Let us take a mixture of ethanol and acetone as an example. When ethanol is mixed with acetone, the hydrogen bonding between ethanol molecules gets disturbed as actone molecules get in between the ethanol molecules. This results in $\Delta_\text{mix} H\gt0 $ as heat must be supplied to make the solution from its constituents (as attraction is less) but I am not sure about $\Delta_\text{mix} V$. This page says it is greater than zero because volume expands on mixing. But I think it should be less than zero because the observed vapour pressure is more than that predicted on the basis of theoretical calculations, so a greater amount of solution will be present in the form of vapour, hence volume of solution will be less (than the sum of the volumes of its constituents). Where am I wrong?

  • $\begingroup$ An intuitive explanation can be found on the page you referenced. The lessened attraction between the constituents effectively means that the average "intermolecular distance" would be greater after mixing than it was before, hence the increased volume. The vapor pressures before and after mixing are not directly relevant here. $\endgroup$ – voffch Mar 8 at 13:59
  • $\begingroup$ I agree with the increased intermolecular distance, but more vapour pressure means more amount of vapour over solution. As total amount of solution + vapour is constant, this means less solution, hence less volume of solution, so there are two conflicting processes which one dominates? $\endgroup$ – drake01 Mar 8 at 14:20
  • $\begingroup$ @drake01 By the volume change is meant the total volume after mixing, at the same temperature versus sum of volumes of the same amount of liquids before mixing. It should be obvious that evaporation changes the volume, as the there is less of liquids. $\endgroup$ – Poutnik 2 days ago

I'm a little late to the party here, but wanted to chime in.

The link you posted isn't working anymore, but you appear to be making too many assumptions about the system being measured. Nothing about the measurement of the volume change implies there is necessarily an open space above the liquid for vapor to form. Imagine an open beaker of ethanol with a small disk floating on top so that no molecules can escape upwards, but it can move up and down with the change in liquid level. This would also keep the pressure of the liquid constant (although it doesn't really matter since the liquid is essentially incompressible) .

If you were able to inject a certain amount of acetone into this beaker without opening it to the atmosphere for very long, the resulting volume of liquid in the beaker would be greater than the sum of its parts. Hopefully it makes sense that if there is no space for liquid to form, the positive volume change of mixing can show itself.

As far as the idea that more vapor would form to decrease the volume of liquid, realize that evaporation is a time-dependent process, so likely the mixed liquid would reach an effective equilibrium state where it's volume change could be observed before the extra liquid has time to evaporate. It is true that it would evaporate faster if the effective vapor pressure (technically "fugacity" or "chemical potential") were higher, as the driving force for mass transfer would be greater, but this is also very dependent on the way the system is specified. Technically if you left the container open in a large room forever, it would all evaporate in either case.


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