As I understand it, evaporation is thought of as a phase transition from a liquid into a gas. Individual molecules get enough energy to break surface tension, and flung up into the gas. A gas that's saturated with water vapor can precipitate out when the temperature goes below the saturation point. e.g. dew.

Is it useful to think of this as the liquid dissolving in the gas, much like a solid dissolves into a liquid rather than thinking of it as two gasses mixed together? i.e. are these fundamentally related concepts, but just occurring in different states of matter? They seem rather similar in their behavior. Water at room temperature and under atmospheric pressure easily "dissolves" in air, but iron doesn't.

If not, why is it not thought of in this way?

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    $\begingroup$ You would have to consider air to be a solvent in this case. Normally, solvents are liquids, not gases. $\endgroup$ Nov 2, 2020 at 23:19
  • $\begingroup$ Why? What purpose would that serve? Don't forget that the whole reason we use technical terms is to make communication easier. A term (or really, even a way of thinking about something) must pay its cost in usefulness. This doesn't pass the test, IMO. $\endgroup$
    – Luaan
    Nov 3, 2020 at 8:28
  • $\begingroup$ I think it's more useful to think of it as two processes--evaporation, where the liquid becomes gas, and dissolution, where the two gases combine. One requires energy input, the other mostly Brownian motion. $\endgroup$ Nov 3, 2020 at 15:28
  • $\begingroup$ Well, you did thought it, so it can be thought like that, doesn't it? More to the point, water vapor in atmosphere is dissolved in air. You can think of all kinds of solutions soild-solid, liquid in solid, solid in plasma, etc. $\endgroup$
    – Mithoron
    Nov 3, 2020 at 17:07

4 Answers 4


I don't think that's a useful perspective to adopt. Fundamentally, dissolution (or solvation) involves particle of the solute being pulled away from the bulk and surrounded by particles of the solvent. Whether this occurs or not depends on how strong the solute-solvent intermolecular forces are relative to the solute-solute forces which bind the solute together. For a given solute under fixed conditions, different solvents will dissolve it more or less readily.

On the other hand, vaporization involves the atoms/molecules of the liquid gaining a sufficient amount of energy to leave the liquid phase. When this happens, the molecules become essentially free, and in particular they are not surrounded by a solvation shell like dissolved solute particles are. This process is not influenced very much by the nature of the gas which exists above the surface of the liquid, and would proceed even if there were no gas above the liquid at all.

Ultimately, solvation is a process which involves the attractive intermolecular forces from a solvent while vaporization does not. Put differently, vaporization is a process which occurs to a single substance while solvation is a physical reaction between two different substances. I think that's too important a distinction to sweep under the rug.

  • $\begingroup$ The similarity I'm looking at is that liquids have a solubility in gasses, related to temperature and pressure, and can become saturated. The gas has a capacity for dissolving the liquid. We even call rain precipitation. $\endgroup$ Nov 3, 2020 at 4:44
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    $\begingroup$ @SteveSether The point is that the gas isn't actively dissolving the liquid, it's just a spectator. Molecules aren't being pulled out of the liquid phase by the gas molecules, they are escaping of their own accord. Once they are in the vapor phase, they aren't surrounded by a cloud of particles of the other gas, they are essentially free. The identity - or even the presence - of the other gas species is very nearly inconsequential. $\endgroup$
    – J. Murray
    Nov 3, 2020 at 5:14
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    $\begingroup$ @SteveSether You mention saturated air, but the fact that some parcel of air is saturated with H$_2$O vapor is a statement about the pressure and temperature of the H$_2$O vapor which is contained within the air. The fact that nitrogen, oxygen, CO$_2$, etc. are also present in the parcel is essentially irrelevant, so thinking that they are somehow actively dissolving the H$_2$O like a solvent dissolves a solute is not correct. $\endgroup$
    – J. Murray
    Nov 3, 2020 at 5:17
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    $\begingroup$ Maybe describing what dissolving is on a molecular level might prove useful in explaining why air isn't a solvent for water? $\endgroup$ Nov 3, 2020 at 19:22
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    $\begingroup$ @Steve Are you familiar with the concept of partial pressure? I think it's relevant to your question. $\endgroup$
    – PM 2Ring
    Nov 4, 2020 at 2:44

Like J Murray, I'd say its not all that useful of a construct.

In particular, its worth looking at the difference between evaporation and boiling. Usually these are both considered "vaporization." It would be very difficult to consider boiling to be a form of dissolving the liquid into the gas, as it can occur within the body of liquid, far from any existing gas.

Vaporization can also occur without a gas at all. If you expose a liquid to a true vacuum, with "no" gas molecules present, the liquid will vaporize. In fact, for many liquids, like water, it will indeed boil, despite there being no gas to boil into.

It also obeys different sorts of behaviors. Consider the case where you have two different solids dissolving into a liquid. For the most part, their dissolution can be thought of independent reactions. However, in the case of two liquids evaporating into a gas, we must consider them combined, as each one's vaporization may increase the pressure of the container, which changes the vaporization properties of the other.


tl;dr: It's complicated.

You've identified some valid relationships between evaporation and solubility, however, it's not so much that evaporation into air is a form of solubility, but rather both evaporation and solvation are forms of phase change.

I think this idea is best illustrated by looking at solid-state solubility: that is, dissolution of solids into other solids. Below is an image of the iron-carbon solid solution phase diagram.

Fe-C Phase diagram

If you've only done introductory Chemistry you've probably only ever seen phase diagrams of temperature plotted against pressure. We can also consider the "concentration" of different chemical elements or molecules in our phase diagrams as well. Here, we have the carbon concentration plotted on the $x$-axis as a mass concentration. On the $y$-axis we have temperature. This is a fairly busy plot, but there are a few features worth pointing out.

For low carbon concentrations, carbon can dissolve completely into the iron structure: that is, carbon is soluble into the iron solvent. With too much carbon, the material becomes a mixture of pure iron and iron carbide (Fe$_{3}$C). The diagram cuts out at 6.67%-mass carbon, which corresponds to pure Fe$_{3}$C. At higher temperatures different structures become favorable, including different phases of solid iron (that is, different arrangements of iron atoms).

The point here is that solution of one compound into another is a phase change - that's why you can see similarities between evaporation and solution. When dissolving a solid into a liquid, the solid will dissolve up to the solubility limit of the system (incl. temperature and pressure), with any excess remaining as a separate phase. This is a general phenomenon with phase changes.

Now, if you're evaporating a liquid into a vacuum the system will also have some finite "solubility limit" (if you insist on thinking of it that way). Note that this is a property of the system, not the liquid or vacuum! If you add more compounds to the system, you are effectively adding more dimensions to your phase diagram (and more complexity to boot!). In general, things like vapor pressure are relatively insensitive to the chemical composition of a gas, so we can approximate them as being independent. There are exceptions! If the vapor reacts chemically with the gas you end up with another dimension in your phase diagram where you need to account for the (potentially 2-way) reactions within the gas itself and the overall equilibrium of the system.

To try and put it as simply as possible: you can think of a phase diagram as describing the chemical equilibrium of a system, in the broadest possible sense.

This is a topic that might be covered over half a course at undergraduate level. Hopefully this answer gives you some sense of where to look for further reading.


There are situations where that kind of formulation can be useful.

For example, in the oil industry, where vapor-liquid equilibrium is often modelled using the 'black oil' model.

The crude oil and natural gas found in reservoirs consist of a mixture of many different hydrocarbons. At any given temperature and pressure, a fraction of the mixture will be either in the gas phase or the liquid phase.

In the 'black oil' model, quantities of oil and gas are referred to by their volume at standard conditions. In the subsurface, two hydrocarbon phases (liquid and vapor) are considered which contain an amount of the oil and gas at standard conditions.

For example, a unit volume of the liquid phase in an oil reservoir (at reservoir temperature and pressure) will contain an amount of the oil at standard conditions (these volumes are related by the 'oil formation volume factor') and a certain amount of the gas, quantified using the 'solution gas-oil ratio' which is the volume of gas (at standard conditions) per volume of oil (also at standard conditions). The gas stays 'dissolved' in the liquid until the pressure falls below the saturation pressure, at which point the gas starts to be liberated from the liquid phase.

This model is extended in a similar manner to the vapour phase, where the gas is modelled as containing a certain amount of vaporised oil.

This isn't really dissolution in the same way that a solid dissolves in a liquid because it is a model of a complex mixture of many different chemical compounds, but it is a useful construct for the oil industry because the volumes of oil and gas, at a set of standard conditions, are the saleable product.


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