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Q & A: boiling at fixed volume or pressure

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Most recent answer: 02/09/2011
Q:
If water and dry air are sealed into a container, and the water evaporates until equilibrium is reached between the water vapor in the air and the evaporation rate, and temperature remains constant in both the air and water, will the pressure inside of the container increase? That is my question, it probably has a pretty simple answer, but I was wondering because of the somewhat disjointed thought process below,: To determine the true amounts of gas collected over water, one subtracts the partial pressure of water vapor in the gas and then uses the resulting partial pressure to find moles,using PV=nRT, or similar depending on units. It would seem to me that the pressure would increase in side the contairer as more water evaporated. Based on this, wouldn't it bet rue, that if a container could expand to allow for more pressure, then the temperature increase, however minuscule, would be negated by the increase in volume, subsequently tying volume, indirectly, to the equilibrium point. This would make it seem to me that percentages of humidity in the air, partial pressures of water vapor in gasses collected over water, and the equilibrium point of air and water could change if the pressure in the air was higher or lower, or if the container wasn't completely rigid.
- Marc (age 16)
Caribou, Maine, USA
A:
For your first question, assuming the air was dry to begin with, yes the water vapor will increase the total pressure.

Yes, if the cylinder were constant-pressure rather than constant volume, it would expand dramatically when heated. That's because in addition to the usual PV=NRT, you have an increasing N due to the evaporation of the water. At some T, the vapor pressure of water plus the pressure from the other gases will equal the set pressure. At that point, the container will expand until all the water has turned to vapor.

So long as the vapor pressure is less than the total pressure, it is only weakly affected by the total pressure (for fixed temperature) because the gas is nearly ideal and the chemical potential of the liquid is rather insensitive to pressure.

What you may perhaps be noticing is this: In a fixed-pressure container, ther'es a very sharply defined boiling point. It's the temperature for which the vapor pressure equals the total pressure. If there's no other gas around, as a function of T the system abruptly switches from all liquid to all gas. We've discussed that in some old answers. If there's some air around, we've described the behavior above. For a fixed volume container, there's a broad vapor-liquid coexistence regime even if there's no other gas around. Assuming the volume is bigger than needed to hold the liquid, the pressure just becomes the vapor pressure, until all the liquid is evaporated.

Mike W.

(published on 02/09/2011)

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