Calculate the mean free path and relaxation time for a gas with a mean speed $\langle v \rangle = 485 \ m/s$. Assume standard conditions $(STP)$ where the number density $n \approx 2.7 \times 10^{25} \ m^{-3}$ and molecular diameter $d = 2 \ \mathring{A}$.

(N/A) Given:
Mean speed $\langle v \rangle = 485 \ m/s$
Number density $n = 2.7 \times 10^{25} \ m^{-3}$
Diameter $d = 2 \ \mathring{A} = 2 \times 10^{-10} \ m$
$1$. Mean free path $(\bar{l})$:
The formula for mean free path is $\bar{l} = \frac{1}{\sqrt{2} n \pi d^2}$.
$\bar{l} = \frac{1}{\sqrt{2} \times (2.7 \times 10^{25}) \times 3.14 \times (2 \times 10^{-10})^2}$
$\bar{l} = \frac{1}{1.414 \times 2.7 \times 10^{25} \times 3.14 \times 4 \times 10^{-20}}$
$\bar{l} \approx 2.08 \times 10^{-7} \ m$.
$2$. Relaxation time $(\tau)$:
The formula for relaxation time is $\tau = \frac{\bar{l}}{\langle v \rangle}$.
$\tau = \frac{2.08 \times 10^{-7}}{485}$
$\tau \approx 4.29 \times 10^{-10} \ s$.