previous index next PDF
Molecular Collisions
Michael Fowler 6/19/06
Molecular Visualization of the Diffusion Process
In analyzing the gas so far, we’ve ignored collisions between molecules, and in fact for air at ordinary temperatures the relationship between pressure, volume and temperature came out correctly. Furthermore, Maxwell’s speed distribution can be used to find what fraction of the molecules in a planet’s atmosphere are moving at above escape velocity, so we can predict what gases will remain surrounding a planet, given the gravitational force near the surface, and the temperature.
But there are other phenomena for which an understanding of collisions is all-important. For example, if two different gases, say oxygen and nitrogen, in a container at the same pressure and temperature are separated by a partition, how quickly will they mix once the partition is removed? Assuming room temperature, the molecules will be moving at hundreds of meters per second, so one might imagine the mixing will be over in hundredths of a second. But that is not the case at all—observationally, it might take an hour, for a box holding a few liters. This surprisingly slow penetration of one gas by another is called diffusion. (We’re assuming the temperature is kept constant so that convection currents don’t arise—such currents could reduce the time substantially.)
The reason it takes the gases so long to mix becomes evident on tracking one molecule as it enters the other gas. Think of an oxygen molecule moving into nitrogen. We’ll take O2 and N2 to be little 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.) So now visualize the little O2 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.
How far can we expect the O2 to get before it hits an N2? The average distance before a collision is called the mean free path. Let’s try to picture how much room there is to fly between these fixed N2 spheres. 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 d3 (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 N2 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 d3 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 8.75d. So the picture is a gas of spheres of diameter d, placed at random, but separated on average by distances of order 10d. It’s clear that shooting an oxygen molecule into this it will get quite a way.
We now try to estimate just how far an O2 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 O2 center have to get to an N2
center for a hit? Taking both O2,
N2 to be spheres of diameter d,
if an N2 center lies within d
of the O2 center’s path, there will be a hit. So we can think of the O2 as
sweeping out a volume, a cylinder of radius d centered on its path, hitting and deflecting if it encounters an
N2 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 N2’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 670d 3 to
itself, so n is just how many of
those volumes there are in one cubic meter, that is, 
.
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.)
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.
Let us now put in some numbers to find this mean free path:
for O2, N2, 
, so the mean free path 
, or 
The speed of the
molecules at room temperature v is
approximately 500 meters per sec., so the molecule has of order 1010
collisions per second!
Actually, there is one further correction we should
make. We took the N2
molecules to be at rest, whereas in fact they’re moving as fast as the oxygen
molecule, approximately. This means that
even if the O2 is temporarily at rest, it can undergo a collision as
an N2 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 O2 has velocity 
and the N2 
, 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 
. Recall from above
that , or 
and there are of order
1010 collisions per second.
This means that the average distance diffused in one second is 
, say half a centimeter.
The average distance in one hour would be only 60 times this, or 30 cm.,
one foot, and in a day about five feet—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 
.
Why did Newton
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 107
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.