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I'm trying to make sure my understanding of vapor pressure and transients is correct.

Let's assume I have a thermally-controlled sealed environment which started out as the regular air (i.e. atmospheric gasses and their respective ratios) with no gaseous water or ethanol at 101.325 kPa, 30 C. This container will maintain a temperature of 30 C after the start of this experiment.

Inside this environment there are two containers which can be remotely opened.

When I open the first container, exposing the pure water inside with a saturation vapor pressure (First question: is that the right term?) of 4.2455 kPa (from Wikipedia) and wait until it reaches equilibrium (100% humidity), does this mean this is the new state of the container?

  • Pressure: 101.325 + 4.2455 = 105.5705 kPa

  • Partial-Pressure of H2O: 4.2455 kPa

Then I open the second container, exposing pure isopropyl alcohol inside with a saturation vapor pressure of 10.555 kPa (from WolframAlpha) and wait until it reaches equilibrium again, does this mean this is the new state of the container?

  • Pressure: 101.325 + 4.2455 + 10.555 = 116.1255 kPa

  • Partial-Pressure of H2O: 4.2455 kPa

  • Partial-Pressure of Ethanol: 10.555 kPa

So, in the end, the pressure of the vessel increased each time a new liquid was introduced.

In this hypothetically-ideal environment, would the pressures simply rise until it reached that perfectly equilibrium? If so, I expect in a real environment the pressure would rise but there would always be a cold spot somewhere which caused the vapor pressure (in that specific area) to drop, leading to water condensing, leading to the partial pressure being lower as the air mixes, leading to more evaporation; is this correct?

Some notes:

  • Assume that there is enough of each liquid to reach equilibrium without running out of liquid.

  • The sealed environment effectively adds energy to counteract the heat of vaporization of each liquid, and also prevents the gasses from releasing/absorbing energy into/from the external environment or from changing due to pressure changes. It is essentially a magic make-everything-this-one-temperature machine.

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There are a couple of gotchas' with the problem.

The volumes of the liquids and gases are specified. So 1 drop of water in a million gallon tank of dry air isn't going to be enough water to reach saturation.

You stated that "This container will maintain a temperature of 30 C after the start of this experiment." However the problem statement doesn't explicitly state that the contents of the container will be in thermal equilibrium with the container. That does seem to be a reasonable interpretation however. (If the evaporation had taken place adiabatically, then the liquid and gas would have to cool to account for the evaporation.)

There is another significant factor here. The water would absorb ethanol, and the ethanol would absorb water. You can't really figure this out without knowing all the volumes, or at least the relative volumes.

Now if you assume that there is enough liquid to saturate the gas phase, that the system (gas phase and liquid phase) absorbs heat from outside the system, then the vapor pressure of the liquid at $30\ ^\circ\mathrm{C}$ is added to the internal pressure of the gas phase before the evaporation takes place.

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  • $\begingroup$ "If the evaporation had taken place adiabatically, then the liquid and gas would have to cool to account for the evaporation" - Yes, my intention with that statement was to say that the container would keep the liquid at that temperature, adding energy to compensate for evaporation. And yes, my assumption was that there was enough water to reach equilibrium without running out of water. $\endgroup$ – iAdjunct Dec 30 '17 at 21:35
  • $\begingroup$ "...is added to the internal pressure of the gas phase before the evaporation takes place" - are you saying that the pressure change leads the actual water vaporization? So, once the first canister is opened, in [comparatively] short order, the chamber would rise to 105.5705 kPa, even though there may not actually be H2O vapor in the air? $\endgroup$ – iAdjunct Dec 30 '17 at 21:37
  • $\begingroup$ I didn't think about the water/ethanol cross-absorption; I'll have to research that part further. Before the ethanol container is opened, is my original question combined with my previous comment a correct understanding of the system? $\endgroup$ – iAdjunct Dec 30 '17 at 21:40
  • $\begingroup$ I didn't look it up, but I'll assume that the vapor pressure of water at 30C is 4.2455 kPa. So once the gas phase is saturated with water vapor, then the pressure will rise to 101.325 + 4.2455 = 105.5705 kPa. How long it takes once the first canister is opened is another matter entirely... $\endgroup$ – MaxW Dec 30 '17 at 22:10
  • $\begingroup$ water/ethanol cross-absorption - I think you'd end up with one of the three situations (1) all ethanol migrates to water (2) All water migrates to ethanol or (3) two volumes of ethanol-water at the same concentration. With one of the possibilities would come true would depend on the (relative) volumes. $\endgroup$ – MaxW Dec 30 '17 at 22:14

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