Exercises — Mean free path, mean speed, RMS speed — derivations

Energy scales as $\frac{1}{2}m{v}^2$ — it depends on the average of $v^2$. So the energy-carrying speed is the root-mean-square speed $v_{rms}=\sqrt{3k_{B}T/m}$.

Collision frequency depends on how fast molecules actually travel past each other — an ordinary average of $v$. That is the mean speed $\bar{v}=\sqrt{8k_{B}T/\pi m}$.

Neither uses $v_p$ (the peak of the distribution), which is just "the single most common speed".

$v_p<\bar{v}<v_{rms}$ Why: squaring before averaging (as $v_{rms}$ does) gives extra weight to the fast molecules in the tail, pulling the number up. The peak $v_p$ ignores the tail entirely, so it sits lowest. The numerical ratios are $\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}\approx 1.41:1.60:1.73$.

$n=N/V$ is the number density — how many molecules per cubic metre. It is the "crowdedness" of the gas.

$\lambda \propto 1/n$, so tripling $n$ makes $\lambda$ one third as large. Three times as many targets ⇒ collisions happen after one third of the distance.

$v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.314)(300)}{0.004}}=\sqrt{1.871\times 10^6}\approx 1368\text{m/s}$ Why so fast? Helium is light ($M$ small, in the denominator), so at the same temperature it moves far faster than nitrogen (~517 m/s). This is why helium leaks out of balloons quickly.

$\bar{v}=\sqrt{\frac{8RT}{\pi M}}=\sqrt{\frac{8(8.314)(273)}{\pi (0.032)}}=\sqrt{\frac{18158}{0.1005}}=\sqrt{1.806\times 10^5}\approx 425\text{m/s}$ Why the $\pi$? It comes straight out of the Gaussian integral used to average $v$ over the Maxwell–Boltzmann distribution — see the parent derivation.

Use the pressure form so we never compute $n$ by hand: $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}=\frac{(1.38\times 10^{-23})(300)}{\sqrt{2}\pi (2.7\times 10^{-10}{)}^2(1.0\times 10^5)}$ Numerator: $4.14\times 10^{-21}$. Denominator: $\sqrt{2}(3.1416)(7.29\times 10^{-20})(1.0\times 10^5)=3.239\times 10^{-14}$. $\lambda \approx 1.28\times 10^{-7}\text{m}\approx 128\text{nm}$ Why big? Small molecule ($d$ small ⇒ $d^2$ tiny in the denominator) ⇒ long flights.

$\frac{v_{rms}}{\bar{v}}=\frac{\sqrt{3RT/M}}{\sqrt{8RT/\pi M}}=\sqrt{\frac{3RT/M}{8RT/\pi M}}=\sqrt{\frac{3\pi }{8}}$ Every $R$, $T$, $M$ cancels — that's the whole point. Numerically: $\sqrt{\frac{3\pi }{8}}=\sqrt{1.178}\approx 1.085$ So $v_{rms}$ is always about $8.5\%$ larger than $\bar{v}$, no matter the gas or temperature. This is why the two are so easy to confuse.

(a) $v_{rms}\propto \sqrt{T}$, so doubling $T$ multiplies $v_{rms}$ by $\sqrt{2}\approx 1.414$.

(b) Use the density form $\lambda =1/(\sqrt{2}n\pi d^2)$. At constant volume the number of molecules and $V$ are unchanged, so $n$ is unchanged, so $\lambda$ does not change at all (factor $1$).

Why the surprise? The pressure form $\lambda =k_{B}T/(\sqrt{2}\pi d^{2}P)$ looks like $\lambda \propto T$ — but at constant $V$, heating also doubles $P$ (since $P\propto T$ at fixed $n$). The $T$ upstairs and $P$ downstairs cancel. $\lambda$ is really about crowding ($n$), not temperature.

At the same $T$ and $P$, both have the same $n=P/k_{B}T$. So $\frac{{\lambda }_{He}}{{\lambda }_{Ar}}=\frac{d_{Ar}^2}{d_{He}^2}={\left(\frac{3.6}{2.1}\right)}^2=(1.714{)}^2\approx 2.94$ Helium's $\lambda$ is about 2.9× longer. Why: the smaller a molecule's diameter, the tinier its collision cross-section $\sigma =\pi d^2$, so it slips much further between hits. Mass never enters $\lambda$ — only size and crowding.

(a) $\bar{v}=\sqrt{\frac{8RT}{\pi M}}=\sqrt{\frac{8(8.314)(273)}{\pi (0.028)}}=\sqrt{2.063\times 10^5}\approx 454\text{m/s}$.

(b) $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}=\frac{(1.38\times 10^{-23})(273)}{\sqrt{2}\pi (3.7\times 10^{-10}{)}^2(1.01\times 10^5)}$. Numerator $=3.767\times 10^{-21}$. Denominator $=\sqrt{2}(3.1416)(1.369\times 10^{-19})(1.01\times 10^5)=6.143\times 10^{-14}$. $\lambda \approx 6.13\times 10^{-8}\text{m}\approx 61\text{nm}$

(c) The cleanest, drop-free way to get $Z$ is to divide the distance a molecule actually travels per second by the average distance between its collisions: $Z=\frac{\text{distance travelled per second}}{\text{distance per collision}}=\frac{\bar{v}}{\lambda }$ Why plain $\bar{v}$ here, no $\sqrt{2}$? The $\sqrt{2}$ is already baked into $\lambda$ (it shortened the free path because targets move). If you also multiplied the top by $\sqrt{2}$ you would count that same effect twice. Put differently: the collision rate is (relative speed)$\times n\sigma =\sqrt{2}\bar{v}n\sigma$, and $\lambda =1/(\sqrt{2}n\sigma )$ so $1/\lambda =\sqrt{2}n\sigma$; forming $Z={\bar{v}}_{rel}n\sigma =\sqrt{2}\bar{v}n\sigma$ equals $\bar{v}/\lambda$ — the two $\sqrt{2}$'s cancel to leave $\bar{v}/\lambda$. $Z=\frac{454}{6.13\times 10^{-8}}\approx 7.4\times 10^9{\text{s}}^{-1}$ A molecule collides about 7 billion times per second — the swarm is that busy. The average time between collisions is $1/Z\approx 1.35\times 10^{-10}$ s.

Effusion rate $\propto \bar{v}\propto \sqrt{T}$. So $\frac{{\text{rate}}_2}{{\text{rate}}_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{4}=2$ The rate doubles. Why not quadruple? Speed grows with $\sqrt{T}$, not $T$, because kinetic energy $\frac{1}{2}m{\bar{v}}^2\propto T$ ⇒ $\bar{v}\propto \sqrt{T}$. See Diffusion and effusion.

Invert the boxed formula for $d$: $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}\Rightarrow d^2=\frac{k_{B}T}{\sqrt{2}\pi P\lambda }$ $d^2=\frac{(1.38\times 10^{-23})(300)}{\sqrt{2}\pi (1.0\times 10^5)(7.0\times 10^{-8})}=\frac{4.14\times 10^{-21}}{3.111\times 10^{-2}}=1.331\times 10^{-19}{\text{m}}^2$ $d=\sqrt{1.331\times 10^{-19}}\approx 3.65\times 10^{-10}\text{m}=0.365\text{nm}$ Why this matters: measuring a bulk quantity ($\lambda$, from viscosity or diffusion experiments) lets us weigh and size invisible atoms — one of the great triumphs of kinetic theory. See Viscosity and thermal conductivity of gases.

(a) From $v_{rms}=\sqrt{3k_{B}T/m}$, square both sides, then solve for the per-molecule (single-particle) mass $m$: $m=\frac{3k_{B}T}{v_{rms}^2}=\frac{3(1.38\times 10^{-23})(273)}{485^2}=\frac{1.1305\times 10^{-20}}{2.3522\times 10^5}=4.806\times 10^{-26}\text{kg}$

(b) To identify the gas you need the molar mass, which is the per-molecule mass times Avogadro's number: $M=m{N}_A=(4.806\times 10^{-26})(6.022\times 10^{23})=0.02894\text{kg/mol}\approx 28.9\text{g/mol}$. That is essentially air / N₂ ($\approx 28$–$29$ g/mol).

(c) Same inversion as L5.1: $d^2=\frac{k_{B}T}{\sqrt{2}\pi P\lambda }=\frac{(1.38\times 10^{-23})(273)}{\sqrt{2}\pi (1.01\times 10^5)(9.0\times 10^{-8})}=\frac{3.767\times 10^{-21}}{4.038\times 10^{-2}}=9.33\times 10^{-20}{\text{m}}^2$ $d\approx 3.05\times 10^{-10}\text{m}\approx 0.31\text{nm}$ The moral: from one measured speed and one measured length, kinetic theory hands you the mass, identity, and size of an atom you can never see.

Given $v_{rms}$, find the per-molecule mass $m$ (NOT the molar mass) ::: $m=3k_{B}T/v_{rms}^2$; then molar mass $M=m{N}_A$ Given $\lambda$, $T$, $P$, find $d$ ::: $d=\sqrt{k_{B}T/(\sqrt{2}\pi P\lambda )}$ Universal ratio $v_{rms}/\bar{v}$ ::: $\sqrt{3\pi /8}\approx 1.085$ Collision frequency ::: $Z=\bar{v}/\lambda$ (the $\sqrt{2}$ is already inside $\lambda$)