The Distribution of Molecular Speeds — Question 8

Question 8

A mixture of species has a weighted speed distribution

An equilibrium ideal mixture at temperature $T$ contains number fractions $x_A=.250$ and $x_B=.750$. Species B has molecular mass $4m_A$. The mean speed of pure A at this temperature is 800 m/s. Use $\overline v=\sqrt{8k_BT/(\pi m)}$ and $\langle v^2\rangle=3k_BT/m$.

1. Find the mixture mean speed and rms speed, averaging over molecules.

2. Compare the true mixture mean with the result obtained by inserting the number-averaged mass into the single-species mean-speed formula. Explain the discrepancy.

Question 8 - Solution

A mixture of species has a weighted speed distribution

(a) Method and calculation

$$\overline v_B=400\,\mathrm{m/s},\quad\overline v_{\mathrm{mix}}=.250(800)+.750(400)=500\,\mathrm{m/s}.$$$$\langle v^2\rangle_A=\frac{3\pi}{8}(800)^2,\quad\langle v^2\rangle_{\mathrm{mix}}=\frac{3\pi}{8}[.250(800)^2+.750(400)^2].$$$$v_{{\mathrm{rms,mix}}}=\sqrt{105000\pi}\approx574.341\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

$$\overline m=.250m_A+.750(4m_A)=3.25m_A.$$$$\overline v_{\mathrm{shortcut}}=800/\sqrt{3.25}\approx443.760\,\mathrm{m/s}.$$The actual density is $f_{\mathrm{mix}}=.250f_A+.750f_B$; a nonlinear function of an averaged mass does not equal the corresponding average over species.

Physical check

Both species share temperature and mean energy per molecule, while their speed scales differ.