JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2019
- Hard
- MCQ
An ideal gas is enclosed in a cylinder at pressure of \(2\,atm\) and temperature, \(300\,K.\) The mean time between two successive collisions is \(6 \times 10^{-8}\, s.\) If the pressure is doubled and temperature is increased to \(500\,K,\) the mean time between two successive collisions will be close to
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4\times 10^{-8}\,s\)
Step-by-step Solution
Detailed explanation
\({\text { The mean time between two collision } \propto \frac{P}{\sqrt{T}}} \)
\({\qquad \frac{\Delta t_{1}}{\Delta t_{2}}=\frac{P_{1}}{P_{2}} \times \frac{\sqrt{T_{2}}}{\sqrt{T_{1}}}} \)
\({\Rightarrow \quad \frac{6 \times 10^{-8}}{\Delta t_{2}}=(\sqrt{\frac{3}{5}}) \times 2} \)
\({\Rightarrow \quad \Delta t_{2}=6 \times 10^{-8} \times \frac{1}{2} \sqrt{\frac{5}{3}} \approx 4 \times 10^{-8} \text { sec }}\)
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