I'm fully aware of the various questions asked about Raoult's Law, and I'm asking this after having gone through all of them and not finding a suitable answer to my question.

I have studied, that at a certain temperature, the equilibrium vapour pressure of a liquid remains constant, irrespective of the amount of liquid taken. From this, I conclude that if I take a mixture of two liquids, the equilibrium vapour pressure should just be the sum of their individual equilibrium vapour pressures, since both liquids attain equilibrium when the partial pressures of their vapours are equal to their equilibrium vapour pressures. However, Raoult's Law states that the vapour pressure depends on the amount of each liquid in the solution.

This question almost answered my question, but I do not understand why the concentration of the liquid matters when it's in a solution. I understand that this is not due to any intermolecular forces, since the mixture in consideration is assumed to be ideal.

We had a long discussion in class about this, and in the end we concluded that the equilibrium vapour pressure would depend on the surface area of the liquid, and in a mixture the surface is not completely available to a single component of the mixture. I'm not sure if this explanation is correct, and I would like your opinions about this.

This question talks about a different explanation, but I believe it talks about solutions of solids in liquids. Is a similar explanation valid for mixtures of liquids? I can't seem to get my head around the explanation that the liquid molecules occupy a fraction of the surface which is equal to their mole fraction, which reduces the equilibrium vapour pressure of each component of the mixture.


1 Answer 1


When a liquid is in solution, the rate of its escaping molecules per unit surface area is decreased. But the rate of its condensing molecules, for the given vapour pressure and the unit surface area, is not decreased. The result is that at equilibrium, the vapour pressure is lower.

Therefore, the total equilibrium vapour pressure is weighted sum of equilibrium vapour pressures of pure liquids, where the weights are molar fractions.

The direct sum of partial pressures of pure liquids is applicable, if they are kept separated or if they are not miscible (like water and essential plant oils).

  • $\begingroup$ Why does the rate of escaping molecules decrease? $\endgroup$ Dec 17, 2021 at 5:05
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    $\begingroup$ Imagine students running out of school. Does the rate of boys running out depend on percentage of boys? Is this rate the same if it is boys only school as well as if there is just 10% of boys? Molecules do not escape 10 times faster after realization there is 10 times less of them. $\endgroup$
    – Poutnik
    Dec 17, 2021 at 5:09
  • $\begingroup$ Of course, the area argument is only a starting point for understanding—it's highly simplistic. Vapor pressure lowering is, as you wrote, dependent on mole fraction. Suppose you have two solvents, A and B, where the surface area of A is twice that of B. Then in a 1:1 mixtue of A and B, the fractions of surface taken up by A and B are 2/3 and 1/3, respectively, rather than 1/2 each. Thus, by the area argument, the vapor pressures of A and B would be reduced to 2/3 and 1/3 of their pure vapor pressures, respectively. But that's incorrect. Ignoring non-ideality, they're each reduced to 1/2. $\endgroup$
    – theorist
    Dec 18, 2021 at 4:59
  • $\begingroup$ @theorist by surface area of A being twice that of B, do you mean the surface area of the molecule? In that case, even if the molecules of A were to occupy twice the surface occupied by the molecules of B, the number of molecules of A and B at the surface would be equal, right? $\endgroup$ Dec 18, 2021 at 6:58
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    $\begingroup$ @UmeshKonduru Ultimately what drives this is the entropy of mixing of the two liquids, which lowers their chemical potentials and thus their vapor pressures. Poutnik knows this, but it's difficult to give a mechanistic explanation of a statistical, emergent property like entropy, so the area argument is often given to students before they are ready to study this at a more advanced level, which is why Poutnik provided it (plus you were asking about area arguments). So it's fine, but it's also nice to know that it is a simplistic picture. $\endgroup$
    – theorist
    Dec 18, 2021 at 7:55

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