Kinetic Theory of Gases: Step-by-Step Worked Examples (4)

Question

Calculate the rms speed of nitrogen ($N_2$, molar mass $28\,\text{g/mol}$) molecules at $27^\circ$C. Also find the temperature at which the rms speed becomes double this value.

Solution — Step by Step

Final answers: $v_{\text{rms}} \approx 517$ m/s; $T_2 = 1200$ K.

Why This Works

The factor of 3 in the formula comes from the equipartition theorem — each translational degree of freedom contributes $\tfrac{1}{2}k_B T$ of energy, and there are three such DOFs. Equating $\tfrac{3}{2}k_B T$ to $\tfrac{1}{2}m\langle v^2\rangle$ gives the rms speed.

Notice the speed depends on $T$ (always in Kelvin) and $M$ — not on pressure or volume. Heavier molecules move slower at the same temperature, which is why hydrogen escapes the atmosphere faster than nitrogen.

Alternative Method

Use $v_{\text{rms}} = \sqrt{3k_B T / m}$ where $m$ is the mass of one molecule. For $N_2$, $m = 28 \times 1.66 \times 10^{-27}$ kg. Same answer, but requires Avogadro’s number — slower in an exam.

Common Mistake

Forgetting to convert molar mass to kg/mol. Using $M = 28$ instead of $0.028$ throws the answer off by a factor of $\sqrt{1000} \approx 31.6$. Always check units before computing.