Pressure, Temperature, and RMS Speed — Question 9

Question 9

One measured velocity component can determine an isotropic temperature

A zero-drift equilibrium gas has one-component rms speed $\sqrt{\langle v_x^2\rangle}=250$ m/s. Its molecular mass is $6.40\times10^{-26}$ kg. Use isotropy and $m\langle v_x^2\rangle=k_BT$, with $k_B=1.381\times10^{-23}$ J/K.

1. Find the three-dimensional rms speed and temperature.

2. An analyst interprets 250 m/s as the full rms speed. Find the inferred temperature and its ratio to the correct temperature. Explain why zero drift alone is not enough for the isotropy step.

Question 9 - Solution

One measured velocity component can determine an isotropic temperature

(a) Method and calculation

$$\langle v^2\rangle=3\langle v_x^2\rangle\ \Rightarrow\ v_{\mathrm{rms}}=250\sqrt3=433.013\,\mathrm{m/s}.$$$$T=\frac{6.40\times10^{-26}(250)^2}{1.381\times10^{-23}}\approx289.645\,\mathrm K.$$

(b) Interpretation and further calculation

$$T_{\mathrm{wrong}}=\frac{m(250)^2}{3k_B}=T/3\approx96.5484\,\mathrm K.$$$$\langle\mathbf v\rangle=0\not\Rightarrow\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle.$$A distribution may be symmetric about zero yet broader along one axis; equilibrium isotropy supplies the extra information.

Physical check

The rms component is smaller than the full rms speed by $\sqrt3$, not by three.