$\Delta S = \frac{\Delta H}{T} \quad \text{--- (Equation 5)}$$

Substituting the expression for $\Delta S$ from Equation 5 into Equation 4 yields the Clapeyron Equation:

$$\frac{dP}{dT} = \frac{\Delta H}{T \cdot \Delta V} \quad \text{--- (Equation 6)}$$

Equation 6 is applicable to all closed phase transitions taking place at constant pressure.

Application to Fusion and Vaporization Systems

If Phase-1 is solid while Phase-2 is liquid ($\text{solid} \rightleftharpoons \text{liquid}$), Equation 6 frames the fusion curve:

$$\frac{dP}{dT} = \frac{\Delta_{\text{fus}}H}{T_f \cdot \Delta V} \quad \text{--- (Equation 7)}$$

Where $\Delta_{\text{fus}}H$ is the latent heat of fusion and $T_f$ is the melting point.

For liquid-vapor equilibrium ($\text{liquid} \rightleftharpoons \text{vapor}$):

$$\frac{dP}{dT} = \frac{\Delta_{\text{vap}}H}{T \cdot V_v} \quad \text{--- (Equation 8)}$$

Assuming the gas phase behaves ideally, the molar volume of the vapor can be written using the ideal gas law ($V_v = \frac{RT}{P}$). Substituting this approximation alters the expression:

$$\frac{dP}{dT} = \frac{\Delta_{\text{vap}}H \cdot P}{RT^2} \quad \text{--- (Equation 9)}$$ $$\frac{1}{P}\left(\frac{dP}{dT}\right) = \frac{\Delta_{\text{vap}}H}{RT^2} \quad \text{--- (Equation 10)}$$

Using calculus identity $\frac{1}{P}dP = d(\ln P)$, we get the standard differential Clausius-Clapeyron Equation:

$$\frac{d(\ln P)}{dT} = \frac{\Delta_{\text{vap}}H}{RT^2} \quad \text{--- (Equation 11)}$$

Integrated Form of the Clausius-Clapeyron Equation

Rearranging Equation 11 to solve across changing state variables:

$$d(\ln P) = \left(\frac{\Delta_{\text{vap}}H}{RT^2}\right)dT$$

Integrating between definite temperature boundaries ($T_1$ to $T_2$) and corresponding vapor pressures ($P_1$ to $P_2$):

$$\int_{P_1}^{P_2} d(\ln P) = \frac{\Delta_{\text{vap}}H}{R} \int_{T_1}^{T_2} \frac{1}{T^2}dT$$

Evaluating this yields the integrated form in terms of natural logarithm ($\ln$):

$$\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta_{\text{vap}}H}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right] \quad \text{--- (Equation 12)}$$

Converting the expression to a common logarithm (base 10) for standard scientific calculations:

$$2.303 \log_{10}\left(\frac{P_2}{P_1}\right) = \frac{\Delta_{\text{vap}}H}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right] \quad \text{--- (Equation 13)}$$

$$\log_{10}\left(\frac{P_2}{P_1}\right) = \frac{\Delta_{\text{vap}}H}{2.303 R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right] \quad \text{--- (Equation 14)}$$