JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
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 ..........
- A \(2500\)
- B \(3600\)
- C \(4900\)
- D \(4200\)
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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