A 25 times 10^{-3} , m^3 volume cylinder is f | vedclass

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(A) The collision frequency $Z$ is given by $Z = \frac{v_{avg}}{\lambda}$,where $\lambda$ is the mean free path.The mean free path is $\lambda = \frac

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$\sim 10^{10}$

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(A) The collision frequency $Z$ is given by $Z = \frac{v_{avg}}{\lambda}$,where $\lambda$ is the mean free path.The mean free path is $\lambda = \frac{1}{\sqrt{2} \pi \sigma^2 n}$,where $n = \frac{N}{V} = \frac{N_A}{V}$ is the number density.Substituting $n$,we get $\lambda = \frac{V}{\sqrt{2} \pi \sigma^2 N_A}$.Thus,$Z = \frac{v_{avg} \sqrt{2} \pi \sigma^2 N_A}{V}$.Given: $V = 25 	imes 10^{-3} \, m^3$,$\sigma = 0.3 	imes 10^{-9} \, m$,$v_{rms} = 200 \, m/s$,$N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Using $v_{avg} = \sqrt{\frac{8}{3\pi}} v_{rms} \approx 0.921 	imes 200 \approx 184.2 \, m/s$.$Z = \frac{184.2 	imes \sqrt{2} 	imes 3.14 	imes (0.3 	imes 10^{-9})^2 	imes 6.022 	imes 10^{23}}{25 	imes 10^{-3}}$.$Z \approx \frac{184.2 	imes 1.414 	imes 3.14 	imes 0.09 	imes 10^{-18} 	imes 6.022 	imes 10^{23}}{0.025} \approx \frac{439.5}{0.025} 	imes 10^5 \approx 1.75 	imes 10^{10} \, s^{-1}$.Therefore,the collision rate is $\sim 10^{10} \, s^{-1}$.

$A$ $25 \times 10^{-3} \, m^3$ volume cylinder is filled with $1 \, mol$ of $O_2$ gas at room temperature $(300 \, K)$. The molecular diameter of $O_2$ is $0.3 \, nm$ and its root mean square speed is $200 \, m/s$. What is the average collision rate (per second) for an $O_2$ molecule?

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