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Can steam (water vapour) exist in a vacuum, and if so does it look and behave the same as in air? Let's assume the actual case is a kettle boiling in a vacuum.

Note that I'm talking about a gas-less vacuum, not truly empty space devoid of all particles except the water.

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Water vapour at the Earth's surface consists of molecules of water mixed up with molecules of oxygen and nitrogen. At room temperature 100% relative humidity is about 20g of water per kilogram of air (roughly a cubic metre), so about 4 molecules per 100 are water.

To get water vapour in a vacuum you just remove the air molecules and leave the 20g per cubic metre of water molecules. Put this way it may sound a bit trite, but to a first approximation air is an ideal gas so the molecules don't interact except for hard sphere collisions. That means taking the air molecules away doesn't have a large effect on the water molecules. The properties of 20g/m$^3$ of water dispersed in air and 20g/m$^3$ in a vacuum are actually pretty similar.

If you put a saucer of water into a closed box of air it will evaporate until it reaches a concentration of about 20g/m$^3$. Similarly if you put a saucer of water into a closed box containing a vacuum it will evaporate until it reaches the same concentration of 20g/m$^3$. In both cases the liquid water will be in equilibrium with the water vapour.

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Excellent answer. This is precisely the kind of answer I was looking for. –  Polynomial Nov 19 '12 at 13:44
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