Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a small molecule like $\mathrm{N}_{2}$ or $\mathrm{H}_{2} \mathrm{O}?$

$$\textrm{The average volume per molecule for an ideal gas at room temperature and atmospheric pressure is Volume of one mole over the number of atoms in this mole which is the Avogadro number $6.022 \times 10^{23}$, at the previous conditions, therefore :}$$

$$v= \dfrac{V}{A}$$

$$\textrm{where $v$ is the volume of the molecule, $V$ is the volume of one mole $ ~0.02444 \mathrm{ ~m^{3}}$ and $A$ is the Avogadro number. Thus,}$$

$$v=\dfrac{0.02444}{6.022 \times 10^{23}}=4.06 \times 10^{-26} \mathrm{~m^{3}}$$

$$\textrm{The cube root of $v$ provides an estimate of the inter molecule distance:}$$

$$d=\sqrt[3]{v}=\sqrt[3]{4.06 \times 10^{-26}}=3.436 \times 10^{-9} \mathrm{~m}$$

$$d= 3.436 \mathrm{~nm}$$

$$\textrm{The size of a water molecule is about $3 \times 10^{-10} \mathrm{~m}$, therefore in a gas of water vapour, the distance is about ten times the molecular size }$$