The Distribution of Molecular Speeds — Question 2

Question 2

Mode mean and rms speed represent different features of one distribution

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 $C=4a^{3/2}/\sqrt\pi$, $\int_0^\infty v^3e^{-av^2}dv=1/(2a^2)$ and $\langle v^2\rangle=3/(2a)$. Take $M=.0200$ kg/mol, $T=360$ K and $R=8.314$ J/(mol K).

1. Derive the most probable speed by differentiation, then calculate the mean and rms speeds.

2. Find the speed standard deviation and its ratio to the mean. Establish the ordering of the three characteristic speeds.

Question 2 - Solution

Mode mean and rms speed represent different features of one distribution

(a) Method and calculation

$$\frac{d\ln f}{dv}=\frac2v-2av=0\ \Rightarrow\ v_{\mathrm{mp}}=a^{-1/2}=\sqrt{2RT/M}\approx547.087\,\mathrm{m/s}.$$$$\overline v=C/(2a^2)=\sqrt{8RT/(\pi M)}\approx617.321\,\mathrm{m/s},\quad v_{\mathrm{rms}}=\sqrt{3RT/M}\approx670.042\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

$$\sigma_v^2=\langle v^2\rangle-\overline v^2=(3-8/\pi)RT/M.$$$$\sigma_v\approx260.519\,\mathrm{m/s},\quad\frac{\sigma_v}{\overline v}=\sqrt{3\pi/8-1}\approx.422016.$$$$v_{\mathrm{mp}}<\overline v<v_{\mathrm{rms}}.$$The asymmetric tail raises the mean above the mode; squaring speeds gives fast particles additional weight.

Physical check

The rms-minus-mean difference is not the standard deviation; the variances must be subtracted before taking a square root.