- #1
zenterix
- 708
- 84
- 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?
$$=G^\circ+RT\ln{\left ( \frac{P}{P^\circ }\right )}-G(l,P)\tag{4}$$
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?