# Why do we need the rms, mean, and most probable velocities?

In the kinetic theory of gases, we have rms (root mean square), mean, and mp (most probable) velocities. I understand the concept well. But my question is why do we have three different kinds of velocities? Why don't we stick onto any one of them? What is the significance of these velocities?

And it would be helpful if someone could explain how:

$$\bar{v} = \sqrt{\frac{8RT}{\pi M}}$$

• Because you need those different averages to calculate different properties of your gas?
– Karl
Sep 14 '18 at 15:47

Alright, here's my best attempt at it.

## Root Mean Squared $v_\mathrm{rms}$

The root-mean-square speed (r.m.s. speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds,$v$, of the molecules. That is, for a sample consisting of $N$ molecules with speeds $v_{1}, v_2,...,v_N$, we square each speed, add the squares together, divide by the total number of molecules (to get the mean, denoted by angle brackets $\langle\rangle$), and finally take the square root of the result:

\begin{align} v_{rms} &= \langle v^2\rangle^{1/2} \\&= \left(\frac{v_1^2+v_2^2+...+v_N^2}{N}\right)^{1/2} \end{align}

## Mean Speed $\bar{v}$

The r.m.s speed might at first sight seem to be rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass $m$ travelling at a speed $v$ is $E_\mathrm{k} = \frac{1}{2}mv^2$, which implies that the mean kinetic energy of a collection of molecules, $\langle E_\mathrm{k} \rangle$, is the average of this quantity, or $\frac{1}{2}mv_{rms}^2$. It follows from the relation $\frac{1}{2}mv_\mathrm{rms}^2= \langle E_\mathrm{k}\rangle$, that

$$v_\mathrm{rms} = \left(\frac{2\langle E_\mathrm{k}\rangle }{m}\right)^{1/2}$$

Therefore, wherever $v_\mathrm{rms}$ appears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s speed is quite close in value to another and more readily visualized measure of molecular speed, the mean speed, $\bar{v}$, of the molecules:

$$\bar{v} = \frac{v_1+v_2+...+v_N}{N}$$

For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed. The precise relation is

$$\bar{v} = \left(\frac{8}{3 \pi}\right)^{1/2}v_\mathrm{rms} = 0.921v_\mathrm{rms}$$

For elementary purposes, and for qualitative arguements, we don't need to distinguish between the two measures of average speed, but for precise work the distinction is important.

## Most Probable $v_\mathrm{mp}$

$v_\mathrm{mp}$ is the most probable speed of the molecules, and is the maximum value for the velocity distribution curve. When putting all of of these equations together we see the relation

$$v_\mathrm{rms} > \bar{v} > v_\mathrm{mp}$$

That $v_\mathrm{mp}$ is the smallest of the three speeds is due to the asymmetry of the curve.

## References

Atkins, P.; de Paula, J. Elements of Physical Chemistry 6th ed. W.H. Freeman and Company. New York, NY. 2013.

Chang, R. Physical Chemistry for the Chemical and Biological Sciences. University Science Books. Sausalito, CA. 2000.