Discussion Forum

The mass of a hydrogen molecule is 3.32 x10^-27 kg. If 10²³hydrogen molecules strike per second, a fixed wall of area 2 cm² at an angle of 45° to the normal and rebound elastically with a speed of 10³ m/s, then the pressure on the wall is nearly

Answer:

Explanation:

Momentum imparted due to first collision

$\Rightarrow 2mv\sin45$ =$\sqrt{2mv}$      

$\therefore$    Pressure on surface

 = $\frac{n\sqrt{2}mv}{Area}=\frac{10^{23}\times \sqrt{2}\times 3.32\times 10^{-27} \times10^{3}}{(2\times  10^{-2})^{2}}$

$p =2.35\times 10^{3} N/m^{2}$