JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
\(0.056 \,kg\) 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 \(R =2\) \(cal \,mole\) \(^{-1} K ^{-1}\) )
- A \(12\)
- B \(18\)
- C \(17\)
- D \(122\)
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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