Kinetic Theory of Gases: Tricky Questions Solved (3)

Question

A vessel contains $1 \text{ mole}$ of $\text{O}_2$ (molar mass $32 \text{ g/mol}$) and $2 \text{ moles}$ of $\text{H}_2$ (molar mass $2 \text{ g/mol}$) at temperature $T = 300 \text{ K}$. Find (a) the ratio of average kinetic energies per molecule of $\text{O}_2$ to $\text{H}_2$, (b) the ratio of rms speeds of $\text{O}_2$ to $\text{H}_2$, and (c) the total internal energy of the mixture (treat both as diatomic). Take $R = 8.314 \text{ J/mol K}$.

Solution — Step by Step

Final answers: (a) $\mathbf{1:1}$, (b) $\mathbf{1:4}$, (c) $U \approx \mathbf{1.87 \times 10^4 \text{ J}}$.

Why This Works

The big idea: temperature is a measure of average KE per molecule. Two gases at the same $T$ have the same KE per molecule, regardless of mass. But heavier molecules move slower to carry that same KE, so $v_{rms} \propto 1/\sqrt{M}$.

For internal energy, we count all degrees of freedom. Diatomic gases have 3 translational + 2 rotational = 5 active modes at room temperature. Vibrational modes “freeze out” below ~$1000 \text{ K}$, so we don’t count them here.

Alternative Method

For (b), since $\tfrac{1}{2}m v_{rms}^2 = \tfrac{3}{2}k_B T$ and $T$ is the same, $m_1 v_1^2 = m_2 v_2^2$, so $v_1/v_2 = \sqrt{m_2/m_1}$. Same answer, derived from KE equality directly.

Common Mistake