Finding Average Gas Properties from the MBS Distribution

The Maxwell-Boltzmann speed (MBS) distribution tells the probability of finding a molecule with a particular speed in a gas. In this section, you will learn how to find overall properties of the gas as a whole: the average speed, the average kinetic energy, and the most probable speed.

Part 1: simplifying the MBS distribution and finding normalization constant

We begin with the MBS distribution (remember that exp(x)=ex):

\begin{displaymath}P(u)=4\pi\left(\frac{m}{2\pi k_{\rm B}T}\right)^{3/2}u^2\exp\left(-mu^2/2k_{\rm B}T\right)\end{displaymath}

Let's define constants c and c' so that

P(u)=c'u2 exp(-cu2)

The whole point is just to make this function easier to deal with. The constant c is defined to make the exponent of P(u) correct:

\begin{displaymath}c = \frac{m}{2k_{\rm B}T}\end{displaymath}

We can also deduce the definition of c' by comparing with the form of P(u). But we don't even have to! In the section on probability distributions, we found that a probability distribution must be normalized. Since the MBS is a continuous distribution, the normalization condition is

\begin{displaymath}\int_0^\infty P(u)\ du = 1 = \int_0^\infty c'u^2e^{-cu^2}\ du\end{displaymath}

We can now evaluate this definite integral: (See the mathematics section below for step-by-step instructions on how to do this.)

\begin{displaymath}1 = c'I_2(c) = c' \frac{\sqrt{\pi}}{4}c^{-3/2}=c' \frac{\sqrt{\pi}}{4}\left(\frac{2k_{\rm B}T}{m}\right)^{3/2}\end{displaymath}

Solving this for c' gives

\begin{displaymath}c'=\frac{4}{\sqrt{\pi}}\left(\frac{m}{2k_{\rm B}T}\right)^{3/2}=4\pi\left(\frac{m}{2\pi k_{\rm B}T}\right)^{3/2}\end{displaymath}

This is the same constant shown in your textbook and in the top equation on this page....BUT WE DERIVED IT, just from knowing that all probability distributions must be normalized!

Part 2: average speed and average kinetic energy

Once we know the probability distribution, we have all the information we need to find average values. The average value of a quantity is just the value in each state, weighted by the probability of finding the system in this state. For each speed u, the probability of finding a molecule with this speed is simply P(u). So the average speed, $\bar{u}$ is

\begin{displaymath}\bar{u} = \int_0^\infty uP(u)\ du = c'I_3(c)\end{displaymath}

Note that the extra power of u is why this integral involves I3(c) rather than I2(c) as found in the normalization integral. Again, the mathematics section permits us to evaluate this integral:

\begin{displaymath}\bar{u} = \frac{c'}{2c^2} = \sqrt{\frac{8k_{\rm B}T}{\pi m}}\end{displaymath}

We can also find the average kinetic energy of molecules in the gas, by finding the weighted average of $\frac{1}{2}mu^2$, the kinetic energy. Since the mass of each molecule in the gas is the same, the average value of the kinetic energy is just (m/2) times the average value of u2:

\begin{displaymath}\overline{u^2}=\int_0^\infty u^2 P(u)\ du = c'I_4(c)\end{displaymath}

We can simplify this integral, and plug in for c' and c:

\begin{displaymath}\overline{u^2}=c'\frac{3\sqrt{\pi}}{8}c^{-5/2}=\frac{3k_{\rm B}T}{m}\end{displaymath}

It is common to define $u_{\rm rms}$, the "root mean squared" speed. This means that we first compute the SQUARE of the speed, u2, find the MEAN value of this quantity, $\overline{u^2}$, and then take the square ROOT. So, $u_{\rm rms}=\sqrt{\overline{u^2}}$.

\begin{displaymath}u_{\rm rms}=\sqrt{\frac{3k_{\rm B}T}{m}}\end{displaymath}

(Please note: $\overline{u^2}\ne\bar{u}^2$. Review this section and make sure that you see why they are different.)
From $\overline{u^2}$ we can also find the average kinetic energy per molecule:

\begin{displaymath}\overline{KE}=\frac{3}{2}k_{\rm B}T\end{displaymath}

Part 3: most probable speed

The most probable speed is the speed u for which P(u) has a maximum. For a function to have a maximum or minimum, its first derivative must equal 0:


The expression exp(-cu2) never becomes zero, so it can be divided out, leaving

2 c'u (-cu2+1) = 0

We can see clearly that u=0 is a minimum, not a maximum, so the only possible solution is cu2=1, or

\begin{displaymath}u_{\rm mp}=\frac{1}{\sqrt{c}}=\frac{2k_{\rm B}T}{m}\end{displaymath}

Exercises to Illustrate Average Gas Properties

  1. Find ump for a gas of argon atoms at 500K and at 1000K.  Then run the JAVA applet, and compare your computed values of ump with the purple MBS distribution curves.  What feature on each curve corresponds to ump?
  2. Find ump , urms and $\bar{u}$ for CO2 gas at room temperature (300K).  Rank these speeds from fastest to slowest.  Would this ordering hold for other temperatures?  How about for different gases?
  3. For helium and radon gases, find $\bar{u}$ and the average kinetic energy at room temperature.  Explain any similarities or differences you find between the two systems.
  4. Suppose that a system does not obey the MBS distribution but instead obeys P(u)=d'exp(-du).  Find an expression for d' in terms of d.  How would you find $\bar{u}$ for this system?
  5. Thought question:  Explain clearly why $\overline{u^2}\ne\bar{u}^2$.  From the shape of the MBS distribution, which is bigger, $\bar{u}$ or  ump?  (You know the answer from comparing numerical formulas; now give an explanation based on the shape of the MBS curve.  Run the applet for MBS curves at various temperatures.)


The Big Picture

The average system properties (like average speed, rms speed, or most probable speed) are very important, because they enable us to make definitive statements about the system as a whole, even though its components are behaving randomly.

© Andrew M. Rappe