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(B) The mean time between two successive collisions is given by $	au = \frac{\lambda}{v_{avg}}$,where $\lambda$ is the mean free path and $v_{avg}$ i

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$4 	imes 10^{-8} \, s$

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(B) The mean time between two successive collisions is given by $	au = \frac{\lambda}{v_{avg}}$,where $\lambda$ is the mean free path and $v_{avg}$ is the average speed.Since $\lambda \propto \frac{T}{P}$ and $v_{avg} \propto \sqrt{T}$,we have $	au \propto \frac{T/P}{\sqrt{T}} \propto \frac{\sqrt{T}}{P}$.Given $P_1 = 2 \, atm$,$T_1 = 300 \, K$,and $	au_1 = 6 	imes 10^{-8} \, s$.For the new state,$P_2 = 2 P_1 = 4 \, atm$ and $T_2 = 500 \, K$.Using the ratio $\frac{	au_2}{	au_1} = \frac{P_1}{P_2} \sqrt{\frac{T_2}{T_1}}$:$\frac{	au_2}{6 	imes 10^{-8}} = \frac{2}{4} \sqrt{\frac{500}{300}} = \frac{1}{2} \sqrt{\frac{5}{3}}$.$	au_2 = 6 	imes 10^{-8} 	imes 0.5 	imes 1.29 \approx 3.87 	imes 10^{-8} \, s$.This is closest to $4 	imes 10^{-8} \, s$.

Similar Questions

Write the equation for the mean free path of gas molecules.

At $S.T.P.$,the mean free path of a gas molecule is $1500 \ d$,where '$d$' is the diameter of the molecule. What will be the mean free path at $373 \ K$ at constant volume?

If $n$ is the number density and $d$ is the diameter of the molecule,then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by :

Statement-$1$: Real gas approaches ideal gas behaviour for low pressures and high temperatures.
Statement-$2$: At low pressure,the density of a gas is very low.

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