In an answer here, by Dr Kim Aaron, he describes a hypothetical scenario in which there's a piston cylinder system with water. A vacuum pump then sucks out the air. Water evaporates until the pressure equals the saturation pressure.

Now if we force in some air above the piston, it will push the piston down, this leads to a partial pressure in excess of the saturation pressure, and so there's condensation until the pressure reduces till $P_{sat}$ again.

Now, what's claimed is if the piston separating the air and water vapour at some point "disappears", the two will mix and there'll be nothing else that happens. If we force in more air into the mixture, it'll lead to more condensation of the vapor until the total pressure equals the saturation pressure.

So even though the partial pressure of the vapor is not changing, only the total pressure is, this is leading to condensation. Which makes me wonder then,

  1. Is total pressure and not partial pressure the relevant quantity here

  2. Since water vapor is condensing to keep the total pressure constant as we increase the pressure of the other gas, what happens when once the pressure of the other gas equals or exceeds the vapour pressure. Has all the water vapour now condensed like in the case of the piston being pushed all the way down or does it decrease to some minimum value?


1 Answer 1


I don’t agree with that section of the linked answer (”Now the magic piston disappears. What happens? Well, the molecules of air and molecules of water vapor just mix up. They are free to mingle. But other than that, nothing much different happens. The mixture of air and water vapor is all at the vapor pressure of the water.”)

If the piston disappears, the boiling temperature will stay the same, yes, as the pressure is unchanged if the water and vapor are considered ideal in mixing.

The water vapor and the the air will both expand into their new total volume and mix, yes.

But this will reduce the partial pressure of the water vapor, as the same amount now fills a larger space, so water will evaporate to reattain equilibrium.

I also don’t agree with another section of the answer (”If we force in more air, it just takes up more room and forces more of the water vapor to turn back into liquid.”) I show here how pressurizing a condensed phase instead increases its vapor pressure (because the stored strain energy penalizes the condensed phase), although the difference is extremely small and usually ignored for water.

I think the author is getting mixed up by the fact that we use pressure as a surrogate for the chemical potential or molar Gibbs free energy, which is the true arbiter of how matter moves, shifts, and changes phase as the temperature and pressure are altered. (The Second Law implies that the Gibbs free energy must be minimized under these conditions.) But it’s important to carefully keep track of what pressure’s being used to avoid the confusion that seems to have occurred here.

  • $\begingroup$ So, if we push more air into the cylinder that mixes with the vapour, since the partial pressure of the water vapor is unchanged (since it's not increasing the amount of condensation), but we're increasing the total pressure, the implication would be nothing except the fact that the liquid would now be in a compressed liquid state? $\endgroup$
    – xasthor
    Jan 27 at 18:09
  • $\begingroup$ Also, in some sense then can the evaporation/condensation be viewed in an independant way? I.e for a given temperature, the extent of evaporation will be such that the partial pressure of vapour = the saturation pressure. The presence of any other gas or pressurisation is independant of this $\endgroup$
    – xasthor
    Jan 27 at 18:14
  • $\begingroup$ I agree with both comments (assuming equilibrium and ideal mixing). $\endgroup$ Jan 27 at 18:44
  • $\begingroup$ Okay. One thing comes to mind though: if we see it in an independant sense, then the minimum pressure exerted is going to be $P_{sat}$. If there's any additional pressure it's going to be a compressed liquid. In this case how can it ever boil, since for boiling the pressure would have to be equal to $P_{sat}$ and for a given temperature the total pressure is always going to be more than this if there's an additional positive pressure. $\endgroup$
    – xasthor
    Jan 27 at 21:12
  • $\begingroup$ You’ve (re-)discovered that quasistatic boiling can’t occur in a closed container. One can achieve boiling only by heating the liquid fast enough that evaporation is too slow to maintain the saturation vapor pressure. $\endgroup$ Jan 27 at 21:28

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