whereas the average volume V/Nis more of a three-dimensional average vol-
ume. However, equation 19.38 gives us an idea of how far any one gas particle
will travel on average before colliding with another gas particle. If we use the
ideal gas law to substitute for volume (VNkT/p) in equation 19.38, we get
k
dT
(^2) p (19.39)
This expression for shows how the mean free path varies with other observ-
able values of a gas. At higher temperatures (with everything else in equation
19.39 remaining constant), increases. This is consistent with our understand-
ing that if the temperature increases at constant p, the volume of the gas must
increase, meaning that there will be more room between individual gas particles.
If the pressure increases (with everything else in equation 19.39 remaining con-
stant), must decrease. Again, this is consistent, since an increase in pressure will
force the gas particles closer together (that is, the volume will decrease) and in-
dividual gas particles will collide more frequently and over a shorter distance.
Example 19.5
Assume that the hard-sphere radius of a krypton atom is 1.85 Å. Estimate the
mean free path of krypton atoms at 20.0°C and 1.00 bar pressure.
Solution
Recall that 1 Å equals 10^10 m. Since the diameter is twice the radius, we will
use d3.70 Å 3.70 10 ^10 m. Setting up the variables in equation 19.39,
we have
k
d
T
(^2) p
The units don’t automatically work out; instead, we must remember that
100 J 1 Lbar, and that a liter is defined as (0.1 m)^3. Adding these conver-
sion factors:
1
1
L
0
0
ba
Jr
(0.
1
1
L
m)^3
Now the units work out properly, and we calculate as
9.41 10 ^8 m 941 Å
In this case, an average krypton atom travels nearly 300 times its diameter be-
fore it collides with another atom. This would be equivalent to a pool ball
traveling the length of about 10 pool tables before it struck another ball (on
average).Mean free paths are estimates: they are derived using average values, and
they assume that a gas particle is a hard sphere, although many gas molecules
aren’t even spherical. But mean free paths do provide useful quantitative val-
ues for understanding how gas particles interact.Example 19.6
Assume that a nitrogen molecule acts as a hard sphere with radius 1.60 Å. If
you have a vacuum chamber that is 1.00 m on a side, what pressure would it(1.381 10 ^23 J/K)(293.15 K)
(3.7 10 ^10 m)^2 (1.00 bar)(1.381 10 ^23 J/K)(293.15 K)
(3.70 10 ^10 m)^2 (1.00 bar)668 CHAPTER 19 The Kinetic Theory of Gases