In kinetic gas theory, we have the following result for the order of speeds $v_{rms}>v_{av}>v_{mp}$ i.e: the ratio of $1:1.128:1.224$ for the speeds $v_{mp}:v_{av} : v_{rms}$.

Definitions (source) and wiki:

$$v_{avg} = \int_0^{\infty} v f(v) dv = \sqrt{ \frac{8RT}{M}}$$

$$v_{rms} = \sqrt{\int_0^{\infty} v^2 f(v) dv}=\sqrt{\frac{3RT}{M}} $$

$$v_{mp} = \sqrt{ \frac{2RT}{M}}$$

Where $M$ is molecular mass of gas, $R$ is universal gas constant, $T$ is temperature in kelvins, $f(v)$ is probability density function

The most probable speed is characterized by the setting $\frac{df}{dv}=0$ where $f$ is the probability density function.

Are there any intuitive explanations to understand the above statistical result?

  • $\begingroup$ For the pdf, f( ), we are finding v=argmax(f), $\int vfdv$ and $\sqrt {\int v^2 fdv}$. I ‘think* i see why it doesnt drop in one. Not positive. Checkin back soon $\endgroup$
    – Al Brown
    Aug 28, 2021 at 17:45

1 Answer 1


The most probable speed, $v_{mp}$ is by definition the position of the maximum of the curve $f(v)$. Since this curve is asymmetric and skewed to the right, the mean is greater than the position of the maximum (they are equal for symmetric curves, but it couldn't be symmetric in our case, since $v$ is always positive, i.e., bounded from the left, but not from the right.)

The variance of the speed is by definition a non-negative quantity: $$ Var(v) = \int_0^{+\infty}(v-v_{avg})^2f(v)dv = v_{rms}^2-v_{avg}^2 >0 $$ which explains why $v_{rms}>v_{avg}$.

  • $\begingroup$ Late comment but how did the integral turn into v_{rms}^2 - v_{avg}^2? $\endgroup$
    – Babu
    Aug 28, 2021 at 17:01
  • $\begingroup$ @Buraian if you expand the square brackets you will immediately see. This is a standard thing for variance. $\endgroup$
    – Roger V.
    Aug 28, 2021 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.