Translational Kinetic Energy — Question 10

Question 10

Mean energy gives a bound on the fraction of very fast molecules

A molecular population has known rms speed $v_{\mathrm{rms}}=500$ m/s. No distribution shape is specified. Let $f$ be the fraction with speed at least $v_*=1500$ m/s. Kinetic energies are nonnegative.

1. Derive an upper bound on $f$ using the mean square speed.

2. Give a discrete speed population that attains this bound, and explain why an equilibrium Maxwell gas would generally have a smaller fraction.

Question 10 - Solution

Mean energy gives a bound on the fraction of very fast molecules

(a) Method and calculation

$$\langle v^2\rangle\ge f v_*^2\ \Rightarrow\ f\le\frac{v_{\mathrm{rms}}^2}{v_*^2}.$$$$f\le\frac{500^2}{1500^2}=\frac19\approx11.111\%.$$

(b) Interpretation and further calculation

$$f=1/9\text{ at }v=1500\,\mathrm{m/s},\quad 1-f=8/9\text{ at }v=0.$$$$\langle v^2\rangle=(1/9)(1500)^2=250000\,\mathrm{m^2/s^2}.$$This saturates the inequality by giving no energy to the slower group and putting the entire fast group at threshold. A continuous equilibrium distribution spreads energy elsewhere, so equality is not attained.

Physical check

A moment bound is not a prediction of the actual tail probability.