Question

The mass of hydrogen molecule is $3.32\times {10}^{-27}$ kg. If ${10}^{23}$ hydrogen molecules strike a fixed wall of area 2 $c{m}^2$ at an angle of ${45}^{\circ }$ to the normal and rebound elastically with a speed of 103 m/s. Calculate the pressure exerted on the wall.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The molecules strike the wall along AO and rebound along OB such that

$\angle AON=\angle BON={45}^{\circ }$
Because the collision of molecules with the wall is elastic, hence
$\left|{\overline{P}}_1\right|=\left|{\overline{P}}_2\right|=p=mv=3.32\times {10}^{-27}\times {10}^3$
$=3.32\times {10}^{-24}Kgm{s}^{-1}.$
Now change in momentum normal to the wall per molecule
$(\Delta p{)}_n=|{\overline{P}}_{2n}-{\overline{P}}_{1n}|=2_{p_n}=2pcos45\circ =\sqrt{2}p$
Number of collision per sec or collision frequency, $f={10}^{23}$ per sec.
Hence, change in momentum per sec = change in momentum per molecule × collision frequency
or $\frac{\Delta p}{\Delta t}=(\Delta p{)}_n\times f$
According to Newton's second law of motion, force exerted by the molecules on the wall
$F=\frac{\Delta p}{\Delta t}=(\Delta p{)}_n\times f=\sqrt{2}pf=1.414\times 3.32\times {10}^{-24}\times {10}^{23}=0.4695N\text{Now, area of the wall}A=2c{m}^2=2\times {10}^{-4}{m}^2\therefore \text{Pressure on the wall,}P=\frac{F}{A}=\frac{0.4695}{2\times {10}^{-4}}=2.347\times {10}^{3}N/m^2.$