An ideal gas is at a temperature of $40^{\circ} \mathrm{C}$ and a pressure of 2.00 atmospheres. If the diameter of an $\mathrm{N}_{2}$ molecule is $0.300 \mathrm{~nm}$, then what is the mean free path of the $\mathrm{N}_{2}$ molecule in the gas?
A) $6.02 \times 10^{-8} \mathrm{~m}$
B) $5.33 \times 10^{-8} \mathrm{~m}$
C) $4.75 \times 10^{-8} \mathrm{~m}$
D) $5.00 \times 10^{-8} \mathrm{~m}$
E) $4.01 \times 10^{-8} \mathrm{~m}$

Question

Quizplus · Accepted Answer

The mean free path ($\lambda$) of a molecule in a gas can be calculated using the formula $\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}$, where $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin, $d$ is the diameter of the molecule, and $P$ is the pressure in Pascals. Given the diameter and converting the temperature to Kelvin ($40^{\circ}C = 313K$) and pressure to Pascals ($2.00 \, \text{atm} \approx 2.00 \times 101325 \, \text{Pa}$), the calculation yields a mean free path close to the value in choice B.

An Ideal Gas Is at a Temperature Of 40∘C40^{\circ} \mathrm{C}40∘C And a Pressure of 2

An Ideal Gas Is at a Temperature Of $40^{\circ} \mathrm{C}$ And a Pressure of 2