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If I boil water at room temperature in a vessel using a vacuum pump, will the water instantly liquefy and fall back into a liquid pool if I reintroduce air into the system?

Basically, if this is a large system, would I see a cloud form and fall? Would this happen all at once or over time?

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  • $\begingroup$ How much water? How large a room? How quickly is the air withdrawn? Vacuum boiling usually withdraws not just air but water-vapor from the region where the water is boiling. There's a few ways to approach this problem depending on the specifics of the scenario. $\endgroup$
    – userLTK
    Commented Nov 10, 2017 at 21:57
  • $\begingroup$ Let's call it a small volume of water in a large volume of vacuum. And let's say the vacuum is created instantly, and then the system is sealed. $\endgroup$
    – GrantG
    Commented Nov 11, 2017 at 17:41
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    $\begingroup$ You can do this with an oral (baby) syringe. Fill to 1/3 or so, plug the business end, pull back, watch the bubble form--release, and they disappear. It's easier with 99% isopropyl. $\endgroup$
    – JEB
    Commented Nov 11, 2017 at 19:33

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If heat were being supplied to the boiling water to hold its temperature constant at room temperature, then the pressure of the water vapor in the gas phase would be equal to the equilibrium vapor pressure of water at that temperature. This would be roughly 17.5 torr. If you then stopped the boiling, the pressure in the gas phase would remain 17.5 torr. Now, if you introduced air at room temperature into the gas space at 1 atm. (760 torr) and mixed the gas very rapidly, there would be no change in the partial pressure of water vapor (provided the size of the container did not change). It would remain at 17.5 torr, while the air pressure would adjust to 760 - 17.5 = 743.5 torr. Essentially none of the water vapor would condense.

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  • $\begingroup$ Is it only boiled into water vapor? Isnt it technically steam at this point? Once normal pressure is reintroduced, I would think it'd go back to vapor, and any extra may fall out. $\endgroup$
    – GrantG
    Commented Nov 11, 2017 at 17:39
  • $\begingroup$ You are aware that pure water vapor and steam are one-and-the same entity, right? $\endgroup$ Commented Nov 11, 2017 at 19:56
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Let's call it a small volume of water in a large volume of vacuum. And let's say the vacuum is created instantly, and then the system is sealed.

Once the air is removed, the water evaporates and becomes gas, so some of the pressure returns. There's an equilibrium of liquid water, vacuum, sealed container and temperature where some, or all of the water would evaporate. With a small amount of water and a large room, all of it would probably evaporate unless the temperature was close to freezing. To calculate how much you can look at the vapor pressure of water.

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As water evaporates it cools, so some of the water would turn to ice in your scenario and the temperature in your container would drop some but if we assume the temperature remains above freezing, the ice would eventually all sublimate.

Video of water boiling, then freezing in a 60 degree F vacuum chamber.

When the air returns, what would happen depends on the initial humidity of the air and if the combined humidity of the boiled water and the initial water was over 100% relative humidity - called super-saturation. If it was low relative humidity to begin with, the water-vapor might remain below 100% relative humidity and remain in the air as invisible gas. If the relative humidity is over 100%, when the air returns, the water-vapor from the boiled water should condense out of the air back to water fairly quickly. Some could get absorbed in any absorbing material and it wouldn't necessarily return to the floor, it would condense on any surface including walls and ceiling. (like the drops on the wall and ceiling after a hot shower - which is a very similar principal of temporarily super-saturated air when the water from the shower is warmer than the air in the bathroom).

You might get some fog or mist in the air like a hot shower too, foggy mirrors for example. Related question on why showers have this effect.

Will the water instantly liquefy and fall back into a liquid pool if I reintroduce air into the system?

Basically, if this is a large system, would I see a cloud form and fall? Would this happen all at once or over time?

Instantly is a funny word in physics, but no, it wouldn't be instant. Water likes to stick to other water. It would take some time (but not a lot of time) for the water to recombine. I think it would happen fairly quickly but not "instantly". The trick with seeing it is that most of the water could condense on surfaces in tiny droplets. Observing how quickly they form would take some careful looking, but I think it would be fairly fast - a few seconds perhaps, but that's a guess. That's a good question on the timing that might need to be clarified or re-asked for a better answer.

Would it fall back into the pool? Probably not, especially with a relatively small amount of water to begin with - as noted above, you'd get drops on the walls, floor and ceiling.

Would you see a cloud form? Probably not. Clouds are mostly ice not water, and require updrafts and falling air pressure. A sudden increase in air pressure would be quite different. You might see some mist, but I don't think it would resemble a cloud and it would all happen fairly quickly. I'd guess over a couple seconds.

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    $\begingroup$ Keep in mind that any pump used to maintain a constant low pressure would be removing water vapor as it works. If no pump was working to maintain the constant low pressure, the boiling water would raise the pressure as it boils. $\endgroup$ Commented Dec 1, 2021 at 21:13

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