The Distribution of Molecular Speeds — Question 6

Question 6

Changing from speed to energy changes the location of the density peak

For a classical equilibrium gas, write the speed density as $f(v)=Cv^2e^{-av^2}$ for $v\ge0$, where $a=m/(2k_BT)$. Use normalized $C=4a^{3/2}/\sqrt\pi$ and define translational kinetic energy $E=mv^2/2$. The energy density satisfies $g(E)dE=f(v)dv$.

1. Derive $g(E)$ and find its most probable energy.

2. Compare that mode with the energy corresponding to $v_{\mathrm{mp}}$ and with the mean energy. Explain why applying $E=mv^2/2$ to the speed mode does not locate the energy mode.

Question 6 - Solution

Changing from speed to energy changes the location of the density peak

(a) Method and calculation

$$\frac{dv}{dE}=\frac1{mv}=\frac1{\sqrt{2mE}},\quad g(E)=\frac2{\sqrt\pi(k_BT)^{3/2}}\sqrt E\,e^{-E/(k_BT)}.$$$$\frac{d\ln g}{dE}=\frac1{2E}-\frac1{k_BT}=0\ \Rightarrow\ E_{\mathrm{mode}}=\tfrac12k_BT.$$

(b) Interpretation and further calculation

$$\tfrac12mv_{\mathrm{mp}}^2=k_BT,\quad\langle E\rangle=\tfrac32k_BT.$$$$E_{\mathrm{mode}}<E(v_{\mathrm{mp}})<\langle E\rangle.$$Equal-width speed bins become unequal-width energy bins. The Jacobian $dv/dE$ changes the density and therefore its peak.

Physical check

Probabilities are invariant under a change of variables; probability-density modes generally are not.