Further discussion of mean free path
Teaching Guidance for 14-16
√N steps gives the average mean free path for a bromine molecule in its wandering amongst air molecules. That would be similar to the stride of an air molecule amongst air molecules. You can therefore conclude that an air molecule moves about 10 -7 m between one collision and the next at atmospheric pressure.
This estimate is one of the great classical approaches to estimating the mean free path of an air molecule, first used in the nineteenth century. There are other methods, depending on measurements of viscosity, Van der Waals constants, etc. The result differs somewhat according to the method chosen. For air at room temperature, older estimates gave the mean free path of an air molecule as (0.8 – 1.0) 10 -7 m but the modern value lies between (0.6 – 0.7) 10 -7 m. In the following discussion 10 -7 m will be assumed.
The number of collisions that an air molecule makes per second will be different from the experimental estimate for bromine molecules because oxygen and nitrogen molecules move faster. Combining an estimate of molecular speed with the mean free path estimate, in one second an air molecule travels 500 m of straightened out path (a distance which contains 500/ 10 -7 mean free paths). So the molecule makes 500 x 10 7 collisions per second.
Footnote on simplifications
This is very rough calculation. Any attempt to make a correction for bromine molecules being bigger would place a very unscientific emphasis on precision in one particular place in a method that is imprecise overall. Judging from the relative densities of liquids and relative molecular masses, bromine molecules have a diameter about 1.2 times that of air molecules.
That makes the ‘average diameter’ for a bromine molecule hitting an air molecules 1.1 times on average as great as for air molecules colliding. Because any mean free path depends on the collision
target area (cross-section), a bromine molecule probably has a mean free path among air molecules about 1/(1.1)
or 1/1.2 or (0.83) times the mean free path of an air molecule among air molecules.
This approach makes no attempt to decide what kind of average should be used for the resultant in a random walk treatment. Does the estimate ‘half brown’ fit best with the root mean square average of the random walk of bromine molecules, or should you use the plain arithmetical average? Since progress is estimated in a vertical direction alone, should you take some component of velocity, or of a mean free path?
Unless you give up this simple experiment, in which students make a guess, and resort to colorimetry and density measurements, these questions remain unanswered. Nor would it be sensible to try to answer them here; that would miss the point of proceeding quickly in a simple story so that you do not lose your students on the way.