Kinetic Theory of Gases: Application Problems (5)

Question

Calculate the rms speed of nitrogen molecules ($M = 28\text{ g/mol}$) at $T = 300\text{ K}$. Take $R = 8.314\text{ J/(mol·K)}$.

Solution — Step by Step

Why This Works

The factor of 3 comes from the equipartition theorem: each translational degree of freedom carries $\frac{1}{2}k_B T$ of kinetic energy, and there are three translational degrees, so total translational KE per molecule is $\frac{3}{2}k_B T$. Setting this equal to $\frac{1}{2}m v_{\text{rms}}^2$ and replacing $m$ with $M/N_A$ gives the formula above.

This is why heavier gases (oxygen, argon) move slower at the same temperature than lighter gases (helium, hydrogen). At room temperature, hydrogen molecules zip around at about $1900\text{ m/s}$.

Alternative Method

If we don’t have $R$, we can use Boltzmann’s constant directly:

$v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}$

where $m$ is the mass of one molecule. For nitrogen, $m = 28 \times 1.66 \times 10^{-27}\text{ kg} = 4.65 \times 10^{-26}\text{ kg}$. Plug in $k_B = 1.38 \times 10^{-23}\text{ J/K}$ — same answer.

Common Mistake

Forgetting the kg/mol conversion. Using $M = 28$ instead of $0.028$ gives a speed about 32 times too small. Always sanity-check: gas molecules at room temperature should be a few hundred m/s.