Pressure, Temperature, and RMS Speed — Question 3

Question 3

An anisotropic population exerts unequal normal stresses

A zero-drift gas has density $.900\,\mathrm{kg/m^3}$ and component mean squares $\langle v_x^2\rangle=90000$, $\langle v_y^2\rangle=40000$, $\langle v_z^2\rangle=10000$ m$^2$/s$^2$. The normal kinetic stresses are $p_j=\rho\langle v_j^2\rangle$.

1. Find the three normal stresses and the rms speed.

2. If collisions isotropize the velocities while conserving total translational kinetic energy and density, find the resulting common pressure. Can the original state be assigned this single mechanical pressure?

Question 3 - Solution

An anisotropic population exerts unequal normal stresses

(a) Method and calculation

$$p_x=81000\,\mathrm{Pa},\quad p_y=36000\,\mathrm{Pa},\quad p_z=9000\,\mathrm{Pa}.$$$$\langle v^2\rangle=140000\,\mathrm{m^2/s^2},\quad v_{\mathrm{rms}}=374.166\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

$$\langle v_j^2\rangle_{\mathrm{final}}=140000/3,\quad p_f=.900(140000)/3=42000\,\mathrm{Pa}.$$$$p_f=(p_x+p_y+p_z)/3.$$This trace average characterizes the conserved kinetic energy; it was not the original stress on every wall.

Physical check

The factor $1/3$ for each direction requires isotropy, not merely zero mean velocity.