Biological Physics: Energy, Information, Life

3.2. Decoding the ideal gas law[[Student version, December 8, 2002]] 77

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2

detection rate (arbitrary units)

reduced velocity ̄u

Figure 3.7:(Experimental data with fit.) Speeds of atoms emerging from a box of thallium vapor, at two different
temperatures. Open circles:T= 944 K. Solid circles:T= 870 K. The quantity ̄uon the horizontal axis equals
u
√
m/ 4 kBT;both distributions have the same most-probable value, ̄umax=1. Thusumaxis larger for higher
temperatures, as implied by Idea 3.21. The vertical axis shows the rate at which atoms hit a detector after passing
through a filter like the one sketched in Figure 3.6 (times an arbitrary rescaling factor). Solid line: theoretical
prediction (see Problem 3.5). This curve fits the experimental data withnoadjustable parameters. [Data from
(Miller & Kusch, 1955).]

3.2.3 The Boltzmann distribution

Let’s use the ideas of Section 3.2.1 to understand the experimental data in Figure 3.7. We are
exploring the idea that while each molecule’s velocity cannot be predicted, nevertheless there is
adefinite prediction for thedistributionof molecular velocities. One thing we know about that
probability distribution is that it must fall off at large velocities: Certainly there won’t be any
gas molecules in the room moving at a million meters per second! Moreover, the average speed
must increase as we make the gas hotter, since we’ve argued that the average kinetic energy is
proportional toT(see Idea 3.21 on page 74). Finally, the probability of finding a molecule moving
to the left at some velocityvxshould be the same as that for finding it moving to the right at−vx.
One probability distribution with these properties is the Gaussian (Equation 3.8), where the
spreadσincreases with temperature and the mean is zero. (If the mean were nonzero, there’d be a
net, directed, motion of the gas, that is, wind blowing.) Remarkably, this simple distribution really
does describe any ideal gas! More precisely, the probabilityP(vx)offinding that a given molecule
at a given time has itsx-component of velocity equal tovxis a Gaussian, like the form shown in
Figure 3.2, but centered on zero. Each molecule is incessantly changing its speed and direction.
What’s unchanging is not the velocity of any one molecule, but rather the distributionP(vx).
Wecan replace the vague idea that the varianceσ^2 ofvxincreases with temperature by something
more precise. Because the mean velocity equals zero, Your Turn 3b says that the variance ofvxis
〈vx^2 〉.According to Equation 3.19, the mean kinetic energy must equal^32 kBT.Sowemust take

σ^2 =kBT/m. (3.24)

Section 1.5.4 on page 23 gave the numerical value ofkBTat room temperature askBTr≈ 4. 1 ×
10 −^21 J.That’s pretty small, but so is the massmof one gas molecule. Thus the spread of velocities