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Will a vacuum chamber that is maintained at 0.1 Atmospheres with an open water container inside ever reach 100% humidity of the remaining air, so that no more water evaporates or will water continously evaporate and be sucked out of the chamber by the vacuum pump irrespective of the humidity level of the air?

I see 2 scenarios:

  1. Water evaporation/Boiling can only happen up to 100% humidity of the (little) air in the container. When this is reached, the water stays in its liquid form.

  2. Water boils and evaporates irrespective of the presence of air and irrespective of its humidity. The evaporation increases the gasseous volume and the vacuum pump removes this extra vapour while fighting to keep the pressure at 0.1 Atmospheres.

The reason I ask is because I want to dry something in a vacuum and I want to know if I just keep the vacuum at 0.1 Atm or if I should pulsate with Vacuum, then fill with dry air, then vacuum again and then fill with dry air again....

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  • $\begingroup$ Not sure on the, say, drying out a phone. However, pump/purge cycles are a really good idea to get water vapor out of a vacuum system. But one is usually going well below a tenth of an atmosphere (more like a millionth of an atmosphere) on the pump cycles. $\endgroup$
    – Jon Custer
    Aug 22, 2018 at 14:30

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From the wiki article of Relative Humidity:

Water vapor is independent of air

The notion of air "holding" water vapor or being "saturated" by it is often mentioned in connection with the concept of relative humidity. This, however, is misleading—the amount of water vapor that enters (or can enter) a given space at a given temperature is almost independent of the amount of air (nitrogen, oxygen, etc.) that is present. Indeed, a vacuum has approximately the same equilibrium capacity to hold water vapor as the same volume filled with air; both are given by the equilibrium vapor pressure of water at the given temperature.There is a very small difference described under "Enhancement factor" below, which can be neglected in many calculations unless high accuracy is required.

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Both methods dry but consider the following: 1) some materials bond strongly to water so you may be unsuccessful at drying, this would be the case in chemistry experiments. 2) let say you are drying your cell phone because you dropped it in a puddle, the vacuum pump at is at 0.1 atm because it has reached its limit, i.e not much more gas is going in or out of the pump. Initially though a lot of gas and water vapour will be sucked out. Any liquid water should be fully vaporized and adding the dry air at the end further drys. At 0.1 atm there is H2O gas in your system, but the dry air dilutes it.

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  • $\begingroup$ Thanks for the reply. So for drying a phone this would quickly evaporate the water, suck it out of the system and hence dry it. What about if I have a one gallon container of water in the system? Am I right in assuming that this would generate water vapour, which increases the pressure and the pump would suck out this vapour together with some of the remaining air... more evaporates, pump removes it.. so in the end you would have almost no air inside, just 0.1 ATM of Water vapour.... and as long as I keep it at 0.1ATM, the water would further evaporate? $\endgroup$ Aug 22, 2018 at 13:56
  • $\begingroup$ As long as the pump is running everything is sucked out leaving a water vapour gas. Just be careful that if you chill it you will maybe see condensation at some temperature and eventually frost ( solid) . There is a table somewhere on the net that lists water condensation pressure as a function of temp. $\endgroup$ Aug 22, 2018 at 16:49
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Water boils when the vapor pressure is greater than the ambient pressure.

Water evaporates when the vapor pressure is greater than partial pressure of water in the atmosphere.

At room temperature, the vapor pressure of water is around 0.03atm. So 0.1atm is insufficient to cause boiling. If, once the vessel reaches 0.1atm the pump stops moving air, that will cause it to saturate and evaporation will slow.

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