Estimate the mean free path for a water molecule in water vapour at $373 \; K$. The density of water is $1000 \; kg \; m^{-3}$. The density of water vapour at $100 \; ^{\circ}C$ and $1 \; atm$ pressure is $0.6 \; kg \; m^{-3}$. The volume of a molecule multiplied by the total number gives,what is called,molecular volume.

The mean free path $l$ is given by the formula $l = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the molecular diameter and $n$ is the number density.
For a water molecule,the diameter $d \approx 2 \times 10^{-10} \; m$.
The number density $n$ is given by $n = \frac{\rho}{m}$,where $\rho = 0.6 \; kg \; m^{-3}$ and $m$ is the mass of one water molecule $(H_2O)$.
The mass of one water molecule $m = \frac{18 \times 10^{-3} \; kg}{6.022 \times 10^{23}} \approx 3 \times 10^{-26} \; kg$.
Thus,$n = \frac{0.6}{3 \times 10^{-26}} = 2 \times 10^{25} \; m^{-3}$.
Substituting these values into the formula:
$l = \frac{1}{\sqrt{2} \times 3.14 \times (2 \times 10^{-10})^2 \times 2 \times 10^{25}}$
$l = \frac{1}{1.414 \times 3.14 \times 4 \times 10^{-20} \times 2 \times 10^{25}}$
$l = \frac{1}{35.47 \times 10^5} \approx 2.8 \times 10^{-7} \; m$.