We continue to examine the motion of molecules in an ideal gas. Figure 19-4 shows the path of a typical molecule as it moves through the gas, changing both speed and direction abruptly as it collides elastically with other molecules. Between collisions, the molecule moves in a straight line at constant speed. Although the figure shows the other molecules as stationary, they are also moving.
Fig. 19-4 A molecule traveling through a gas, colliding with other gas molecules in its path. Although the other molecules are shown as stationary, they are also moving in a similar fashion.
One useful parameter to describe this random motion is the mean free path λ of the molecules. As its name implies, λ is the average distance traversed by a molecule between collisions. We expect λ to vary inversely with N/V, the number of molecules per unit volume (or density of molecules). The larger N/V is, the more collisions there should be and the smaller the mean free path. We also expect λ to vary inversely with the size of the molecules—with their diameter d, say. (If the molecules were points, as we have assumed them to be, they would never collide and the mean free path would be infinite.) Thus, the larger the molecules are, the smaller the mean free path. We can even predict that λ should vary (inversely) as the square of the molecular diameter because the cross section of a molecule—not its diameter—determines its effective target area.
The expression for the mean free path does, in fact, turn out to be
To justify Eq. 19-25, we focus attention on a single molecule and assume—as Fig. 19-4 suggests—that our molecule is traveling with a constant speed v and that all the other molecules are at rest. Later, we shall relax this assumption.
We assume further that the molecules are spheres of diameter d. A collision will then take place if the centers of two molecules come within a distance d of each other, as in Fig. 19-5a. Another, more helpful way to look at the situation is to consider our single molecule to have a radius of d and all the other molecules to be points, as in Fig. 19-5b. This does not change our criterion for a collision.
Fig. 19-5 (a) A collision occurs when the centers of two molecules come within a distance d of each other, d being the molecular diameter. (b) An equivalent but more convenient representation is to think of the moving molecule as having a radius d and all other molecules as being points. The condition for a collision is unchanged.
As our single molecule zigzags through the gas, it sweeps out a short cylinder of cross-sectional area πd2 between successive collisions. If we watch this molecule for a time interval Δt, it moves a distance v Δt, where v is its assumed speed. Thus, if we align all the short cylinders swept out in interval Δt, we form a composite cylinder (Fig. 19-6) of length v Δt and volume (πd2)(v Δt). The number of collisions that occur in time Δt is then equal to the number of (point) molecules that lie within this cylinder.
Fig. 19-6 In time Δt the moving molecule effectively sweeps out a cylinder of length v Δt and radius d.
Since N/V is the number of molecules per unit volume, the number of molecules in the cylinder is N/V times the volume of the cylinder, or (N/V)(πd2v Δt). This is also the number of collisions in time Δt. The mean free path is the length of the path (and of the cylinder) divided by this number:
This equation is only approximate because it is based on the assumption that all the molecules except one are at rest. In fact, all the molecules are moving; when this is taken properly into account, Eq. 19-25 results. Note that it differs from the (approximate) Eq. 19-26 only by a factor of 
.
We can even get a glimpse of what is “approximate” about Eq. 19-26. The v in the numerator and that in the denominator are—strictly—not the same. The v in the numerator is vavg, the mean speed of the molecules relative to the container. The v in the denominator is vrel, the mean speed of our single molecule relative to the other molecules, which are moving. It is this latter average speed that determines the number of collisions. A detailed calculation, taking into account the actual speed distribution of the molecules, gives 
and thus the factor 
.
The mean free path of air molecules at sea level is about 0.1 μm. At an altitude of 100 km, the density of air has dropped to such an extent that the mean free path rises to about 16 cm. At 300 km, the mean free path is about 20 km. A problem faced by those who would study the physics and chemistry of the upper atmosphere in the laboratory is the unavailability of containers large enough to hold gas samples that simulate upper atmospheric conditions. Yet studies of the concentrations of Freon, carbon dioxide, and ozone in the upper atmosphere are of vital public concern.
(a) What is the mean free path λ for oxygen molecules at temperature T = 300 K and pressure p = 1.0 atm? Assume that the molecular diameter is d = 290 pm and the gas is ideal.
Solution: The Key Idea here is that each oxygen molecule moves among other moving oxygen molecules in a zigzag path due to the resulting collisions. Thus, we use Eq. 19-25 for the mean free path, for which we need the number of molecules per unit volume, N/V. Because we assume the gas is ideal, we can use the ideal gas law of Eq. 19-9 (pV = NkT) to write N/V = p/kT. Substituting this into Eq. 19-25, we find
This is about 380 molecular diameters.
(b) Assume the average speed of the oxygen molecules is v = 450 m/s. What is the average time t between successive collisions for any given molecule? At what rate does the molecule collide; that is, what is the frequency f of its collisions?
Solution: To find the time t between collisions, we use this Key Idea: Between collisions, the molecule travels, on average, the mean free path λ at speed v. Thus, the average time between collisions is
This tells us that, on average, any given oxygen molecule has less than a nanosecond between collisions.
To find the frequency f of the collisions, we use this Key Idea: The average rate or frequency at which the collisions occur is the inverse of the time t between collisions. Thus,
This tells us that, on average, any given oxygen molecule makes about 4 billion collisions per second.
CHECKPOINT 3 One mole of gas A, with molecular diameter 2d0 and average molecular speed v0, is placed inside a certain container. One mole of gas B, with molecular diameter d0 and average molecular speed 2v0 (the molecules of B are smaller but faster), is placed in an identical container. Which gas has the greater average collision rate within its container?
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