JEE Main 12 Jan 2019 Evening Question 9
Question: 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 [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 4\times {10^{-8}}s $
B) $ 3\times {10^{-6}}s $
C) $ 0.5\times {10^{-8}}s $
D) $ 2\times {10^{-7}}s $
Show Answer
Answer:
Correct Answer: A
Solution:
Mean time between two successive collisions
$ \tau \propto \frac{1}{ \begin{aligned} & velocity\times numberotparticles \\ & per,unit,volume \\ \end{aligned}} $
$ \Rightarrow $ $ \tau \propto \frac{1}{\sqrt{T}}\frac{T}{P}=\frac{\sqrt{T}}{P}\therefore {\tau_1}={\tau_2}\frac{P _2}{P _1}\sqrt{\frac{T _1}{T _2}} $
$ \Rightarrow $ $ {\tau_2}=\frac{6\times {10^{-8}}}{2}\sqrt{\frac{500}{300}}=\sqrt{15}\times {10^{-8}}\simeq 4\times {10^{-8}}s $
