Kinetic Theory

JEE Main level, numerical answer, medium

46 of 168

Question 46 of 168

0.056kg0.056kg of Nitrogen is enclosed in a vessel at a temperature of 127∘C127^{\circ}C. The amount of heat required to double the speed of its molecules is_____ kcalkcal.
(Take R=2calmoleK−1−1R=2calmole{K-1}^{-1} )

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Answer: 12

The RMS speed of the molecule is given by vRMS=3RTM{v}_{RMS}=\sqrt{\dfrac{3RT}{M}}. Therefore, to double the speed the temperature must become 44 times.
Therefore, the change in the temperature is ∆T=3T0∆T=3{T}_{0}.
Now the process is isochoric (enclosed vessel) therefore, the heat required will be,
H=nCv∆T⇒H=5628×52R×3×(273+127)=12000cal=12kcalH=n{C}_{v}∆T \Rightarrow H=\dfrac{56}{28}\times \dfrac{5}{2}R\times 3\times \left(273+127\right)=12000cal=12kcal

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