Q 11-12-141JEE MainJEE Main 2019 (8 Apr, Shift 1)Easy
If $10^{22}$ gas molecules each of mass $10^{-26}\ \text{kg}$ collide with a surface (perpendicular to it) elastically per second over an area $1\ \text{m}^2$ with a speed $10^4\ \text{m/s}$, the pressure exerted by the gas molecules will be of the order of
Answer: (A) $2\ \text{Pa}$
Each elastic, normal collision transfers momentum $2mv$. Force per unit area:
$$P = \frac{n\cdot2mv}{A} = 10^{22}\times2\times10^{-26}\times10^4 = 2\ \text{Pa}$$
Solution by Sreeraj P, M.Sc Physics