In the case of a single substance (water), looking at the phase diagram is enough to conclude what happens upon heating:

Water phase diagram (source: WolframAlpha)

But what if I have water and air (or some other gas) in the same sealed container? Would the air saturate at some point and prevent further water from turning into gas?

(Wolfram have a "Heating Water and Air in a Sealed Container" demonstration)


According to Wikipedia:

Superheated water is stable because of overpressure that raises the boiling point, or by heating it in a sealed vessel with a headspace, where the liquid water is in equilibrium with vapour at the saturated vapor pressure.

But is there a quantification of this "equilibrium"?

  • $\begingroup$ en.wikipedia.org/wiki/Boiling_liquid_expanding_vapor_explosion $\endgroup$
    – valerio
    Commented May 19, 2016 at 16:29
  • $\begingroup$ Notice that if you say that there is only water and the pressure is different from 0 you are actually saying that there is also air in the container. Otherwise it would be water in a vacuum, i.e. water at 0 pressure. $\endgroup$
    – valerio
    Commented May 19, 2016 at 16:32
  • $\begingroup$ Related and maybe a duplicate: Is there a state beyond gas? $\endgroup$ Commented May 19, 2016 at 16:51
  • $\begingroup$ @JohnRennie, that's not at all what I'm asking. Updated question accordingly. $\endgroup$
    – Sparkler
    Commented May 19, 2016 at 17:00
  • $\begingroup$ I might be mistaken, but I thought that the "pressure" axis in that chart represents the partial pressure of the water vapor, not the total pressure. If true, then the air (or any other gaseous substance) in the container is irrelevant. $\endgroup$ Commented May 19, 2016 at 17:29

2 Answers 2


You have a 2 liter rigid container, featuring 1 liter of liquid water and, above it, one liter of a mixture of air and water vapor all at 1 atm. The temperature is 20 C, and the partial pressure of the water vapor in the head space is the equilibrium vapor pressure, so that the system is at equilibrium. This is the initial thermdynamic equilibrium state of the system. Start out by determining the partial pressure of air in the head space, the mass of air in the container, and the mass of water. In the ensuing calculations, it is permissible to assume that the partial pressure of the water vapor in the head space is equal to the equilibrium vapor pressure at the liquid temperature and the air is not soluble in the liquid water.

Vapor pressure of water at 20 C = 17.5 torr = 0.023 atm

Partial pressure of air in container at 20 C = 0.977 atm

From ideal gas law, moles of water vapor in head space = 0.00096

Mass of water in head space = 0.017 grams

Total mass of water in container = 1000.017 grams

Moles of air in head space = 0.04066

Mass of air in head space = 1.18 grams

Now you raise the temperature of the system to 50 C and let it equilibrate. What is the partial pressure of water vapor in the head space, and the mass split between the liquid water and water vapor? And what is the partial pressure of the air in the head space? What is the total pressure.


Vapor pressure of water at 50 C = 92.5 torr = 0.121 atm

From ideal gas law, mass density of water vapor in head space = 0.0822 g/l = 0.0000822 g/cc

Let x = mass of water in vapor phase

Mass of water in liquid phase = 1000.017-x

Volume of water in container (cc) = $(1000.017-x)+\frac{x}{0.0000822}=2000$

Solving for x : x = 0.0822 grams

Liquid water remaining = 999.93 grams

Volume of vapor = 1.00007

From ideal gas law, partial pressure of air = 1.077 atm

Total pressure = 1.197 atm.

Now try 100 C, 150 C, 200 C, etc.


Let's say that you have a vessel containing water and air and you start heating it. The temperature of the water and air inside will start to rise and so will the pressure, because the air would like to expand (but volume is fixed and water is almost incompressible). Since the boiling point of a substance depends on both pressure and temperature (for example water can boil at ambient temperature in a vacuum), this will prevent water from boiling until its vapor pressure overcomes the air pressure. This is how a pressure cooker works.

When vapor pressure overcomes the air pressure, if nucleation can occur (and this is almost always the case in everyday ife) the water will start to boil, part of it will be converted into vapor and the pressure will rise again. Now the external pressure will be equal to the vapor pressure and water won't be able to boil anymore.

At a certain point, if the container doesn't break because of the high pressure, you will reach water's critical point at $374$ °C - $218$ atm. Above this point, water will cease to exist as two separate phases and a single gaseous phase will be present.

  • $\begingroup$ is there a chart for that? $\endgroup$
    – Sparkler
    Commented May 19, 2016 at 20:52
  • $\begingroup$ I suggest you use a PV diagram like this one :ohio.edu/mechanical/thermo/Intro/Chapt.1_6/pure_fluid/…. It is better than a PT diagram like the one you posted because it is easier to see what is going to happen if the volume is constant. Just remember that P is the total pressure inside the vessel (it is the same for water and air+vapor). $\endgroup$
    – valerio
    Commented May 19, 2016 at 21:43

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