Describe the distribution of molecular speeds in an ideal
gas

Find the average and most probable molecular speeds in an ideal
gas

Particles in an ideal gas all travel at relatively high speeds,
but they do not travel at the same speed. The rms speed is one kind
of average, but many particles move faster and many move slower.
The actual distribution of speeds has several interesting
implications for other areas of physics, as we will see in later
chapters.

The Maxwell-Boltzmann Distribution

The motion of molecules in a gas is random in magnitude and
direction for individual molecules, but a gas of many molecules has
a predictable distribution of molecular speeds. This predictable
distribution of molecular speeds is known as the
Maxwell-Boltzmann distribution, after
its originators, who calculated it based on kinetic theory, and it
has since been confirmed experimentally (Figure
\(\PageIndex{1}\)).

To understand this figure, we must define a distribution
function of molecular speeds, since with a finite number of
molecules, the probability that a molecule will have exactly a
given speed is 0.

We define the distribution function \(f(v)\) by saying that the
expected number \(N(v_1, v_2)\) of particles with speeds between
\(v_1\) and \(v_2\) is given by

[Since N is dimensionless, the unit of
f(v) is seconds per meter.] We can write this
equation conveniently in differential form: \[dN = Nf(v)dv.\]

In this form, we can understand the equation as saying that the
number of molecules with speeds between v and \(v + dv\)
is the total number of molecules in the sample times
f(v) times dv. That is, the probability
that a molecule’s speed is between v and \(v + dv\) is
f(v)dv.

We can now quote Maxwell’s result, although the proof is beyond
our scope.

Maxwell-Boltzmann Distribution of Speeds

The distribution function for speeds of particles in an ideal
gas at temperature \(T\) is

The factors before the \(v^2\) are a normalization constant;
they make sure that \(N(0,\infty ) = N\) by making sure that
\(\int_0^{|infty} f(v)dv = 1.\) Let’s focus on the dependence on
v. The factor of \(v^2\) means that \(f(0) = 0\) and for
small v, the curve looks like a parabola. The factor of
\(e^{-m_0v^2/2k_BT}\) means that \(\lim_{v\rightarrow \infty} f(v)
= 0\) and the graph has an exponential tail, which indicates that a
few molecules may move at several times the rms speed. The
interaction of these factors gives the function the single-peaked
shape shown in the figure.

Example \(\PageIndex{1}\): Calculating the
Ratio of Numbers of Molecules Near Given Speeds

In a sample of nitrogen \(N_2\) with a molar mass of 28.0 g/mol)
at a temperature of \(273^oC\) find the ratio of the number of
molecules with a speed very close to 300 m/s to the number with a
speed very close to 100 m/s.

Strategy

Since we’re looking at a small range, we can approximate the
number of molecules near 100 m/s as \(dN_{100} = f(100 \, m/s)dv\).
Then the ratio we want is

All we have to do is take the ratio of the two f
values.

Solution

Identify the knowns and convert to SI units if necessary. \[T =
300 \, K, \, k_B = 1.38 \times 10^{-23} J/K\] \[M = 0.0280 \,
kg/mol \, so \, m = 4.65 \times 10^{-26} \, kg\]

Figure \(\PageIndex{2}\) shows that the curve is shifted to
higher speeds at higher temperatures, with a broader range of
speeds.

With only a relatively small number of molecules, the
distribution of speeds fluctuates around the Maxwell-Boltzmann
distribution. However, you can view this simulation to see the
essential features that more massive molecules move slower and have
a narrower distribution. Use the set-up “2 Gases, Random Speeds”.
Note the display at the bottom comparing histograms of the speed
distributions with the theoretical curves.

We can use a probability distribution to calculate average
values by multiplying the distribution function by the quantity to
be averaged and integrating the product over all possible speeds.
(This is analogous to calculating averages of discrete
distributions, where you multiply each value by the number of times
it occurs, add the results, and divide by the number of values. The
integral is analogous to the first two steps, and the normalization
is analogous to dividing by the number of values.) Thus the average
velocity is

as in
Pressure, Temperature, and RMS Speed. The most
probable speed, also called the peak
speed \(v_p\) is the speed at the peak of the
velocity distribution. (In statistics it would be called the mode.)
It is less than the rms speed \(v_{rms}\). The most probable speed
can be calculated by the more familiar method of setting the
derivative of the distribution function, with respect to
v, equal to 0. The result is

In the factor \(e^{-mv^2/2k_BT}\), it is easy to recognize the
translational kinetic energy. Thus, that expression is equal to
\(e^{-K/k_BT}\). The distribution f(v) can be
transformed into a kinetic energy distribution by requiring that
\(f(K)dK = f(v)dv\). Boltzmann showed that the resulting formula is
much more generally applicable if we replace the kinetic energy of
translation with the total mechanical energy E.
Boltzmann’s result is

The first part of this equation, with the negative exponential,
is the usual way to write it. We give the second part only to
remark that \(e^{E/k_BT}\) in the denominator is ubiquitous in
quantum as well as classical statistical mechanics.

Problem-Solving Strategy: Speed
Distribution

Step 1. Examine the situation to determine
that it relates to the distribution of molecular speeds.

Step 2. Make a list of what quantities are
given or can be inferred from the problem as stated (identify the
known quantities).

Step 3. Identify exactly what needs to be
determined in the problem (identify the unknown quantities). A
written list is useful.

Step 4. Convert known values into proper SI
units (K for temperature, Pa for pressure, \(m^3\) for volume,
molecules for N, and moles for n). In many cases,
though, using R and the molar mass will be more convenient
than using \(k_B\) and the molecular mass.

Step 5. Determine whether you need the
distribution function for velocity or the one for energy, and
whether you are using a formula for one of the characteristic
speeds (average, most probably, or rms), finding a ratio of values
of the distribution function, or approximating an integral.

Step 6. Solve the appropriate equation for the
ideal gas law for the quantity to be determined (the unknown
quantity). Note that if you are taking a ratio of values of the
distribution function, the normalization factors divide out. Or if
approximating an integral, use the method asked for in the
problem.

Step 7. Substitute the known quantities, along
with their units, into the appropriate equation and obtain
numerical solutions complete with units.

We can now gain a qualitative understanding of a puzzle about
the composition of Earth’s atmosphere. Hydrogen is by far the most
common element in the universe, and helium is by far the
second-most common. Moreover, helium is constantly produced on
Earth by radioactive decay. Why are those elements so rare in our
atmosphere? The answer is that gas molecules that reach speeds
above Earth’s escape velocity, about 11 km/s, can escape from the
atmosphere into space. Because of the lower mass of hydrogen and
helium molecules, they move at higher speeds than other gas
molecules, such as nitrogen and oxygen. Only a few exceed escape
velocity, but far fewer heavier molecules do. Thus, over the
billions of years that Earth has existed, far more hydrogen and
helium molecules have escaped from the atmosphere than other
molecules, and hardly any of either is now present.

We can also now take another look at evaporative cooling, which
we discussed in the chapter on temperature and heat. Liquids, like
gases, have a distribution of molecular energies. The
highest-energy molecules are those that can escape from the
intermolecular attractions of the liquid. Thus, when some liquid
evaporates, the molecules left behind have a lower average energy,
and the liquid has a lower temperature.