I got some confusion going on with this explanation of Raoult's Law.

This is what my book say:

Because the ideal solution has a higher entropy than the pure solvent, the energy of a vapor increases as its molecules expand into a larger volume, that is, if they are at a lower pressure. Thus, solutes in a solution reduce the vapor pressure of the solvent.

Also, I should mention both solutions have the same $\Delta S$ based on the same $\frac{\Delta H}{T}$.

The point is I am confused about the last part of this explanation about how molecules expand. First off, this is all in a closed container, otherwise there's no vapor pressure. If so, the volume should be the same there is no larger or smaller volume.

  • $\begingroup$ It is possible that adding more paragraph the idea turn to be more clear. I don't get the author point (at least with this info.) He/she should be talking about molar quantities (or some fixed amount of substance). The volume shouldn't be important, because of the necessity stated in previous sentence. The grosse idea should be. There are two cases. Pure and mixed solution in liquid phase, both in contact with respective gas phases. Both cases need to have the same molar Gibbs energy. Comparing the Gibbs free energy of the solution and pure state, it should be predictable the change in $\endgroup$ Mar 30, 2015 at 2:32
  • $\begingroup$ pressure due to the constrain of same molar Gibbs energy of the two phases. But I think that you should add more data to your question or read another text. $\endgroup$ Mar 30, 2015 at 2:36

1 Answer 1


Basically a material in a mixture will have a lower vapor pressure than if it were pure. We know that the vapor pressure of a liquid can be determined by: $$\Delta G = \Delta G_v^o + RT \ln(P_x)$$ Given that $\Delta G$ is $0$ at equilibrium, we get: $$\Delta G_v^o = \Delta H_v^o -T\Delta S_v^o = -RT\ln(P_x)$$ Being an ideal mixture $\Delta H$ is $0$ and the equilibrium vapor pressure is determined by: $$-T\Delta S_v^o = -RT\ln(P_x)$$ $$\Delta S_v^o = R\ln(P_x)$$ Meaning that a higher difference in molar entropy ($\Delta S_g^o$) between the phases means there will be a higher equilibrium vapor pressure ($P_x$). Since gases have higher molar entropy ($S_g^o$) the process of vaporization creates an increase in entropy (positive $\Delta S$). However since a mixture has a higher molar entropy ($S_m^o$) in the initial state than the pure material ($S_p^o$) the phase change to vapor has a lower change in molar entropy: $$S_g^o>S_m^o>S_p^o$$ and thus: $$(S_g^o-S_p^o)<(S_g^o-S_m^o)$$ meaning the $\Delta S_g$ will be higher for a pure material than a mixture.


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    $\begingroup$ Can you tell what the physical implication of vapour pressure being related to entropy is? $\endgroup$
    – user5764
    Jan 16, 2016 at 4:53

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