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I know for sure that if you fill a baloon with water (without leaving any air inside) and you start heating it, the baloon won't expand until the temperature reaches 100 °C. I think it is because while the temperature is below 100 °C, the vapor pressure of the liquid is <760 mmHg so any expansion of the baloon is prevented by the atmospheric pressure. However, let's suppose to put some water in a water bottle and close it. Initially, there was no water vapor in the bottle, there was only air. But, after a certain amount of time, water will evaporate until the pressure of the water inside the bottle reaches the vapor pressure of the water at that given temperature. So, my question is: why can water evaporate if there is some air over it but it cannot without air, although in both cases it has to overcome the atmospheric pressure? What is the difference between a closed container with no air inside (only liquid) and a closed-one with some air inside?

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    $\begingroup$ I'm not sure I'm following you here, I just want to remind that water always contains dissolved air, which solubility decreases as the temperature rises so there is always water+air in the balloon if water hasn't been deaerated beforehand. $\endgroup$ – andselisk Nov 25 '17 at 9:27
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    $\begingroup$ You are right... I should have stated that water has actually been deaerated before being put in the baloon. $\endgroup$ – Marco Pagonia Nov 25 '17 at 10:10
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The difference is the empty space in the latter case and lack of it in the former. By "empty" I mean "filled with air", but that's irrelevant; it could have been truly empty with the same result. Air pressure is also irrelevant; it could have been 10 atm, and that still wouldn't prevent water from evaporating.

See, when several gases occupy the same volume, they do not even know about each other. They do not care if another gas enters the same volume and starts to share it with them. But they care a great deal about the volume itself, and try to increase it with all their might; that's what we call pressure.

So it goes.

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  • $\begingroup$ So, in a bottle which contains some water and some air, water can evaporate because the pressure of the air INSIDE the bottle is irrelevant, while water in a baloon (completely filled with water) cannot evaporate because the pressure of the air OUTSIDE the baloon prevents water from evaporating? $\endgroup$ – Marco Pagonia Nov 25 '17 at 10:35
  • $\begingroup$ @MarcoPagonia No, the water only feels the pressure on itself. In a half-filled bottle, twice as much water is in the air than in a three-quarter-filled bottle, and this linearly goes to zero the more water and less airspace is in it. Fill a bassin to the top and put an even lid on it, and no water will ever evaporate. $\endgroup$ – Karl Nov 25 '17 at 10:46
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    $\begingroup$ @MarcoPagonia Yes, that's right. $\endgroup$ – Ivan Neretin Nov 25 '17 at 10:54
  • $\begingroup$ @IvanNeretin No, the pressure outside the ballon is irrelevant, unless it is so low the resulting pressure inside the ballon is below the vapour pressure of water. $\endgroup$ – Karl Nov 25 '17 at 11:08
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    $\begingroup$ That's precisely the extent to which the outside pressure is relevant. $\endgroup$ – Ivan Neretin Nov 25 '17 at 11:48
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The individual water molecule does not "know" about the external pressure. If there already is a free space it can evaporate into, it will do so, to the point where the partial pressure in the free room is equal to the vapour pressure of the water body.

With an infinitesimally small free space above the water (like in your ballon), the amount of water that evaporates is also infinitesimally small.

If your water bottle was a cylinder with a piston that is permeable to air (not necessarily to water), that piston would of course sink to the water surface over time.

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I think the other answers don't touch the crux of your confusion.

The problem is, you mix up two forms of vaporization:

Evaporation (surface phenomenon): water will fill the entire available volume with its molecules, regardless of pressure due to other vapors.

Boiling (volume phenomenon): bubbles will only form and survive if vapor pressure overcomes pressure.

So…

why can water evaporate if there is some air over it but it cannot without air

Because evaporation needs an available volume, which, in the latter case, it doesn't have.

although in both cases it has to overcome the atmospheric pressure?

Wrong. Evaporation doesn't need to overcome any pressure (water molecules on the surface thoughtlessly jump into air, and that's it*). Evaporation also doesn't increase the vapor phase volume -- that's why your baloon would not expand if you filled it with air and water and left it at room temperature.

What is the difference between a closed container with no air inside (only liquid) and a closed-one with some air inside?

The former does not have available volume for molecules in the liquid state to jump into. The latter does. Therefore, the former doesn't see evaporation. The latter does.

However, any container will accomodate boiling bubbles. That's why your baloon expands. That's why badly designed pressure vessels burst.

Always keep in mind the important distinction between these two modes of vaporization.

* water molecules in the air also jump back to the liquid phase's surface; it's just that, below vapor pressure, more liquid evaporates than vapor condenses back.

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The total evaporation is the same regardless of how much empty space there is above the water. Once it evaporates, it fills the space above the water. Once there is no more empty space, the vapor has nowhere to go except back into the water. Thus, in a close container, an equilibrium will be reached in which as much water is evaporating as is condensing, and there will be no net evaporation. The amount of water vapor that's present in that equilibrium depends on how much empty space there is.

The term "vapor pressure" is a term that is used to describe this process, and draws on normal pressure as an analogy. As more water evaporates, the vapor pressure increases. With normal pressure, the larger the volume, the more gas you need to reach a fixed volume. Similarly, the larger the volume, the more vapor you need to reach a fixed vapor pressure. Thus, we can compare the vapor to volume ratio of the balloon and bottle and find that we get the same number in both cases, and we can think of this constant as being a "pressure". When this "pressure" reaches a critical value, equilibrium is reached, and net evaporation ceases. Given this formulation, your statement "water will evaporate until the pressure of the water inside the bottle reaches the vapor pressure" is true for the balloon as well, it's just that the vapor pressure is reached much more quickly.

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  • $\begingroup$ The total evaporation is the same regardless of how much empty space there is above the water. Could you please clarify this, since evidently 1 m³ of empty space will require more water vapor to reach the same partial pressure than 1 mL of empty space would? $\endgroup$ – André Chalella Nov 27 '17 at 2:24
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    $\begingroup$ In both cases, the number of water molecules evaporating per second is the same. However, with only 1mL of space, molecules that evaporate will quickly condense back into a liquid. With 1mL of space, however, the rate of condensation will start small, and increase as the partial pressure increases. Thus, the first case has smaller net evaporation, but both have the same total evaporation. $\endgroup$ – Acccumulation Nov 27 '17 at 2:31
  • $\begingroup$ I understand now. But I think such terminology is confusing. I took it to mean total volume evaporated before you explained, not rate of evaporation. Is it mainstream? If not, I suggest a change in the wording to become clearer. $\endgroup$ – André Chalella Nov 27 '17 at 2:42

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