Raoult's Law explained with thermodynamics/ free energy

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


This is what my book say: 



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> 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.
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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. 




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.


Q.E.D.


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