Question

Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?

Answer 1

Here,
P = 10⁵ Pa
T = 300 K
For H₂
M = 2×10^-3 kg

$\begin{array}{l}\left(\text{a}\right)\text{Mean speed is given by}\\ <v>=\sqrt{\frac{8RT}{\pi M}}\\ =\sqrt{\frac{8\times 8.3\times 300\times 7}{2\times {10}^{-3}\times 22}}\\ =1780{\text{ms}}^{-1}\\ {\text{Let us consider a cubic volume of 1 m}}^3.\\ V=1{\text{m}}^3\\ \text{Momentum of 1 molecule normal to the striking surface before collision =}mu\sin {45}^0\\ \text{Momentum of 1 molecule normal to the striking surface after collision =}-mu\sin {45}^0\\ \text{Change in momentum of the molecule =}2mu\sin {45}^0=\sqrt{2}mu\\ \text{Change in momentum of n molecules =}2mnu\sin {45}^0=\sqrt{2}mnu\\ \text{Let}\Delta \text{t be the time taken in changing the momentum.}\\ \text{Force per unit area due to one molecule =}\frac{\sqrt{2}mu}{\Delta t}=\frac{\sqrt{2}mu}{\Delta t}\\ \text{Observed pressure due to collision by n molecules =}\frac{\sqrt{2}mnu}{\Delta t}={10}^5\\ n=\raisebox{1ex}{$\frac{\sqrt{2}mnu}{\Delta t}$}\!\left/ \!\raisebox{-1ex}{$\frac{\sqrt{2}mu}{\Delta t}$}\right.=\frac{{10}^5}{\sqrt{2}mu}\\ 6.0\times {10}^{23}\text{molecules}=\text{2}\times {\text{10}}^{-3}\text{kg}\\ \text{1 molecule =}\frac{\text{2}\times {\text{10}}^{-3}}{6\times {10}^{23}}=3.3\times {10}^{-27}\text{kg}\\ \Rightarrow n=\frac{{10}^5}{\sqrt{2}\times 3.3\times {10}^{-27}\times 1780}=1.2\times {10}^{28}\end{array}$