# Why is the most probable KE for a monoatomic ideal gas exactly 1/3 of its average KE?

The most probable value of the kinetic energy for a monoatomic ideal gas can be calculated by setting the first derivative of the probablity density function equal to $0$, this turns out to be $\frac{1}{2}kT$ which is $\frac{1}{3}$ of the average value, $\frac{3}{2}kT$ what is the physical reason behind this?

## 1 Answer

There is no physical reason. It's bare mathematics. The probability distribution in terms of energy is asymmetric, for it is bounded by 0 from below and not bounded at all from above. Asymmetric distributions are known to have (generally speaking) different values for mean, mode, and median. Think of the exponential distribution: its average value is some positive number, but the mode (the most probable value) is $\bf 0$.

Moreover, 3 is not even some intrinsic characteristic of the system. If you rewrite the distribution in terms of velocity (the way it is commonly done), you'll have an altogether different situation. All of a sudden, the ratio of mean to mode is... $2\over\sqrt\pi$, how would you like that?