Pressure, Temperature, and RMS Speed — Question 1

Question 1

Repeated elastic impacts produce an average wall force

A molecule of mass $m=5.00\times10^{-26}$ kg moves in a rectangular box with $x$-length .100 m. Its normal speed component has magnitude 400 m/s. Wall collisions are elastic, and no intermolecular collisions occur. The wall area is .0200 m$^2$.

1. Derive the mean force on one $x$ wall from impulse and return time, then find its pressure contribution.

2. If the box length doubles while that speed component and wall area are fixed, find the new pressure contribution. Explain why the round-trip distance is $2L$, not $L$.

Question 1 - Solution

Repeated elastic impacts produce an average wall force

(a) Method and calculation

$$I=2m|v_x|=4.00\times10^{-23}\,\mathrm{N\,s},\quad\Delta t=2L/|v_x|=5.00\times10^{-4}\,\mathrm s.$$$$\overline F=I/\Delta t=mv_x^2/L=8.00\times10^{-20}\,\mathrm N,\quad p_1=\overline F/A=4.00\times10^{-18}\,\mathrm{Pa}.$$

(b) Interpretation and further calculation

$$p_1=\frac{mv_x^2}{AL}=\frac{mv_x^2}{V}\ \Rightarrow\ p_1(2L)=2.00\times10^{-18}\,\mathrm{Pa}.$$$$\Delta t_{\mathrm{same\ wall}}=L/|v_x|+L/|v_x|.$$The intervening impact is on the opposite wall and does not contribute force to the selected wall.

Physical check

The force averages discrete impulses over many wall returns; it is not a continuous microscopic force between collisions.