1) Evaluate the density $D$ of collisions (= number of collisions per $m^3$ et $s$) of a gas $A$ whose molecular diameter is $d$ and molarity $[A]\frac{mol}{m^3}$ at temperature $T$ and pressure $P$ .
Let $N$ be the number of molecules in a volume $V\,m^3$
The number of collisions per second will be $\frac{1}{2}N z$. Indeed, it takes 2 molecules to a collision. The number of collisions per second will thus be only half the number of molecules multiplied by the frequency $z$ of collision of a molecule.
Then the number of collisions per $s$ and $m^3$ will be equal to $\frac{1}{2}\frac{N}{V}z$
$D =$
$\frac{1}{2}\frac{N}{V}z=$
$\frac{1}{2}\frac{n\cdot N_{Av}}{V}z$
where
$n$ est le nombre of moles
$N_{Av}=6,023\cdot 10^{23}$ is Avogadro's number
$D =$
$\frac{1}{2}[A]\cdot N_{Av}z$
with (see → here)
$z$=
$\frac{P}{k\cdot T}\pi d^2 \sqrt2 \sqrt{\frac{8RT}{\pi\cdot M}}=$
$\frac{P\cdot N_{Av}}{R\cdot T}\pi d^2 \sqrt{\frac{16RT}{\pi\cdot M}}=$
$4[A]\cdot N_{Av}\pi d^2 \sqrt{\frac{RT}{\pi\cdot M}}$
and so
Density of collisions =
$2 [A]^2\cdot N_{Av}^2 d^2 \sqrt{\frac{\pi \cdot RT}{M}}$
2) Use this formula to calculate the density of collisions of nitrogen at $20^oC$ and $1 \;bar$ (diameter of a molecule = $3,64\cdot 10^{-10}m$)
