So I'll preface this by saying this is not some evil homework assignment, this came about from some idle talk of colonizing the moon by a couple of software engineers when we realized we had no idea what the math would look like for this.

I'm sure we've all seen the physics demonstration of water boiling at room temperature in a near vacuum. From what I understand, the higher the pressure, the higher the boiling point, and vice-versa, i.e., the lower the pressure, the lower the boiling point.

I also understand that boiling is the process of the liquid changing state to gas. It stands to reason that as the liquid becomes gas, in a closed system, the water vapor would in fact contribute to the ambient temperature where equilibrium for any given temperature and pressure would ultimately be achieved between liquid and gaseous state water... but I'm not sure about the contributing factors to this.

For example, let's say we had a sealed chamber under the surface of the moon, where gravity is 1/6th of Earth's gravity. Not sure if volume/ shape is important here either, but let's say the chamber is a sphere 1,000,000 liters in volume, and quarter filled with pure liquid water (250,000L liquid water) and nothing else, i.e., no gas of any kind.

I've read that the surface of the moon varies from -173C to 100C depending on how much radiant energy from the sun it's absorbed, which (in theory) should average out to about -35C, so if we say that the chamber and the existent water in the chamber maintain a relatively constant temperature of -35C, I'm wondering how to figure out what the resultant pressure at the surface of the water would equalize out to be.

How could I calculate the gas pressure of the water at the surface of the water in 1/6 gravity when equilibrium has been reached? How much water of the original 250KL liquid be turned into gas?

The corollary to this would be, how hot would the water have to be to get us to Earth sea-level pressure of water vapor... or at least to the point where human bodies don't explode (I know they don't really explode).

How can this be calculated?

  • $\begingroup$ There isn't much to calculate. You can look it up. Search for "cryogenic water vapor pressure", which leads to e.g. here: its.caltech.edu/~atomic/snowcrystals/ice/ice.htm. The average temperature will probably be on the hot side for humans and the humidity will always be near 100%, unpleasant at any temperature. You will need a very powerful AC to condensate the water vapor, if you are dead set on having something like a big "pond" in there. The good news is that you can use the lunar night for this purpose, just like you can use the lunar day for heating. $\endgroup$
    – CuriousOne
    Commented Mar 28, 2016 at 19:45

1 Answer 1


The equilibrium vapor pressure of ice at -35C is about 20 Pa, which is about 0.0002 atm. Almost all the water would turn to ice, except for the vapor in the head space. The radius of the sphere would be about 6.2 meters, and this would not be enough to allow a significant hydrostatic head, especially at 1/6 g. The pressure at the surface of the ice would be about 0.0002 atm.


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