The Clausius-Clapeyron equation (CC equation) can be used to find the (saturated) vapor pressure of a substance ie. the gas pressure at which the two phases (vapor + liquid or vapor + solid) reach equilibrium. However, what happens when the vapor is mixed together with other gases, such as water vapor mixed together with air? The reason I am asking this is because the derivation of the CC equation in my book relies on the assumption that the two phases are at equal temperature and pressure, whereas in the presence of other gases, the temperature of the vapor and liquid/solid will obviously be equal at equilibrium, but the pressure is not (since the other gases contribute with some pressure as well, and in general the partial pressure of the vapor will be smaller than the total pressure). This same problem arises if you consider the the equilibrium between, say, a solid and a liquid, so long as one of the phases is in a mix (like if the liquid is mixed with other liquids). If the CC equation changes as a result of this, then how does, say, the vapor pressure as a function of temperature change too?
If the system is at lower pressures where the ideal gas law is close to valid, then the partial pressure of the volatile species (mole fraction times total pressure) can be used in place of the vapor pressure with the Clausius Clapeyron equation. At higher pressures, the free energy of the liquid is affected by the higher overall pressure, as is the partial molar free energy of the volatile species in the vapor. So the CC equation cannot be used for these situations. To quantify such cases, one needs to study and apply the thermodynamics of mixtures. See Chapters 10 and beyond in Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.