I am a mathematician now studying an introductory chemistry course.

Consider a liquid in a closed container, at (say) room temperature.

Then some of the liquid will vaporize. Then some of the resulting gas would condensate, and vice-versa, until a dynamic equilibrium will be achieved.

Let $L(t),g(t)$ be the mass of the liquid and the gas substances respectively, at time $t$.

Assuming that a fraction $c_1$ of the liquid is converted to gas, and a fraction $c_2$ of the gas is converted to liquid, we arrive at the following system of differential equations:

$$ \dot g =c_1L-c_2g, \dot L =c_2g-c_1L, $$ or in matrix form $$ \begin{pmatrix} \dot{g} \\\dot{L} \end{pmatrix}=\begin{pmatrix} -c_2 & c_1 \\\ c_2 & -c_1 \end{pmatrix} \begin{pmatrix} g \\\ L \end{pmatrix}. $$

(We also have the initial conditions $L(0)=L_0, g(0)=0$ where $L_0$ is the initial mass of the liquid. The initial amount of gas is zero).

Question: Is this (admittedly very naive) model considered in the literature? Is it a reasonable model for describing the phenomena?

  • 1
    $\begingroup$ Short answer is yes. These are standard reaction rate equations that are used for any chemical kinetics description. Read the kinetics section of any intro chem text for more info $\endgroup$
    – Andrew
    May 13 at 23:26

2 Answers 2


What you're trying to develop is a kinetic model of a gas-liquid equilibrium. There are are such models, but I'm afraid yours won't work.

The reason your model won't work is that it says the rate of evaporation is proportional to the mass of the liquid, and the rate of condensation is proportional to the mass of the gas.

But the rate of evaporation is instead proportional to the surface area of the liquid-gas interface and the tendency of the liquid molecules to escape into the gas phase from that surface. The latter is in turn determined by the nature of the liquid and the temperature (and, to a much lesser degree, the pressure, since changing the pressure changes the chemical potential of the liquid).

For instance, if you compare two sealed cylinders of the same diameter containing a liquid-gas mixture at equilibrium (same substance in both cylinders), increasing the mass of the liquid simply raises the position of the liquid-gas interface in the cylinder. It will have no effect on the rate of evaporation at that interface.

And the rate of condensation depends on the rate at which gas molecules are hitting the surface and the fraction of those striking the surface that condense. The former will depend on the the surface area of the liquid-gas interface, the temperature and molecular mass (temperature and molecular mass together determine molecular speed), and the concentration of the gas. The latter (the fraction hitting the surface that condense) is knowns as the condensation coefficient and, surprisingly, there remains disagreement about its value. See: D. T. Jamieson, Condensation Coefficient of Water, Nature, v. 202, p. 583 (1964).

I do applaud your effort to apply your mathematical background to what is a new field for you. But to make useful mathematical models, it is first necessary to understand the phenomenon physically. So I would focus on developing your physical intution for the field.

Indeed, if you want to explore this more deeply, you might want to take a look through the articles that cited that 1964 Nature paper. That list is easily obtained by searching for the paper in Google Scholar (there's about 200, but you should be able to filter that down quite a bit):


Note that, while you're considering an equilibrium system, most of the above appear to be for non-equilibrium systems. This makes sense, since being able to model rates of evaporation and condensation is important for commerical and industrial applications, and is also a key part of atmospheric chemistry and the study of water cycles.

Also, just a note about terms: You wrote "Then some of the resulting gas would condensate". Condensate is a noun—it's the substance that condenses. What you want is the verb, which is "condense".

  • 2
    $\begingroup$ Thanks, that is a perfect answer. It clarifies that I used wrong physical assumptions. In particular , I really liked your comparison between two cylinders containing different masses. I guess I was too hasty to apply math without waiting for more advanced stages in the course, where kinetic models are discussed. I now look forward to studying them. Thanks for your patience and clear explanations. $\endgroup$ May 14 at 6:19

I fail to see how this explains anything about the equilibrium between liquid and vapor. At equilibrium the free energy change between gas and liquid is zero and is independent of the amount of material as long as both phases exist. If the chemical potential of the vapor [fugacity]exceeds that of the liquid, vapor will condense, if that of the liquid exceeds the vapor, liquid will evaporate.


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