Kinetic Theory of Gases: Common Mistakes and Fixes (1)

Question

Find the rms speed of oxygen ($M = 32$ g/mol) molecules at $T = 300$ K. Take $R = 8.314$ J/(mol K).

Solution — Step by Step

Why This Works

The rms speed comes from equipartition: average translational KE per molecule is $\frac{3}{2} k_B T$, so $\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T$, giving $v_{\text{rms}} = \sqrt{3 k_B T / m} = \sqrt{3RT/M}$.

We use $M$ (molar mass) and $R$ (gas constant) for convenience — they’re the macroscopic versions of $m$ and $k_B$.

Alternative Method

Use $v_{\text{rms}} = \sqrt{3 P / \rho}$ if you’re given pressure and density instead of temperature. For an ideal gas, $P = \rho R T / M$, so the two forms are equivalent.

Common Mistake

Students confuse $v_{\text{rms}}$, $v_{\text{avg}} = \sqrt{8RT/(\pi M)}$, and $v_{\text{mp}} = \sqrt{2RT/M}$. The order is always $v_{\text{mp}} < v_{\text{avg}} < v_{\text{rms}}$, in the ratio $\sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1.41 : 1.60 : 1.73$. Memorise this ratio for MCQs.