At 1 atm and room temperature, is water vapor in a metastable state? As I understand it, water vapor is water in a gaseous state when the stable phase of water under this temperature and pressure is liquid, not gaseous. I understand that water vapor can be the result of evaporation, where water molecules near the surface water-air can get enough energy from thermal fluctuations to escape the attraction by other water molecules from the liquid and go freely into the air. It is also my understanding that when the air is saturated in water vapor, the rate at which water molecules escape the liquid is equal to the rate at which water vapor molecules condensate into the liquid but that is beside the original question I believe.

If the answer is yes, does this mean that thermal fluctuations help the system to reach a thermodynamics equilibrium which corresponds to a coexistence of a stable and metastable states of water as opposed as to a stable state of water only? If so, then I don't really understand why the thermodynamcis equilibrium does not imply a stable state only, but forces a metastable state to exist. I would like some explanation here.

If the answer is no, does this mean that in the system liquid water + air, the thermodynamics equilibrium corresponds to saturated air with water vapor as well as a liquid phase (assuming there was enough liquid water to saturate the air), so that the water vapor pressure in the liquid matches the one of water vapor in the air. If so, then I don't really understand what temperature-pressure phase diagrams of water represent. They would seem not to necessarily represent the equilibrium state of water under the indicated T and P. Or maybe they do, but are valid only for a system entirely composed of water and not water + air. This last guess seems to be correct, I am right?

  • $\begingroup$ Yes. The phase diagram for water is for pure water. And the pressure at a given temperature is the equilibrium vapor pressure. For a mixture of air and water, the saturation partial pressure of water vapor in the gas phase is extremely close to the equilibrium vapor pressure. $\endgroup$ – Chet Miller Sep 5 '17 at 22:04
  • $\begingroup$ When you say "yes", you mean yes to which question I asked? The last one, the one of the title or another one? And when you say "the saturation partial pressure of water vapor in the gas phase is extremely close to the equilibrium vapor pressure." I'm not sure what is meant by saturation partial pressure. If that's the pressure of saturation, then I don't understand why it would differ from the equilibrium vapor pressure. Feel free to elaborate. $\endgroup$ – AccidentalBismuthTransform Sep 6 '17 at 19:23
  • 1
    $\begingroup$ The yes refers to what I said in my next sentence. Regarding saturation partial pressure what I meant was that, when the gas phase is saturated with water vapor (i.e., dew point), the partial pressure of the water vapor in the gas phase is equal to the equilibrium vapor pressure. $\endgroup$ – Chet Miller Sep 6 '17 at 19:27
  • 1
    $\begingroup$ Water vapor in a metastable state exists as supercooled vapor, but hardly at 1 atm and room temperature. BTW, is this question about water, or the system air/water? $\endgroup$ – aventurin Sep 6 '17 at 21:37
  • $\begingroup$ @aventurin The question is about the system air/water. So all hints that my last guess is the correct answer, i.e. that the answer to the title question is no, and that the thermodynamics equilibrium of the system air/water is reached when the vapor pressure of water vapor in water and the one in air matches. If the system was only water (say in a closed non rigit plastic bag), there would be no water vapor under this pressure and temperature. Feel free to write an answer because the current one is not satistfying. $\endgroup$ – AccidentalBismuthTransform Sep 7 '17 at 6:12

Metastable water vapor exists as supercooled vapor or in supersaturated air e.g. in the absence of cloud condensation nuclei.

At room temperature and a pressure of 1 atm supersaturated air exists, but I doubt that it is possible to prepare pure supercooled water vapor at this conditions.

The usual phase diagram of water describes water as a pure substance, not as an n-ary mixture with gases. According to this phase diagram, water is liquid at $20\mathrm{°C}$ and $1~\mathrm{atm}$ ($1013.25~\mathrm{hPa}$).

enter image description here

However, when you follow the line $T = 20\mathrm{°C}$, you will find an intersection with the curve that separates the vapor and liquid phase. The pressure at this point ($23~\mathrm{hPa}$) is called the vapor pressure at $20\mathrm{°C}$. This is approximately the same as the partial pressure of water in saturated air at the same temperature. So you can think of air molecules as mere spectators that do not contribute to the "water pressure".

Conclusion: The answer would be no under the additional assumption that the air is saturated with water vapor. However if air is supersatured with water vapor then the water vapor would be in a mestastable state and the answer would be yes.

  • $\begingroup$ Nice job. If I understand well, the answer to my title question is a no. In the system air + water, water vapor is in a stable state at room temperature and 1 atm. If you could just add "The answer to your title question is a no" or something like that, I'd gladly accept your answer. $\endgroup$ – AccidentalBismuthTransform Sep 8 '17 at 21:55
  • 1
    $\begingroup$ I don't dare to say no, since, as said, the system water + air at 1 atm and room temperature could contain air supersaturated with water vapor. This would make a metastable system. The answer would be no under the additional assumption that the air is saturated with water vapor. $\endgroup$ – aventurin Sep 8 '17 at 22:03
  • 1
    $\begingroup$ Ok then if you could add "The answer would be no under the additional assumption that the air is saturated with water vapor. However if air is supersatured with water vapor then the water vapor would be in a mestastable state and the answer would be yes.", it would be great. I'll accept your answer because I'm already satisfied. Thanks a lot! $\endgroup$ – AccidentalBismuthTransform Sep 8 '17 at 22:15

A liquid necessarily requires many water molecules to collect together.

You could ask how many are required before it is considered a liquid?

If the answer is "more than one", then yes, water vapor molecules in the air are colliding with each other all the time.

Only when the conditions of the collision lead to combination due to Hydrogen bonds is a liquid formed.

These bonds can exist momentarily prior to being broken - in which case the liquid became a gas again. I'm not sure what you mean by meta-stable in this context, but it could be described as bi-stable.

Behavior of water in air is indeed different from in a vacuum.

A vacuum can be empty space and therefore low pressure, but if water vapor is compressed and cooled in a steel box then a higher pressure can be achieved in the absence of air, and it will liquefy.

Ideal PT diagrams for water should be in the absence of air. It's a fair approximation to behavior in the presence of air.

Note is actually quite a special liquid (or maybe we just spent more time studying it). From the first observation that the solid floats on the liquid it continues to deliver mysteries.

  • $\begingroup$ I'm not sure how this answers my question(s). Anyway by metastable I mean a state like diamond at 1 atm and room temperature. It isn't in a stable state, but slowly converting to graphite. It's in a metastable state. $\endgroup$ – AccidentalBismuthTransform Sep 6 '17 at 19:26
  • $\begingroup$ OK my bad, I looked up en.wikipedia.org/wiki/Metastability You could say that the water vapour is a higher energy stable state for the water within your system, then claim Metastability for that. Thus the answer to your first question is yes. $\endgroup$ – JMLCarter Sep 7 '17 at 0:50
  • $\begingroup$ The second question, yes the stable (liguid) and metastable (water vapour) states are in equilibrium; with water moving between the two in equal quantity. This is possible because energy is not uniformly distributed throughout the closed system. Gravity pulls the low energy molecules closer together, increasing the possiblity of a collision releasing a higher energy molecule again. $\endgroup$ – JMLCarter Sep 7 '17 at 0:55
  • $\begingroup$ I don't think this is correct. From the comments I'm getting, it seems that water vapor isn't in a metastable state. On the top of that I also doubt gravity increases the probability of a collision releasing a higher energy molecule again. If you can back up this claim with some math, I'd be extremely glad to learn from it. $\endgroup$ – AccidentalBismuthTransform Sep 7 '17 at 6:17
  • $\begingroup$ If more than one state of differing energy is involved it must be metastable. It's true the water vapour is not metastable, but that statement excludes the liquid water as a state. $\endgroup$ – JMLCarter Sep 7 '17 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.