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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2019
  • Medium
  • MCQ

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

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Verified Solution

Answer & Solution

Correct Answer

(D) None of this

Step-by-step Solution

Detailed explanation

Pressure is defined as normal force per unit area.

Force is calculated as change in \( momentum/time.\)

By this answer is \(2 \mathrm{N} / \mathrm{m}^{2}\)

None of the option matches so this question must be Bonus.

Detailed solution is as following:

\(Magnitude\, of\, change\, in\, momentum \,per\, collision\) \(=2\) \(mv\)

Pressure \(=\frac{\text { Force }}{\text { Area }}=\frac{N(2 m v)}{1}\)

\({=\frac{10^{22} \times 2 \times 10^{-26} \times 10^{4}}{1}} \)

\({=2 \mathrm{N} / \mathrm{m}^{2}}\)

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