Nitrogen fills a vacuum chamber at 300 K…

Step 1. Number density (a). $n = \dfrac{P}{k_BT} = \dfrac{1.010 \times 10^{5}\ \text{Pa}}{1.38 \times 10^{-23}\ \text{J/K} \times 300\ \text{K}} = 2.44 \times 10^{25}\ \text{m}^{-3}$.

Step 2. Mean free path (b). $\pi d^2 = 4.301 \times 10^{-19}\ \text{m}^2$. The bottom line is $\sqrt2 \times 4.301 \times 10^{-19}\ \text{m}^2 \times 2.440 \times 10^{25}\ \text{m}^{-3} = 1.484 \times 10^{7}\ \text{m}^{-1}$, so

Step 3. Collision rate (c). $\bar v = \sqrt{\dfrac{8RT}{\pi M}}$, with $8RT = 8 \times 8.314\ \text{J/(mol K)} \times 300\ \text{K} = 19954\ \text{J/mol}$ and $\pi M = \pi \times 0.028\ \text{kg/mol} = 0.0880\ \text{kg/mol}$. So $\bar v = 476\ \text{m/s}$, so a molecule collides $\dfrac{\bar v}{\lambda} = 7.07 \times 10^{9}$ times each second.

Step 4. Spacing (d). $n^{-1/3} = 3.448 \times 10^{-9}\ \text{m} = 3.45\ \text{nm}$. The mean free path is $19.5$ times larger. A molecule passes many neighbours before it hits one, because each neighbour is a very small target.

Step 5. Pumped chamber (e). At a fixed temperature $\lambda \propto 1/P$:

Below about this pressure, molecules cross the chamber from wall to wall without hitting each other.