Question

mean free path and mean free time of gas molecule
n = # atoms
v = vol.
(\\(\frac{n}{v}\\)) = number density.
= # atom per unit vol.
(\\(\frac{n}{v}\\)) × vol. = 1
within the one - collision
(\\(\frac{n}{v}\\)) × \\(\pi(2r)^2\\)d = 1
cylindrical vol.
d = \\(\frac{v}{4\pi r^2n}\\)
d = \\(\frac{k_{b}t}{4\pi r^2p}\\)
pv = nrt
pv = nk_{b}t ⇒ \\(\frac{v}{n}=\frac{k_{b}t}{p}\\)

Explanation:

Step1: Recall relevant gas - law and collision concepts

The number density is defined as $\frac{N}{V}$, where $N$ is the number of atoms and $V$ is the volume. The idea is to find the mean - free path $\lambda$. We consider the volume swept by a molecule between collisions.

Step2: Consider the volume swept by a molecule

If we assume a molecule of radius $r$ moving with an average speed $\bar{u}$, the volume it sweeps in a distance $l$ is a cylindrical volume $V_{cylinder}=\pi(2r)^{2}l$. In the time between collisions, the number of molecules in this volume is related to the number density. We know that the number of molecules per unit volume $\frac{N}{V}$ times the volume of the cylinder gives the number of molecules in the cylinder. For one - collision, $\frac{N}{V}\times\pi(2r)^{2}l = 1$. Solving for $l$ (which is the mean - free path $\lambda$), we get $\lambda=\frac{1}{4\pi r^{2}\frac{N}{V}}$.

Step3: Use the ideal gas law

The ideal gas law is $PV = nRT$ and also $PV=Nk_{B}T$, where $k_{B}$ is the Boltzmann constant. From $PV = Nk_{B}T$, we can express $\frac{N}{V}=\frac{P}{k_{B}T}$. Substituting $\frac{N}{V}$ into the mean - free path formula, we have $\lambda=\frac{k_{B}T}{4\pi r^{2}P}$.

Answer:

The mean - free path of a gas molecule is $\lambda=\frac{k_{B}T}{4\pi r^{2}P}$ and the mean - free time $\tau=\frac{\lambda}{\bar{u}}=\frac{k_{B}T}{4\pi r^{2}P\bar{u}}$, where $k_{B}$ is the Boltzmann constant, $T$ is the temperature, $r$ is the radius of the gas molecule, $P$ is the pressure, and $\bar{u}$ is the average speed of the gas molecule.