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 :

[c] 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 \[=\frac{Force}{Area}=\frac{N(2mv)}{1}\]

\[=\frac{{{10}^{22}}\times 2\times {{10}^{-26}}\times {{10}^{4}}}{1}\]

\[=2N/{{m}^{2}}\]