Problem — Ranking root-mean-square speeds

Problem

Ranking. Four ideal-gas samples with temperatures $T$ and molecular masses $m$:

• (A) $T = 300\ \text{K}$, helium ($m = 4\ \text{u}$).
• (B) $T = 300\ \text{K}$, argon ($m = 40\ \text{u}$).
• (C) $T = 600\ \text{K}$, helium.
• (D) $T = 1200\ \text{K}$, helium.

Order the root-mean-square speeds $v_\text{rms} = \sqrt{3k_B T/m}$ from smallest to largest.

Solution

Since $v_\text{rms} = \sqrt{3k_B T/m}$, the factor $3k_B$ is common to all: it suffices to compare the ratio $T/m$ (larger $\Rightarrow$ faster). Computing $T/m$ for each sample (masses in u): $\begin{aligned} \text{A}: \frac{300}{4}=75 \qquad & \text{B}: \frac{300}{40}=7{,}5 \\ \text{C}: \frac{600}{4}=150 \qquad & \text{D}: \frac{1200}{4}=300 \end{aligned}$ Ordering the values from smallest to largest: $7{,}5 < 75 < 150 < 300$, i.e. B $<$ A $<$ C $<$ D. Argon (B) is the slowest because it is heavy; among the helium samples, the one at the highest temperature (D) wins. $\ev{\text{B} < \text{A} < \text{C} < \text{D}}$

Links

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