Methane is at 1.0 × 10⁵ Pa and 300 K. A methane molecule has a diameter of 3.8 × 10⁻¹⁰ m and an…

Step 1. Mean free path. The top line is $k_BT = 1.38 \times 10^{-23}\ \text{J/K} \times 300\ \text{K} = 4.14 \times 10^{-21}\ \text{J}$. The bottom line is $\sqrt{2}\,\pi \times (3.8 \times 10^{-10}\ \text{m})^2 \times 1.0 \times 10^5\ \text{Pa} = 6.42 \times 10^{-14}\ \text{N}$. So $\lambda = \dfrac{4.14 \times 10^{-21}\ \text{J}}{6.42 \times 10^{-14}\ \text{N}} = 6.45 \times 10^{-8}\ \text{m}$, which is $64.5\ \text{nm}$.

Step 2. Collisions. $\nu = \dfrac{630\ \text{m/s}}{6.45 \times 10^{-8}\ \text{m}} = 9.76 \times 10^9$ each second.

Step 3. Time between them. $\tau = \dfrac{1}{\nu} = 1.02 \times 10^{-10}\ \text{s}$, which is $102$ picoseconds.