Problem — Root-mean-square speed and the speed of sound

Problem

Calculate the root-mean-square speed of nitrogen molecules at room temperature ($T = 300$ K, $M_{N_2} = 28$ g/mol). Compare with the speed of sound in air ($\sim 340$ m/s).

Solution

The root-mean-square speed is obtained by equating the mean translational kinetic energy per mole, $\tfrac{3}{2}RT$, to the expression $\tfrac{1}{2}M v_\text{rms}^2$: $v_\text{rms}=\sqrt{\frac{3RT}{M}}$ With $M = 28$ g/mol $= 0{,}028$ kg/mol: $v_\text{rms}=\sqrt{\frac{3\cdot 8{,}31\cdot 300}{0{,}028}}\approx 517\ \text{m/s}$ Note: the (arithmetic) mean speed is a distinct, smaller quantity, $\langle v\rangle=\sqrt{\frac{8RT}{\pi M}}\approx 475\text{–}484\ \text{m/s}$ It is this $\langle v\rangle$ that gives the often-quoted value $\approx 484$ m/s; the $v_\text{rms}$ asked for here is $\approx 517$ m/s. Comparison with sound: the molecules move, on average, faster than the speed of sound ($\sim 340$ m/s), consistent with the fact that sound propagates precisely thanks to collisions between molecules, so it cannot exceed their typical speed. $\ev{v_\text{rms}\approx 517\ \text{m/s} > v_\text{sound}\approx 340\ \text{m/s}}$

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