Problem — Drill on the v_rms of nitrogen

Problem

Calculate the root-mean-square speed of $\text{N}_2$ molecules ($M = 28$ g/mol) at $T = 300$ K. Compare it with the speed of sound in air.

Solution

The root-mean-square speed is $v_\text{rms} = \sqrt{\frac{3RT}{M}}$ I convert the molar mass: $M = 28\ \text{g/mol} = 0{,}028\ \text{kg/mol}$. $v_\text{rms} = \sqrt{\frac{3 \cdot 8{,}31 \cdot 300}{0{,}028}}$ $v_\text{rms} = \sqrt{2{,}67\cdot10^{5}}\ \text{m/s}$ $\ev{v_\text{rms} \approx 517\ \text{m/s}}$ It is of the same order as the speed of sound in air ($\approx 340$ m/s), just a bit larger: sound propagates through collisions between molecules, so its speed is governed precisely by $v_\text{rms}$.

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Topics: Kinetic theory of gases Concepts: Root-mean-square speed Skills: Symbolic setup