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?
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$\begingroup$ The net pressure changes, but the individual partial pressure of the vapour being looked at remains same, doesn't it? $\endgroup$– TRCCommented Jul 23, 2021 at 14:24
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$\begingroup$ @TRC Could you check out the comments I made under Chet Miller's answer? The issue is that in the derivation of the CC equation, you get a term involving the change in the pressure of the liquid (more specifically, a term like vdp, where v is the volume per particle of the liquid and dp is the change in pressure), which for most liquids around us should be zero (since the pressure of the liquid should remain at the ordinary atmospheric pressure). This would give a different result for the equation. $\endgroup$– Physics2718Commented Jul 23, 2021 at 14:53
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$\begingroup$ The vdp term for the liquid leads to the Poynting correction I referred to in my comment. For the vapor, even the pure vapor, the ideal gas law leading to the CC equation is not valid, and you need to use VdP for that too. $\endgroup$– Chet MillerCommented Jul 23, 2021 at 19:18
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$\begingroup$ I'm not familiar with the derivation as of now - I will look it up and try to solve your question. $\endgroup$– TRCCommented Jul 24, 2021 at 3:26
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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.
answered Jul 23, 2021 at 12:35
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$\begingroup$ Thanks for the answer, but I was thinking more in terms of how the derivation of the CC equation/relation would change in the presence of other gases. For instance, if you use the derivation that uses the fact that the changes in the chemical potentials of the liquid and vapor are the same, then the change in the pressure of the liquid will appear, which for constant atmospheric pressure would be zero. The overall result is that the relation will only have the volume of the vapor in it instead of the difference in volumes of the vapor and liquid. For a liquid-vapor transition, then maybe there $\endgroup$ Commented Jul 23, 2021 at 13:11
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$\begingroup$ isn't a huge difference (since the volume of the liquid is so small), but for a liquid-solid transition, that would be a huge difference. Maybe I should have included this in my question, as I was wondering whether the relation would change like this or something. $\endgroup$ Commented Jul 23, 2021 at 13:13
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$\begingroup$ Like I said, the chemical potential of the species in both phases changes. In the liquid phase, it changes according to the Poynting relationship, and in the gas phase, the change is much more complicated because of molecular interactions between species in question with its own molecules, but also between its molecules and the other species. It is not worthwhile for you to speculate about the possibility of minor modifications to the CC relationship until you study multicomponent mixtures. $\endgroup$ Commented Jul 23, 2021 at 14:57
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$\begingroup$ Yeah, I've thought more about it, and perhaps as a first order approximation I don't really need to worry about the presence of other gases in the air to compute the vapor pressure. I guess I can ignore these gases first to derive the CC equation, and only later do I imagine that these gases come in and "fix" the pressure of the liquid to 1 atm (without changing the partial pressure of the vapor). The corrections you describe seem to be way past my current understanding. Do you think this would be a fine enough way to think about it for now? $\endgroup$ Commented Jul 23, 2021 at 21:42
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$\begingroup$ Of course. This is how a mixture of gases behaves in the ideal gas limit. The chemical potential of a species in the gas mixture is equal to that of the pure species at the same temperature and partial pressure of the species in the mixture. $\endgroup$ Commented Jul 23, 2021 at 22:49