statistical mechanics - Why is most probable speed not equal to rms speed for an ideal gas?

The rms speed of ideal gas is $$\mathit{v_{rms}} = \sqrt{\dfrac{3RT}{M}}.$$ The most probable speed is the speed where $\dfrac{dP(\mathit {v})}{dv} =0$ where $P(\mathit{v})$ is the probability distibution. Solving for $\mathit{v}$, we get $$ \mathit{v_p} = \sqrt{\dfrac{2RT}{M}}.$$

Now, $$\mathit{v_p} \neq \mathit{v_{rms}}.$$ Why? Why is it so?

Answer

We're used to thinking of "most probable" and "mean value" as the same thing, but it need not be so. It's worth remembering that the "expectation value" of a six sided die is 3.5, but this is not a very probable result. You might object that this is due to discrete effects, but consider this example: you have two identical Gaussians, with width $\sigma$, but they are separated. One has mean value $m_1$ and the other has mean value $m_2 = m_1 + \delta$. If they're identical and we average between them, we get an expectation value of $(m_1 + m_2) /2 = m_1 + \delta/2$. But $\delta$ could be quite large, in particular perhaps the Gaussians are very separated $\delta >> \sigma$. Then the mean value could occur in a point with arbitrarily small probability of actually being selected!

So as a general principle, the most probable value of a distribution and the average value need not be together. Does that help, or would you rather talk more directly about Maxwell-Boltzmann distributions (of atomic velocity)?