JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2022
- Medium
- Numerical
A vessel contains \(14\,g\) of nitrogen gas at a temperature of \(27^{\circ}\,C\). The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be \(......J\) \(\left(\right.\) Take \(R =8.32\,J\,mol ^{-1} k ^{-1}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(9360\)
Step-by-step Solution
Detailed explanation
\(v _{ rms } \propto \sqrt{ T }\)
\(v _{ rms } \propto \sqrt{300\,K }, v _{ rms _{ f }}=2 v _{ rms _{ i }}\)
\(v _{ rms _{ f }} \propto \sqrt{1200\,K }\)
\(T _{ f }=1200\,K , T _{ i }=300 K , n =\frac{14}{28}=\frac{1}{2}\)
\(Q = nC _{ v } \Delta T =\frac{1}{2} \times \frac{5 R }{2} \times 900\)
\(Q =9360\,J\)
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