JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2021
- Hard
- Numerical
A monoatomic gas of mass \(4.0\, u\) is kept in an insulated container. Container is moving with velocity \(30 \,m / s\). If container is suddenly stopped then change in temperature of the gas \(\left( R =\right.\) gas constant) is \(\frac{ x }{3 R } .\) Value of \(x\) is ..........
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3600\)
Step-by-step Solution
Detailed explanation
Given that mass of gas is \(4 u\) hence its molar
mass \(M\) is \(4 g / mol\)
\(\therefore \frac{1}{2} mv ^{2}= n C _{ v } \Delta T\)
\(\frac{1}{2} m \times(30)^{2}=\frac{ m }{ M } \times \frac{3 R }{2} \times \Delta T\)
\(\therefore \Delta T =\frac{3600}{3 R }\)
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