Understanding derivation of Clausius-Clapeyron equation

[zenterix]

Homework Statement

For an ideal gas undergoing an infinitesimal process at constant $T$ we have

$$dG=VdP\tag{1}$$

$$G(P)-G^\circ=RT\ln{\left ( \frac{P}{P^\circ }\right )}\tag{2}$$

Relevant Equations

Next, consider $\Delta G_{vap}$, the change in Gibbs free energy between gas and liquid phases of a substance at some constant pressure.

$$\Delta G_{vap}=G(g,P)-G(l,P)\tag{3}$$

$$=G^\circ+RT\ln{\left ( \frac{P}{P^\circ }\right )}-G(l,P)\tag{4}$$

The Gibbs free energy of a liquid is almost independent of pressure, so we use the approximation that $G(l,P)\approx G(l,P^\circ)=G^\circ(l)$.

Then

$$\Delta G_{vap}=G^\circ-G(l,P)+RT\ln{\left ( \frac{P}{P^\circ }\right )}\tag{5}$$

$$=\Delta G^\circ_{vap}+RT\ln{\left ( \frac{P}{P^\circ }\right )}\tag{6}$$

At dynamic equilibrium between gas and liquid phases, we have $\Delta G_{vap}=0$.

$$0=\Delta G^\circ_{vap}+RT\ln{\left ( \frac{P}{P^\circ }\right )}\tag{7}$$

$$\ln{\left ( \frac{P}{P^\circ }\right )}=-\frac{\Delta G^\circ_{vap}}{RT}=-\frac{\Delta H^\circ_{vap}-T\Delta S^\circ_{vap}}{RT}\tag{8}$$

In (8), $P$ is the vapor pressure at temperature $T$.

If we write (8) out for two different temperatures we have

$$\ln{\left ( \frac{P_2}{P^\circ }\right )}=-\frac{\Delta H^\circ_{vap}-T_2\Delta S^\circ_{vap}}{RT_2}\tag{9a}$$

$$\ln{\left ( \frac{P_1}{P^\circ }\right )}=-\frac{\Delta H^\circ_{vap}-T_1\Delta S^\circ_{vap}}{RT_1}\tag{9b}$$

If we subtract one from the other we get the Clausius-Clapeyron equation.

What I don't quite understand here is why the terms $\Delta H^\circ_{vap}$ and $\Delta S^\circ_{vap}$ are the same in both (9a) and (9b).

In each of (9a) and (9b) these terms represent enthalpy and entropy of vaporization at standard pressure and, as far as I can tell, the temperature at which the vaporization is happening.

Aren't these terms dependent of temperature?

 

[Borek]

They are. CC equation in its 1/T1-1/T2 form is only an approximation.

 

[Chestermiller]

The usual way of deriving the CC equation is to recognize that, along the equilibrium line, the change in Gibbs free energy between saturated vapor and the saturated liquid is zero. So, if there is a temperature change and equilibrium is maintained, $$dG_{liq}-dG_{vap}=0$$Now, for the liquid, we have $$dG_{liq}=-S_{liq}dT+V_{liq}dP$$and, for the vapor we have $$dG_{vap}=-S_{vap}dT+V_{vap}dP$$So we have $$-S_{vap}dT+V_{vap}dP=-S_{liq}dT+V_{liq}dP$$or $$-(S_{vap}-S_{liq})dT+(V_{vap}-V_{liq})dP=0$$or$$\frac{dP}{dT}=\frac{(S_{vap}-S_{liq})}{(V_{vap}-V_{liq})}$$But, at any one temperature along the equilibrium line, we have $$G_{vap}-G_{liq}=(H_{vap}-H_{liq})-T(S_{vap}-S_{liq})=0$$So, combining the previous two equations, we have, $$\frac{dP}{dT}=\frac{(H_{vap}-H_{liq})}{T(V_{vap}-V_{liq})}$$For the case of an ideal vapor far below the critical temperature, we have $$V_{vap}=\frac{RT}{P}>>V_{liq}$$Under this approximation, the Clapeyron equation reduces to the CC equation: $$\frac{d\ln{P}}{dT}=\frac{(H_{vap}-H_{liq})}{RT^2}$$