The Distribution of Molecular Speeds — Question 5

Question 5

The Maxwell tail is much tighter than a second-moment bound

Let $x=v/v_p$, where $v_p$ is the Maxwell most probable speed. The supplied cumulative distribution is $F(x)=\operatorname{erf}(x)-(2/\sqrt\pi)xe^{-x^2}$. Use $\operatorname{erfc}(2)=.00467773498$, with $\operatorname{erfc}=1-\operatorname{erf}$.

1. Find the fraction faster than $2v_p$.

2. Using $\langle v^2\rangle=3v_p^2/2$, compare the exact tail with the distribution-independent bound $P(v\ge v_*)\le\langle v^2\rangle/v_*^2$. For $10^8$ molecules, give the predicted tail count.

Question 5 - Solution

The Maxwell tail is much tighter than a second-moment bound

(a) Method and calculation

$$P(v>2v_p)=1-F(2)=\operatorname{erfc}(2)+\frac4{\sqrt\pi}e^{-4}.$$$$P\approx.0460117\quad(4.60117\%).$$

(b) Interpretation and further calculation

$$P(v\ge2v_p)\le\frac{(3/2)v_p^2}{4v_p^2}=.375.$$$$\langle N_{>2v_p}\rangle=10^8(.0460117)\approx4.60117\times10^6.$$The moment bound permits many distributions; the Maxwell assumption supplies the additional shape information needed for a sharper prediction.

Physical check

Neither the rms speed nor twice the most probable speed is a maximum molecular speed.