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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2021 · 25 February · Morning Shift · Physics · Kinetic Theory of Gases

A monoatomic gas of mass 4 . 0u is kept in an insulated container....

A monoatomic gas of mass 4.0u4 . 0u is kept in an insulated container. The container is moving with velocity 30ms−130m{s}^{- 1}. If the container is suddenly stopped then a change in temperature of the gas (R=R =gas constant) is x3R\dfrac{x}{3R}. Value of xx is,

Correct answer

3600

Step-by-step solution

Given that mass of gas is 4u4u hence its molar mass MM is 4gmol−14g{mol}^{- 1}

∴12mv2=nCv\DeltaT∴ \dfrac{1}{2}m{v}^{2} = n{C}_{v}\DeltaT

⇒12m×(30)2=mM×3R2×\DeltaT⇒ \dfrac{1}{2}m \times {\left(30\right)}^{2} = \dfrac{m}{M} \times \dfrac{3R}{2} \times \DeltaT

∴\DeltaT=36003R∴ \DeltaT = \dfrac{3600}{3R}

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