74 Chapter 3. The molecular dance[[Student version, December 8, 2002]]
times theaveragevelocity-squared. As in Equation 3.9, we use the shorthand notation〈vx^2 〉for
this quantity.
The force per unit area on the wall is calledpressure,sowehavejust found that
p=m〈vx^2 〉N/V. (3.18)
Eureka. Our simple formula Equation 3.18, which just embodies the idea that gas consists of
molecules in motion, has already explained two key features of the experimentally observed ideal
gas law (Equation 1.11), namely the facts that the pressure is proportional toNand to 1/V.
Skeptics may say, “Wait a minute. In a real gas, the molecules aren’t all traveling along thex
direction!” It’s true. Still, it’s not hard to do a better job. Figure 3.5b shows the situation. Each
individual molecule has a velocity vectorv. When it hits the wall atx=L,its componentvx
changes sign, butvyandvzdon’t. So, the momentum delivered to the wall is again 2mvx.Also,
the time between bounces off this particular wall is once again 2L/vx,eventhough in the meantime
the molecule may bounce off other walls as well, due to its motion alongyandz. Repeating
the argument leading to Equation 3.18 in this more general situation, we find that it needs no
modifications.
Combining the ideal gas law with Equation 3.18 gives
m〈vx^2 〉=kBT. (3.19)
Notice that the gas molecules are flying around at random. So the average〈vx〉is zero: There are
just as many traveling left as there are traveling right, so their contributions to〈vx〉cancel. But the
squareof the velocity can have a nonzero average,〈vx^2 〉.Just as in the discussion of Equation 3.11
above, both the left-movers and right-movers have positive values ofvx^2 ,sothey don’t cancel but
rather add.
In fact, there’s nothing special about thexdirection. The averages〈vx^2 〉,〈vy^2 〉,and〈vz^2 〉are
all equal. That means that their sum is three times as big as any individual term. But the sum
vx^2 +vy^2 +vz^2 is the total length of the velocity vector, so〈v^2 〉=3〈vx^2 〉.Thuswecan rewrite
Equation 3.19 as
1
2 ×
1
3 m〈v
(^2) 〉= 1
2 kBT. (3.20)
Wenow rephrase Equation 3.20, using the fact that the kinetic energy of a particle is^12 mu^2 ,tofind
that:
The average kinetic energy of a molecule in an ideal gas is^32 kBT, (3.21)
regardless of what kind of gas we have. Even in a mixture of gases, the molecules of each type must
separately obey Idea 3.21.
The analysis leading to Idea 3.21 was given by Rudolph Clausius in 1857; it supplies the deep
molecular meaning of the ideal gas law. Alternatively, we can regard Idea 3.21 as explaining the
concept of temperature itself, in the special case of an ideal gas.
Let’s work some numbers to get a feeling for what our results mean. A mole of air occupies
22 liters (that’s 0.022m^3 )atatmospheric pressure and room temperature. What’s atmospheric
pressure? It’s enough to lift a column of water about 10 meters (you can’t sip water through a
straw taller than this). A 10mcolumn of water presses down with a force per area (pressure) equal
to the height times the mass density of water times the acceleration of gravity, orzρm,wg.Thus
atmospheric pressure is
p≈ 10 m×
(
103
kg
m^3)
×
(
10
m
s^2)
≈ 105
kg
ms^2
=10^5 Pa. (3.22)