Problem — Ranking v_rms for four gases

Problem

Ranking $v_\text{rms}$. Four gases at the same temperature $T = 300$ K: (A) He ($M=4$); (B) N$_2$ ($M=28$); (C) O$_2$ ($M=32$); (D) CO$_2$ ($M=44$). Order the root-mean-square speed from highest to lowest.

Solution

At the same temperature all gases have the same mean kinetic energy, so the root-mean-square speed depends only on the molar mass: $v_\text{rms}=\sqrt{\frac{3RT}{M}}\propto \frac{1}{\sqrt{M}}$ The lighter the gas, the faster it is: just order by increasing molar mass. The masses are $M_A = 4 < M_B = 28 < M_C = 32 < M_D = 44$, so the speeds go in reverse order. Helium, much lighter than the others, stands well apart from the group; N$_2$, O$_2$ and CO$_2$ are close to each other because their masses are similar. $\ev{A > B > C > D}$

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Topics: Kinetic theory of gases Concepts: Root-mean-square speed Skills: Ranking reasoning