3.2. Decoding the ideal gas law[[Student version, December 8, 2002]] 75
In the last equality,Pastands forpascal,the SI unit of pressure. SubstitutingV =0. 022 m^3 ,
p≈ 105 kg m−^1 s−^2 ,andN=Nmoleinto the ideal gas law (Equation 1.11 on page 23) shows that
indeed it is approximately satisfied:
(
105
kg
ms^2
)
×
(
0. 022 m^3)
≈
(
6. 0 · 1023
)
×
(
4. 1 · 10 −^21 J
)
.
Wecan go farther. Air consists mostly of nitrogen molecules. The molar mass of atomic nitrogen
is about 14gmole−^1 ,soamole of nitrogen molecules, N 2 ,has mass about 28g.Thusthe mass of
onenitrogen molecule ism=0. 028 kg/Nmole=4. 7 · 10 −^26 kg.
Your Turn 3i
Using Idea 3.21, show that the typical velocity of air molecules in the room where you’re sitting
is about√
〈v^2 〉≈ 500 ms−^1. Convert tomiles/hourto see whether you should drive that fast
(maybe in the Space Shuttle).So the air molecules in your room are pretty frisky. Can we get some independent confirmation
to see if this result is reasonable? Well, one thing we know about air is... there’s less of it on top
of Mt. Everest. That’s because gravity exerts a tiny pull on every air molecule. On the other hand,
the air density in your room is quite uniform from top to bottom. Apparently the typical kinetic
energy of air molecules,^32 kBTr,issohigh that the difference in gravitational potential energy, ∆U,
from the top to the bottom of a room is negligible, while the difference from sea level to Mt. Everest
is not so negligible. Let’s make the very rough estimate that Everest isz=10kmhigh, and that
the resulting ∆Uis roughly equal to the mean kinetic energy:
∆U=mg(10km)≈^12 m〈v^2 〉. (3.23)Your Turn 3j
Show that the typical velocity is then aboutu= 450 ms−^1 ,remarkably close to what we just
found in Your Turn 3i.This new estimate is completely independent of the one we got from the ideal gas law, so the fact
that it gives the same typicaluis evidence that we’re on the right track.
Your Turn 3k
a. Compare the average kinetic energy^32 kBTrof air molecules to the difference in gravitational
potential energy ∆Ubetween the top and bottom of a room. Herez=3mis the height of the
ceiling. Why doesn’t the air in the room fall to the floor? What could you do tomakeit fall?
b. Repeat (a) but this time for a dirt particle. Suppose that the particle weighs about as much
asa50μmcube of water. Why does dirt fall to the floor?In this section we have seen how the hypothesis of random molecular motion, with an average
kinetic energy proportional to the absolute temperature, explains the ideal gas law and a number of
other facts. Other questions, however, come to mind. For example, if heating a pan of water raises
the kinetic energy of the water molecules, why don’t they all suddenly fly away when the temperature
gets to some critical value, the one giving them enough energy to escape? To understand questions
like this one, we need keep in mind that the average kinetic energy is far from the whole story. We
also want to know about the fulldistributionof molecular velocities, not just its mean-square value.