*Michael
Fowler 7/21/08*

Oddly, the first published
calculation of the average speed of a molecule in the kinetic theory of gases
appeared in the *Railway Magazine*, of
all places, in 1836. Why there? Well,
the calculation was by John Herapath—he owned the magazine. He was definitely not part of the scientific
establishment: a previous paper of his
on the kinetic theory had been rejected by the Royal Society. But the calculation of molecular speed was in
fact correct. Another outsider, John
James Waterston, submitted an excellent paper on the kinetic theory to the
Royal Society in 1846, to have it rejected as “nonsense”. This was evidently still the age of the
caloric theory, at least in the Royal Society.
In 1848, Joule (who had worked with Herapath, and was also something of
an outsider) presented a paper at a meeting of British Association where he
announced that of the speed of hydrogen molecules at 60F was about 1 mile per
second, close to correct. Again, though,
this did not excite wide interest…

Finally, in 1857, a pillar of the scientific establishment—Clausius—wrote a paper on the kinetic theory, repeating once more the calculation of average molecular speed (around 460 meters per second for room temperature oxygen molecules). He mentioned the earlier work by Joule, and some more recent similar calculations by Krönig. Suddenly people sat up and took notice! If a highly respected German professor was willing to entertain the possibility that the air molecules in front of our faces were mostly traveling faster than the speed of sound, perhaps there was something to it…

But there were obvious objections to this vision of fast molecules zipping by. As a Dutch meteorologist, C. H. D. Buys-Ballot, wrote: [if the molecules are traveling so fast] how does it then happen that tobacco-smoke, in rooms, remains so long extended in immoveable layers?” (Nostalgia trip for smokers!) He also wondered why, if someone opens a bottle of something really smelly, like ammonia, you don’t smell it across the room in a split second, if the molecules are moving so fast. And, why do gases take ages to intermingle?

These were very good questions, and
forced Clausius to think about the theory a bit more deeply. Buys-Ballot had a point: at that speed, the
smelly NH_{3}’s really would fill a whole room in moments. So what was stopping them? The speed of the molecules follows *directly* from measuring the pressure and
density—you don’t need to know the size of molecules. If the kinetic theory is
right at all, this speed has to be correct. Assuming, then, the speed *is* more or less correct, the molecules
are evidently not going in straight lines for long. They must be bouncing off the other
molecules. Still, the standard kinetic theory pressure calculation assumed each
molecule to be bouncing from wall to wall inside the box, collisions with the
other molecules had always been ignored, because the molecules were so
tiny. Evidently, though, they weren’t.

Clausius concluded that the
molecules must be big enough to get in each other’s way to some extent. He assumed the average speed calculation was
still about right, and that the molecules only interacted when they were really
close. So, watching one molecule, most of the time it’s going in a straight
line, not influenced by the other molecules, then it gets close to another one
and bounces off in a different direction.
He termed the average distance between collisions the **mean free path**.

But how could this be reconciled
with the pressure calculation, the pressure from a single molecule being found
by counting the times per second it bounced off a given wall? Evidently each molecule will now take a lot
longer to get across the container—will that lower the pressure? The answer is no: although molecules now take
a long time to do the round trip, they don’t have to—a molecule bouncing off
the wall can hit another nearby molecule and go straight back to the wall. The pressure on a wall depends on the *density of molecules close to the wall*
(less than a mean free path away), and their velocity distribution. This won’t be too different from the
no-collisions case.

This fixed up the theory in the
sense of answering Buys-Ballot’s objections, but it turned out to do much more.
Maxwell took the concept of mean free path, and used it to prove *the viscosity of a gas should be independent
of density* over a very wide range.
(Viscosity arises when fast molecules move into a slower stream: halving
the density halves the number of fast molecules getting over, but they get
twice as far, so penetrate into even slower streams. For a much more detailed explanation, see my lecture on viscosity.) Maxwell was startled (his word: Phil. Mag,
Jan Jun 1860, p 391) by this result, and thought it probably spelled doom for
the kinetic theory, because it contradicted experimental findings. But the experiments, it turned out, were not
very good, and had found the (wrong) answer they expected. Maxwell did the experiments more carefully
himself, and found agreement with the kinetic theory—the viscosity really *didn’t* depend on density! Usually in physics good experiments knock
down bad theories—this time it was the other way round.

How does the mean free path picture handle mutual diffusion of two gases, say oxygen and nitrogen, when a partition initially separating them is removed?

For a box holding a few liters, it takes of the order an hour or so for the gases to mix. (We’re assuming the temperature is kept constant so that convection currents don’t arise—such currents would reduce the time substantially.) Obviously, the rate of mixing must depend on the mean free path: if it was centimeters, the mixing would be pretty complete in milliseconds. In fact, as we shall see, the mean free path can be deduced from the measured rate of penetration of one gas by the other.

First, though, we’ll show how to
derive the mean free path in units of the diameter of the molecules, taking O_{2}
and N_{2} to be spheres of diameter *d*.
(You’re used to seeing them pictured like dumbbells and that’s true of the two
nuclei, but the surrounding electron cloud is in fact close to spherical.)

Think of one of the oxygen
molecules moving into nitrogen. So now
visualize the little O_{2} sphere shooting into this space where all
these other spheres are moving around. *Temporarily, for ease of visualization,
let’s imagine all the other spheres to be at rest*. This is a pinball machine scenario:

How far can we expect the O_{2}
to get before it hits an N_{2}?
The *average* distance before a
collision is the mean free path. Let’s
try to picture how much room there is to fly between these fixed N_{2}
spheres. (Bear in mind that the picture
above should be three-dimensional!) We
do know that if it were *liquid*
nitrogen, there would be *very little*
room: liquids are just about incompressible, so the molecules must be
touching. Roughly speaking, a molecule
of diameter *d* will occupy a cubical
volume of about *d*^{3} (there
has to be some space left over—we can pack cubes to fill space, but not
spheres.)

We also know that liquid nitrogen
weighs about 800 kg per cubic meter, whereas N_{2} gas at room
temperature (and pressure) weighs about 1.2 kg per cubic meter, a ratio of
670. This means that *on average each molecule in the gas has 670
times more room*—that is, it has a space 670 times the volume *d*^{3} we gave it in the
liquid. So in the gas, the average
center-to-center separation of the molecules will be the cube root of 670,
which is about 8.75*d*. So the picture is a gas of spheres of
diameter *d*, placed at random, but
separated on average by distances of order 10*d*. It’s clear that shooting
an oxygen molecule into this it will get quite a way. Let us emphasize again that this picture is
independent of the actual *size* of *d*: we’re only considering the *ratio* of mean free path to molecular
diameter.

We now estimate just how far an O_{2}
will get, on average, as it shoots into this forest of spheres. Picture the motion of the center of the
oxygen molecule. Before any collision,
it will be moving on a straight-line path.
Just how close does the O_{2} center have to get to an N_{2}
center for a hit? Taking both O_{2},
N_{2} to be spheres of diameter *d*,
if an N_{2} center lies within *d*
of the O_{2} center’s path, there will be a hit.

*Note*: this picture is simplified in that we’re
taking the molecules to be hard spheres. This is not a bad approximation, but
the repulsive potential does not fall quite that suddenly, and consequently the
effective scattering size varies somewhat with speed, i.e. with temperature. This turns out to be important in finding the
temperature dependence of viscosity, for example.

So we can think of the O_{2}
as sweeping out a volume, a cylinder of *radius**d* centered on its path, hitting and deflecting if it encounters an
N_{2} *centered within that
cylinder*. So how far will it get, on
average, before a hit? In traveling a
distance *x*, it sweeps out a volume _{}. Now picture it going
through the gas for some considerable length of time, so there are many
collisions. The volume swept out will
look like a stovepipe, long straight cylindrical sections connected by elbows
at the collisions. The total volume of
this stovepipe (ignoring tiny corrections from the elbows) will be just _{}, *L* being the total length, that is, the total
distance the molecule traveled.

If the density of the nitrogen is *n* molecules per cubic meter, the number
of N_{2}’s in this stovepipe volume will be _{}, in other words, this will be the number of collisions. Therefore, the average distance *between *collisions, the mean free path *l*, is given by:

_{}

So what is *n*? We estimated above that
each molecule has space 670*d ^{ }*

Therefore, the mean free path is given by

_{}

(We can see from this that the average length of stovepipe sections between elbows is 200 times the pipe radius, so neglecting any volume corrections from the elbows was an excellent approximation, and our diagram has the sections far too short compared with the diameter.)

**Notice that this derivation of the mean free path in terms of the molecular diameter depends only on knowing the ratio of
the gas density to the liquid density—it
does not depend on the actual size of the molecules! **

But it *does* mean that if we can somehow measure the mean free path, by
measuring how fast one gas diffuses into another, for example, we can deduce
the size of the molecules, and historically this was one of the first ways the
size of molecules was determined, and so Avogadro’s number was found.

There is one further correction we
should make. We took the N_{2}
molecules to be at rest, whereas in fact they’re moving as fast as the oxygen
molecule, approximately. This means that
even if the O_{2} is temporarily at rest, it can undergo a collision as
an N_{2} comes towards it.
Clearly, what really counts in the collision rate is the *relative* velocity of the molecules.

Defining the average velocity as
the root mean square velocity, if the O_{2} has velocity _{}and the N_{2 } _{}, then the square of the relative velocity

_{},

since _{} must average to zero,
the relative directions being random. So
the average *square *of the relative
velocity is twice the average square of the velocity, and therefore the average
root-mean-square velocity is up by a factor √2, and the collision rate is
increased by this factor. Consequently,
the mean free path is *decreased *by a
factor of √2 when we take into account that *all* the molecules are moving.

Our final result, then, is that the mean free path

_{}.

Finding the mean free path is—literally—the first step in figuring out how rapidly the oxygen atoms will diffuse into the nitrogen gas, and of course vice versa.

What we really want to know is just
how much we can expect the gases to have intermingled after a given period of
time. We’ll just follow the one
molecule, and estimate how far it gets. To begin with, let’s assume for
simplicity that it tales steps all of the same length *l*, but after each collision it bounces off in a random
direction. So after *N* steps, it will have moved to a point

_{},

where each vector _{} has length *l*, but the vectors all point in random
different directions.

If we now imagine many of the
oxygen molecules following random paths like this, how far on average can we
expect them to have drifted after *N *steps? (Note that they could with equal likelihood
be going backwards!) The appropriate
measure is the root-mean-square distance,

_{}

Since the direction after each
collision is completely random, _{}, and the root-mean-square distance

_{}.

If we allow steps of different
lengths, the same argument works, but now *l*
is the root-mean-square path length. *The important factor here is the *_{}.

This means that the average
distance diffused in one second is _{}, say half a centimeter (justified in the next section). The average distance in one *hour* would be *only* 60 times this, or 30 cm., and in a day about a meter and a
half—the average distance traveled is only increasing as the *square root* of the time elapsed!

This is a very general result. For example, suppose we have a gas in which
the mean free path is *l* and the
average speed of the molecules is *v*. Then the average time between collisions_{}. The number of
collisions in time* t* will be _{} so the average
distance a molecule moves in time *t *will
be _{}.

It should be clear from the above
that by carefully observing how quickly one gas diffuses into another, the mean
free path could be estimated. Obviously,
oxygen and nitrogen are not the best candidates: to see what’s going on, a
highly visible gas like bromine diffusing into air would be more practical.
However, there’s a better way to find the mean free path. As we proved in the lecture on viscosity,
the viscosity coefficient _{}, where *n* is the
number density, *m* the molecular mass,
*v* the average speed and *l* the mean free path. The viscosity can be measured quite
accurately, the mean free path in air was found to be _{}, or _{}

In 1865, Josef Loschmidt gave the first good estimate of the size of molecules. He used the viscosity data to find the mean free path, assumed as we did above that the molecules were more or less touching each other in the liquid, then used the geometric argument above to nail down the ratio of molecular size to mean free path. He overestimated by a factor of three or so, but this was much closer to the truth than anyone else at the time.

Here are some numbers: for O_{2},
N_{2}, _{}, _{}, the speed of the molecules at room temperature *v* is approximately 500 meters per sec.,
so the molecule has of order 10^{10} collisions per second.

A famous mystery cleared up by
arguments like this was that _{}, as we discussed earlier in the course, with *B* the bulk modulus. But when *B*
was measured carefully by slowly compressing air, the result was in error by
about 30%! The speed of sound predicted
a higher (stiffer) bulk modulus.

The explanation turned out to be
that in a slow measurement of the bulk modulus, the gas stays at the same temperature—the
heating caused by slow compression leaks away.
But if the compression is rapid, the gas heats up and so the pressure
goes up more than if it had stayed at the same temperature. So the question is whether the compression
and decompression as a sound wave passes through is so rapid that the heated-up
gas doesn’t have time to spread to the cooled regions. For sound at say 1000Hz,
the wavelength is 34 cm. If compression
heats gas locally, the hot molecules will diffuse away in a similar manner to
that discussed above. They will be slightly faster than the average
molecules. In 1/1000 th of a second,
they will have 10^{7} collisions, so will travel about _{}. This tiny distance
compared with the wavelength of the sound wave means that during the
compression/decompression cycles as the wave passes through, the heat has no
chance to dissipate—so, effectively, it’s like compressing a gas in an
insulated container, it’s harder to compress than it would be if the heat
generated could flow away, and the bulk modulus is higher by an amount (around
30%) we shall work out in a forthcoming lecture.