Pressure, Temperature, and RMS Speed — Question 2

Question 2

Pressure depends on mean square speed, not squared mean speed

An isotropic gas has mass density $\rho=1.20\,\mathrm{kg/m^3}$. Half its molecules have speed 200 m/s and half have speed 600 m/s, with random directions. Use $p=\rho\langle v^2\rangle/3$. All molecules have the same mass.

1. Find mean speed, rms speed and pressure.

2. Find the erroneous pressure obtained from $\rho\langle v\rangle^2/3$. Relate the difference to the speed variance.

Question 2 - Solution

Pressure depends on mean square speed, not squared mean speed

(a) Method and calculation

$$\langle v\rangle=(200+600)/2=400\,\mathrm{m/s},\quad\langle v^2\rangle=(200^2+600^2)/2=200000\,\mathrm{m^2/s^2}.$$$$v_{\mathrm{rms}}=\sqrt{200000}=447.214\,\mathrm{m/s},\quad p=80000\,\mathrm{Pa}.$$

(b) Interpretation and further calculation

$$p_{\mathrm{wrong}}=1.20(400)^2/3=64000\,\mathrm{Pa}.$$$$\operatorname{Var}(v)=200000-160000=40000\,\mathrm{m^2/s^2},\quad p-p_{\mathrm{wrong}}=\rho\operatorname{Var}(v)/3=16000\,\mathrm{Pa}.$$The shortcut underestimates pressure by 20%.

Physical check

The discrete speed model is isotropic but need not be a Maxwell equilibrium distribution.