Q.$2.0\ \text{mol}$ of an ideal monatomic gas is confined in a cubical box of side $0.30\ \text{m}$ at $300\ \text{K}$. Using the kinetic theory expression $PV = \tfrac{1}{3}Nm\overline{v^2}$, find the pressure exerted by the gas on the walls of the box.

Given $n=2.0\ \text{mol}$, $T=300\ \text{K}$, $R=8.31\ \text{J/(mol\,K)}$, and $V = (0.30\ \text{m})^3 = 0.027\ \text{m}^3$. Since kinetic theory's $PV=\tfrac{1}{3}Nm\overline{v^2}$ is exactly equivalent to the ideal gas equation $PV=nRT$,

$P = \frac{nRT}{V} = \frac{2.0 \times 8.31 \times 300}{0.027} = \frac{4986}{0.027} \approx 1.85 \times 10^5\ \text{Pa}$ …

Find the box's volume from its side length, then use $P=nRT/V$ (the macroscopic form of …

• Forgetting to CUBE the side length when finding the volume of the cubical box. …