The amount of water vapor that the air in a large region (say a suburb) can hold should be dependent on the average temperature of the liquid water within that region. Not the air temperature of that region. This is because the evaporation rate of water is dependent on the water temperature (not the air temperature) since the water molecules obey a distribution similar to the Maxwell Boltzmann distribution. So if the water temperature rises, the evaporation rate rises and vice versa.

On the other hand, the condensation rate of water in a given region will primarily depend on the partial pressure of the water vapor in the air of that region. So as more water vapor fills the air, the condensation rate increases until we reach an equilibrium point when the evaporation rate equals the condensation rate. At this point, the air is said to be saturated and colloquially we can say that it cannot 'hold any more water vapor'. If we now increase the water temperature in the region, the evaporation rate will briefly exceed the condensation rate, filling the air with more vapor until a new equilibrium is reached. Crucially, this all depends on the water temperature and not the air temperature. At least according to my understanding (which may well be totally incorrect).

Yet I constantly find reliable sources that state in essence that the amount of water vapor air can hold is dependent on the temperature of the air and not the water. For example, "Dropping the temperature of moist air reduces its moisture capacity" (Cengel and boyles, Thermodynamics internation edition). Similarly, wikipedia states "colder air can hold less vapour, so chilling some air can cause the water vapour to condense". The only resolution I have at this point, is to surmise that these statements are actually wrong on a purely technical level but that they almost always provide the correct answer because of the fact that the water temperature in a region is strongly correlated with the air temperature of a region. Hence stating that "a rise in air temperature implies a rise in airs capacity to hold water" is almost always correct because a rise in air temperature almost always implies a rise in water temperature. Am I correct in my understanding? Is it true that reliable texts frequently get this issue wrong or is it my understanding that's faulty?

Any help on this issue would be greatly appreciated!

  • $\begingroup$ Meteo isn't exactly a liquid in a box, I think is "simple" as that. Whatever water evaporates, if the air above can't keep it would lead to rain or fog. The humidity of air ultimately depends on the P, T of the latter, whatever influences them. $\endgroup$
    – Alchimista
    Jun 8, 2021 at 9:48

1 Answer 1


If we now increase the water temperature in the region, the evaporation rate will briefly exceed the condensation rate, filling the air with more vapor until a new equilibrium is reached.

This isn't quite right. If you were studying the liquid/vapor equilibrium in a small, closed container, this would be the right idea. However, the atmosphere is somewhat more complicated.

The temperature of the water vapor is equal to the temperature of the air, not the liquid water. In a large, dynamic system like the atmosphere, the air and water are generally not in thermal equilibrium. As a result, the threshold (in terms of partial pressure) past which water vapor will condense into clouds or fog is a function of the air temperature, not the water temperature.

As an example - warm air is said to be able to hold more water vapor. Technically speaking, this is because the water vapor which is mixed with warm air is itself warm. If that warm air flows over cold water, the air (and the vapor mixed with it) cools down, decreasing the vapor pressure. As a result, this now-cool air becomes supersaturated and condenses into the fog which so frequently surrounds large bodies of water.

As a second example, when warm, moist air rises, it cools via adiabatic expansion. Since the air temperature drops, so does the vapor pressure of the water vapor it's mixed with, which leads to the formation of clouds. This is why storm systems tend to form at cold fronts; the cool air sinks below the warm air in front of it, pushing the latter upward like a snow plow.

You're right that, from a certain point of view, the only relevant player in the game is H$_2$O. The temperature of the liquid water determines the evaporation rate, while the temperature of the water vapor determines the vapor pressure and condensation rate. In a large, complex system like the atmosphere, the temperature of the water vapor can be identified with the temperature of the air with which it's mixed, while being relatively decoupled from the temperature of any nearby bodies of liquid water. In this sense, it is reasonable to talk about the capacity of the air to hold moisture.

  • $\begingroup$ (1/2) Thanks for the great response! Just before I accept your answer, I know that for liquid water, the evaporation rate increases with increasing temperature which is easy to understand. Also , as you say, when studying a liquid/vapor mixture in a closed container, the condensation rate is proportional to the vapors partial pressure which is also easy to intuit since higher pressure implies more collisions with liquid surfaces meaning more condensation. On the other hand, I am not sure why the vapor pressure of the water vapor in the air would decrease with decreasing air/water vapor .. $\endgroup$ Jun 11, 2021 at 20:56
  • $\begingroup$ (1/2) ... temperature and why this decrease in pressure would lead to increased condensation. In the sealed container case, it is an increased vapor pressure that leads to increased condensation, not decreased vapor pressure. Why is there this asymmetry? Thanks again! $\endgroup$ Jun 11, 2021 at 20:57
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    $\begingroup$ @SalahTheGoat I think you may be confusing vapor pressure with the partial pressure of the water vapor. The vapor pressure is not a measure of how much vapor is present; it is a measure of how much vapor would be present if the liquid and gas were in equilibrium. If the partial pressure - which measures how much vapor is actually present - exceeds the vapor pressure, then too much H$_2$O is in the vapor phase. As a result, vapor will condense into liquid (possibly in the form of droplets) until the partial pressure and vapor pressure are equal. $\endgroup$
    – J. Murray
    Jun 11, 2021 at 21:03
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    $\begingroup$ On the other hand, if the partial pressure is less than the vapor pressure, then too much H$_2$O is in the liquid phase, and there will be a net evaporation. Also, it's important to note that in large systems, condensation does not only occur at the interface of a large body of water; if an air parcel becomes oversaturated (e.g. by cooling and thus reducing the vapor pressure), then the (invisible) vapor will condense into (visible, but very small) liquid droplets, i.e. clouds or fog. $\endgroup$
    – J. Murray
    Jun 11, 2021 at 21:06
  • $\begingroup$ Okay everything is crystal clear now. Thanks for the excellent answer! $\endgroup$ Jun 17, 2021 at 15:21

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