question_answer

If \[{{10}^{22}}\]gas molecules each of mass \[{{10}^{-26}}kg\] collide with a surface (perpendicular to it) elastically per second over an area \[1{{m}^{2}}\]with a speed \[{{10}^{4}}m/s,\]the pressure exerted by the gas molecules will be of the order of : [JEE Main 8-4-2019 Morning]

A) \[{{10}^{8}}N/{{m}^{2}}\]                    

B) \[{{10}^{4}}N/{{m}^{2}}\]

C) \[{{10}^{3}}N/{{m}^{2}}\]        

D) \[{{10}^{16}}N/{{m}^{2}}\]

Correct Answer: C

Solution :

Note : Pressure is defined as normal force per unit area. Force is calculated as change in momentum/ time By this answer is \[2N/{{m}^{2}}\] None of the option matches so this question must be Bonus Detailed solution is as following. Magnitude of change in momentum per collision = 2mv Pressure \[=\frac{Force}{Area}=\frac{N(2mv)}{1}\] \[=\frac{{{10}^{22}}\times 2\times {{10}^{-26}}\times {{10}^{4}}}{1}\] \[=2N/{{m}^{2}}\]