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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2022
  • Medium
  • Numerical

\(0.056 \,kg\) of Nitrogen is enclosed in a vessel at a temperature of \(127\,^{\circ} C\). The amount of heat required to double the speed of its molecules is k cal. (Take \(R =2\) \(cal \,mole\) \(^{-1} K ^{-1}\) )

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Answer & Solution

Correct Answer

(A) \(12\)

Step-by-step Solution

Detailed explanation

\(0.056 \,kg N N _{2}=56\, gm\) of \(N _{2}=2\, mole\) of \(N _{2}\)

\(T _{1}=400\, K , v \alpha \sqrt{ T }\) so \(T _{2}=4 T _{1}=1600\, K\)

\(Q =\frac{ f }{2} nR \Delta T\)

\(f=5\)

\(Q =12 \,k \,cal\)

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