Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

80 of 130

Question 80 of 130

If 1022{10}^{22} gas molecules each of mass 10−26{10}^{-26} kg collides with a surface (perpendicular to it) elastically per second over an area 1m21{m}^{2} with a speed 104m/s,{10}^{4}m/s, the pressure exerted by the gas molecules will be of the order of:

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Answer: (A) 2Pa2Pa

Total momentum transferred in one second =n(2mv)=1022(2×10−26×104)=2=n\left(2mv\right)={10}^{22}\left(2\times {10}^{-26}\times {10}^{4}\right)=2
Force experienced by surface =2N=2N
Pressure on surface =ForceArea=21=2Pa=\dfrac{Force}{Area}=\dfrac{2}{1}=2Pa

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