Kinetic Theory

JEE Main level, single correct, medium

41 of 168

Question 41 of 168

A vessel contains 14g14g of nitrogen gas at a temperature of 27∘C27^{\circ}C. The amount of heat to be transferred to the gas to double the r.m.s. speed of its molecules will be : (Take R=8.32Jmol−1k−1R=8.32J{mol}^{-1}{k}^{-1})

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Answer: (C) 9360J9360J

The RMS speed of a gas at a given temperature is given by, vrms=3RTM{v}_{rms}=\sqrt{\dfrac{3RT}{M}}.
Therefore, to double the RMS speed, the temperature should be increased to four times the initial temperature. Therefore,
Tf=1200K,Ti=300K,andn=1428=12{T}_{f}=1200K,{T}_{i}=300K,\text{and}n=\dfrac{14}{28}=\dfrac{1}{2}
The volume of the vessel will remain the same therefore, the process is an isochoric process. Therefore,
Q=nCvΔT=12×5R2×900Q=n{C}_{v}\Delta T=\dfrac{1}{2}\times \dfrac{5R}{2}\times 900
⇒Q=9360J\Rightarrow Q=9360J

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