Evaporation
In this chapter we shall discuss some further applications of kinetic theory. In the previous chapter we emphasized one particular aspect of kinetic theory, namely, that the average kinetic energy in any degree of freedom of a molecule or other object is $\tfrac{1}{2}kT$. The central feature of what we shall now discuss, on the other hand, is the fact that the probability of finding a particle in different places, per unit volume, varies as $e^{-\text{potential energy}/kT}$; we shall make a number of applications of this. The phenomena which we want to study are relatively complicated: a liquid evaporating, or electrons in a metal coming out of the surface, or a chemical reaction in which there are a large number of atoms involved. In such cases it is no longer possible to make from the kinetic theory any simple and correct statements, because the situation is too complicated. Therefore, this chapter, except where otherwise emphasized, is quite inexact. The idea to be emphasized is only that we can understand, from the kinetic theory, more or less how things ought to behave. By using thermodynamic arguments, or some empirical measurements of certain critical quantities, we can get a more accurate representation of the phenomena. However, it is very useful to know even only more or less why something behaves as it does, so that when the situation is a new one, or one that we have not yet started to analyze, we can say, more or less, what ought to happen. So this discussion is highly inaccurate but essentially right—right in idea, but a little bit simplified, let us say, in the specific details. The first example that we shall consider is the evaporation of a liquid. Suppose we have a box with a large volume, partially filled with liquid in equilibrium and with the vapor at a certain temperature. We shall suppose that the molecules of the vapor are relatively far apart, and that inside the liquid, the molecules are packed close together. The problem is to find out how many molecules there are in the vapor phase, compared with the number there are in the liquid. How dense is the vapor at a given temperature, and how does it depend on the temperature? Let us say that $n$ equals the number of molecules per unit volume in the vapor. That number, of course, varies with the temperature. If we add heat, we get more evaporation. Now let another quantity, $1/V_a$, equal the number of atoms per unit volume in the liquid: We suppose that each molecule in the liquid occupies a certain volume, so that if there are more molecules of liquid, then all together they occupy a bigger volume. Thus if $V_a$ is the volume occupied by one molecule, the number of molecules in a unit volume is a unit volume divided by the volume of each molecule. Furthermore, we suppose that there is a force of attraction between the molecules to hold them together in the liquid. Otherwise we cannot understand why it condenses. Thus suppose that there is such a force and that there is an energy of binding of the molecules in the liquid which is lost when they go into the vapor. That is, we are going to suppose that, in order to take a single molecule out of the liquid into the vapor, a certain amount of work $W$ has to be done. There is a certain difference, $W$, in the energy of a molecule in the liquid from what it would have if it were in the vapor, because we have to pull it away from the other molecules which attract it. Now we use the general principle that the number of atoms per unit volume in two different regions is $n_2/n_1 = e^{-(E_2 - E_1)/kT}$. So the number $n$ per unit volume in the vapor, divided by the number $1/V_a$ per unit volume in the liquid, is equal to \begin{equation} \label{Eq:I:42:1} nV_a = e^{-W/kT}, \end{equation} because that is the general rule. It is like the atmosphere in equilibrium under gravity, where the gas at the bottom is denser than that at the top because of the work $mgh$ needed to lift the gas molecules to the height $h$. In the liquid, the molecules are denser than in the vapor because we have to pull them out through the energy “hill” $W$, and the ratio of the densities is $e^{-W/kT}$. This is what we wanted to deduce—that the vapor density varies as $e$ to the minus some energy or other over $kT$. The factors in front are not really interesting to us, because in most cases the vapor density is very much lower than the liquid density. In those circumstances, where we are not near the critical point where they are almost the same, but where the vapor density is much lower than the liquid density, then the fact that $n$ is very much less than $1/V_a$ is occasioned by the fact that $W$ is very much greater than $kT$. So formulas such as (42.1) are interesting only when $W$ is very much bigger than $kT$, because in those circumstances, since we are raising $e$ to minus a tremendous amount, if we change $T$ a little bit, that tremendous power changes a bit, and the change produced in the exponential factor is very much more important than any change that might occur in the factors out in front. Why should there be any changes in such factors as $V_a$? Because ours was an approximate analysis. After all, there is not really a definite volume for each molecule; as we change the temperature, the volume $V_a$ does not stay constant—the liquid expands. There are other little features like that, and so the actual situation is more complicated. There are slowly varying temperature-dependent factors all over the place. In fact, we might say that $W$ itself varies slightly with temperature, because at a higher temperature, at a different molecular volume, there would be different average attractions, and so on. So, while we might think that if we have a formula in which everything varies in an unknown way with temperature then we have no formula at all, if we realize that the exponent $W/kT$ is, in general, very large, we see that in the curve of the vapor density as a function of temperature most of the variation is occasioned by the exponential factor, and if we take $W$ as a constant and the coefficient $1/V_a$ as nearly constant, it is a good approximation for short intervals along the curve. Most of the variation, in other words, is of the general nature $e^{-W/kT}$. It turns out that there are many, many phenomena in nature which are characterized by having to borrow an energy from somewhere, and in which the central feature of the temperature variation is $e$ to the minus the energy over $kT$. This is a useful fact only when the energy is large compared with $kT$, so that most of the variation is contained in the variation of the $kT$ and not in the constant and in other factors.  Now let us consider another way of obtaining a somewhat similar result for the evaporation, but looking at it in more detail. To arrive at (42.1), we simply applied a rule which is valid at equilibrium, but in order to understand things better, there is no harm in trying to look at the details of what is going on. We may also describe what is going on in the following way: the molecules that are in the vapor continually bombard the surface of the liquid; when they hit it, they may bounce off or they may get stuck. There is an unknown factor for that—maybe $50$–$50$, maybe $10$ to $90$—we do not know. Let us say they always get stuck—we can analyze it over again later on the assumption that they do not always get stuck. Then at a given moment there will be a certain number of atoms which are condensing onto the surface of the liquid. The number of condensing molecules, the number that arrive on a unit area per unit time, is the number $n$ per unit volume times the velocity $v$. This velocity of the molecules is related to the temperature, because we know that $\tfrac{1}{2}mv^2$ is equal to $\tfrac{3}{2}kT$ on the average. So $v$ is some kind of a mean velocity. Of course we should integrate over the angles and get some kind of an average, but it is roughly proportional to the root-mean-square velocity, within some factor. Thus \begin{equation} \label{Eq:I:42:2} N_c = nv \end{equation} is the rate at which the molecules arrive per unit area and are condensing. At the same time, however, the atoms in the liquid are jiggling about, and from time to time one of them gets kicked out. Now we have to estimate how fast they get kicked out. The idea will be that at equilibrium the number that are kicked out per second and the number that arrive per second are equal. How many get kicked out? In order to get kicked out, a particular molecule has to have acquired by accident an excess energy over its neighbors—a considerable excess energy, because it is attracted very strongly by the other molecules in the liquid. Ordinarily it does not leave because it is so strongly attracted, but in the collisions sometimes one of them gets an extra energy by accident. And the chance that it gets the extra energy $W$ which it needs in our case is very small if $W \gg kT$. In fact, $e^{-W/kT}$ is the chance that an atom has picked up more than this much energy. That is the general principle in kinetic theory: in order to borrow an excess energy $W$ over the average, the odds are $e$ to the minus the energy that we have to borrow, over $kT$. Now suppose that some molecules have borrowed this energy. We now have to estimate how many leave the surface per second. Of course, just because a molecule has the necessary energy does not mean that it will actually evaporate, since it may be buried too deeply inside the liquid or, even if it is near the surface, it may be travelling in the wrong direction. The number that are going to leave a unit area per second is going to be something like this: the number of atoms there are near the surface, per unit area, divided by the time it takes one to escape, multiplied by the probability $e^{-W/kT}$ that they are ready to escape in the sense that they have enough energy. We shall suppose that each molecule at the surface of the liquid occupies a certain cross-sectional area $A$. Then the number of molecules per unit area of liquid surface will be $1/A$. And now, how long does it take a molecule to escape? If the molecules have a certain average speed $v$, and have to move, say, one molecular diameter $D$, the thickness of the first layer, then the time it takes to get across that thickness is the time needed to escape, if the molecule has enough energy. The time will be $D/v$. Thus the number evaporating should be approximately \begin{equation} \label{Eq:I:42:3} N_e = (1/A)(v/D)e^{-W/kT}. \end{equation} Now the area of each atom times the thickness of the layer is approximately the same as the volume $V_a$ occupied by a single atom. And so, in order to get equilibrium, we must have $N_c = N_e$, or \begin{equation} \label{Eq:I:42:4} nv = (v/V_a)e^{-W/kT}. \end{equation} We may cancel the $v$’s, since they are equal; even though one is the velocity of a molecule in the vapor and the other is the velocity of an evaporating molecule, these are the same, because we know their mean kinetic energy (in one direction) is $\tfrac{1}{2}kT$. But one may object, “No! No! These are the especially fast-moving ones; these are the ones that have picked up excess energy.” Not really, because the moment they start to pull away from the liquid, they have to lose that excess energy against the potential energy. So, as they come to the surface they are slowed down to the velocity $v$! It is the same as it was in our discussion of the distribution of molecular velocities in the atmosphere—at the bottom, the molecules had a certain distribution of energy. The ones that arrive at the top have the same distribution of energy, because the slow ones did not arrive at all, and the fast ones were slowed down. The molecules that are evaporating have the same distribution of energy as the ones inside—a rather remarkable fact. Anyway, it is useless to try to argue so closely about our formula because of other inaccuracies, such as the probability of bouncing back rather than entering the liquid, and so on. Thus we have a rough idea of the rate of evaporation and condensation, and we see, of course, that the vapor density $n$ varies in the same way as before, but now we have understood it in some detail rather than just as an arbitrary formula. This deeper understanding permits us to analyze some things. For example, suppose that we were to pump away the vapor at such a great rate that we removed the vapor as fast as it formed (if we had very good pumps and the liquid was evaporating very slowly), how fast would evaporation occur if we maintained a liquid temperature $T$? Suppose that we have already experimentally measured the equilibrium vapor density, so that we know, at the given temperature, how many molecules per unit volume are in equilibrium with the liquid. Now we would like to know how fast it will evaporate. Even though we have used only a rough analysis so far as the evaporation part of it is concerned, the number of vapor molecules arriving was not done so badly, aside from the unknown factor of reflection coefficient. So therefore we may use the fact that the number that are leaving, at equilibrium, is the same as the number that arrive. True, the vapor is being swept away and so the molecules are only coming out, but if the vapor were left alone, it would attain the equilibrium density at which the number that come back would equal the number that are evaporating. Therefore, we can easily see that the number that are coming off the surface per second is equal to one minus the unknown reflection coefficient $R$ times the number that would come down to the surface per second were the vapor still there, because that is how many would balance the evaporation at equilibrium: \begin{equation} \label{Eq:I:42:5} N_e = nv(1-R) = (v(1-R)/V_a)e^{-W/kT}. \end{equation} Of course, the number of molecules that hit the liquid from the vapor is easy to calculate, since we do not need to know as much about the forces as we do when we are worrying about how they get to escape through the liquid surface; it is much easier to make the argument the other way.