Example 5. Maxwell's theory of gases considers the velocities of molecules in a gas. Suppose we have a
gas in a containter. Let
V = the x component of the velocity (meters/sec) of a molecule in a gas
We treat V as a random variable because the velocities of the molecules are not all the same. In fact,
Maxwell's theory says that V is normally distributed with mean zero and
 =
kT
m
where
m = the mass of a molecule in a gas (kilograms)
T = the absolute temperature of the gas in degrees Kelvin (K) (One adds 273 to the
Celsius temperature to get the absolute temperature)
k = Boltzman's constant = 1.4  10-23 J/K (J = joules)
For example, for a nitrogen (N2) molecule at 300 K one has
14 = atomic weight of nitrogen
m = (2 atoms / molecule)(14)(1.6  10-27 kg) = 4.48  10-26 kg
kT
(1.4  10-23 J/K)(300 K)
=
= 93750 (m/sec)2
m
4.48  10-26 kg
 =
kT
=
m
93750  306 m/s
Suppose we want to know the probability that V for such a nitrogen molecule has absolute value between
200 m/s and 400 m/s. Since the normal distribution is symmetric about the origin this is twice the
probabiltity that V is between 200 m/s and 400 m/s. This is
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