Knudson Flow

We have previously calculated the number of molecules colliding with a wall (or a surface) per unit area per unit time. Recall that we called this number zwall. It was given by the expression,

                 (1)
where <v > is the average speed of the molecules.

What if a small part of the wall isn't there? In other words, there is a hole in the wall. Then the molecules that would have collided with the wall at that spot will escape. As long as the hole is small (we will call it a pin-hole), small enough that there is no bulk mass flow out the hole so that the gas leaks out very slowly, then the pressure and temperature in the vessel will be well defined and we can use Equation 1 to describe the number of molecules that escape through the pin-hole. The number of molecules that escape in time, t, through an area, A, will be given by,

                (2)
If the pressure of the gas in not too high we can approximate the gas as an ideal gas so that
                (3)
and, recalling that the average speed is given by,
                (4)
we get
                (5)
In a typical experiment we would measure the area of the pin-hole and weigh the initial sample of a volatile liquid or solid (and the container) at the beginning of the experiment. Then weigh the sample and container again after they have been thermostated at temperature, T, for a time, t. The mass loss is,
                (6a, b, c)
If we have an independent measure of the pressure, p, we can determine the mass of a molecule and hence the formula weight of the substance. If we know the formula weight of the substance we can determine the vapor pressure of the substance. This method has been used, for example, to measure the vapor pressure of beryllium metal at high temperatures.
 

Molecular Collisions

We will now assume that our molecules have a "size" so that they can collide with each other. (Point masses have no size or volume so they never collide with each other.) In this course we will also assume that the molecules are "hard spheres." That is, they are not "squishy," they don't flex when they collide. One molecule cannot penetrate within another molecule. (That is, the molecules behave more like very hard billiard balls than tennis balls.)

Let the diameter of a molecule be, d, so that the radius of the molecule is d/2. We track the position of the molecule by tracking the location of the center of the molecule. Suppose we freeze the action right at the instant that two molecules collide. At that instant the two molecules are just touching. At that instant the centers of the two molecules are separated by a distance equal to twice the radius, or 2×d/2 = d. The centers of any two molecules can never be closer than a distance d to each other. We define the "collision cross section" of the molecule as,

                (7)
If you were shooting other molecules at our fixed molecule considered to be the target, this σ is the area that the "missile" molecules would have to hit in order to collide with the target molecule. (Some texts use σ in their equations and some texts use πd 2, we will use πd 2.)

Suppose our molecule is traveling through space with speed, <v>. Think of this πd 2 as a cross sectional area traveling through space with speed, <v>. This cross section will sweep out a volume of <v>π d 2 per second. Any other molecule that happens to be in this volume will suffer a collision with the traveling molecule, So, if there are N/V molecules per unit volume, then there will be

                (8)
collisions of our molecule with other molecules per second. (There is a complication, however. The other molecules are not standing still, they are moving too. The average speed, ávñ, that we put into this equation is the average speed relative to our laboratory frame of reference. Really we should put in the average speed relative to all the other moving molecules. So we should rewrite our equation for the number of collisions per second as
                (9)
We don't want to go into too much detail here, but the average relative speed is obtained by replacing the mass of the molecule in the equation for average speed,
                (10)
by the so-called "reduced mass" of the two particles in a collision. The reduced mass, which we will call, m , for two particles, A and B is defined by
                (11)
or
                (12)
If the two molecules are the same, as they are here, the reduced mass is simply
                (13)
If we replace m in the formula for average speed by the reduced mass, m, we get the average relative speed,
                (14)
Then the number of collisions of one molecule (with all the other molecules) per second is,
                (15)
Knowing how many collisions a single molecule makes per second as it moves through the gas allows us to calculate how far, on the average, the molecule moves between collisions. We call this distance the "mean free path," and give it the symbol, l . This number, l , is just the average speed (in the laboratory) divided by the number of collisions per second,
                (16)
Notice that the mean free path does not depend on the speed of the molecule, only on the collision cross section and the number density. If the density is constant the mean free path doesn't even depend on temperature.

There are several more things we have to deal with regarding molecular collisions. We need to talk about collisions between unlike molecules.

Suppose we have two different gases in our sample, say gas A and gas B. Let the diameter of gas a molecules be dA and the diameter of gas B molecules be dB. Then the collision diameter for a collision between a molecule of A and a molecule of B is,

                (17)
and the number of collisions a molecule of A makes with other B molecules is
                (18)
Notice that there is no square root of two here because we are using the actual average relative speed. The number of collisions a molecule of B makes with other A molecules is likewise,
                (19)
We now have enough information to ask how many total A-B collisions there are in the gas per unit volume per second. We will call this number will call, ZAB. It is given by,
                (20)
which is the same as
                (21)
Note, however that the number of A-A collisions is,
                (22)
or
                (23)
where the extra ½ keeps us from counting each A-A collision twice, once when the first A collides with all the other A's and once when all the other A molecules collide with the first A.

These equations will be useful when we try to understand chemical reaction rates resulting from collisions between reacting molecules.

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Copyright 2004, W. R. Salzman
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Last updated 09 Jul 04
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