Exercises — Maxwell-Boltzmann distribution of speeds — most probable, mean, rms

WHAT: order the three known speeds. WHY: the identity $v_{mp}<\bar{v}<v_{rms}$ always holds because squaring (in rms) over-weights the fast tail and the plain mean is pulled right of the peak by the long tail.

• Smallest $=v_{mp}=395\text{m/s}$ (the peak of the curve).
• Middle $=\bar{v}=446\text{m/s}$ (balance point, right of peak).
• Largest $=v_{rms}=484\text{m/s}$.

WHAT: use $v_{rms}=\sqrt{3RT/M}$. WHY this form? We have molar mass, so $M$ and $R$ avoid needing $N_A$. Convert: $M=28\times 10^{-3}=0.028\text{kg/mol}$ (SI — never grams). $v_{rms}=\sqrt{\frac{3(8.314)(300)}{0.028}}=\sqrt{2.673\times 10^5}\approx 517\text{m/s}$

Answer: $\bar{v}$ (mean speed). WHY: effusion counts how many molecules cross a hole per second, which depends on the average speed of the crowd — a timing/counting quantity, not an energy quantity. Energy/pressure problems use $v_{rms}$; collision & escape counting uses $\bar{v}$. See Graham's Law of Effusion.

WHAT: scale using the fixed ratio. WHY these decimals are not magic: divide each formula by $v_{rms}=\sqrt{3RT/M}$ — the $\sqrt{RT/M}$ cancels, leaving pure numbers. $\frac{\bar{v}}{v_{rms}}=\frac{\sqrt{8RT/\pi M}}{\sqrt{3RT/M}}=\sqrt{\frac{8}{3\pi }}=0.9213\dots$ $\frac{v_{mp}}{v_{rms}}=\frac{\sqrt{2RT/M}}{\sqrt{3RT/M}}=\sqrt{\frac{2}{3}}=0.8165\dots$ So $0.921$ and $0.816$ come straight from $\sqrt{8/3\pi }$ and $\sqrt{2/3}$. Now scale: $\bar{v}=0.921\times 517\approx 476\text{m/s},v_{mp}=0.816\times 517\approx 422\text{m/s}$

WHAT: invert $v_{mp}=\sqrt{2RT/M}$ for $T$. WHY: square both sides to free $T$. $v_{mp}^2=\frac{2RT}{M}\Rightarrow T=\frac{M{v}_{mp}^2}{2R}$ With $M=0.004\text{kg/mol}$: $T=\frac{0.004\times (1000{)}^2}{2(8.314)}=\frac{4000}{16.628}\approx 240.6\text{K}$

WHAT: take the ratio. WHY: at equal $T$, the $\sqrt{3RT}$ cancels, leaving only mass: $\frac{v_{rms,\text{He}}}{v_{rms,{\text{O}}_2}}=\sqrt{\frac{M_{O_2}}{M_{He}}}=\sqrt{\frac{32}{4}}=\sqrt{8}\approx 2.83$ He is $2.83\times$ faster. WHY physically: same $T$ ⇒ same average KE $\frac{3}{2}kT$; lighter mass then means higher speed. Same energy, not same speed.

WHAT: set the two $v_{rms}$ equal. WHY: equal $v_{rms}$ ⇒ equal $T/M$ (since $v_{rms}^2=3RT/M$ and $3R$ cancels). $\frac{T_{H_2}}{M_{H_2}}=\frac{T_{O_2}}{M_{O_2}}\Rightarrow T_{H_2}=300\times \frac{2}{32}=18.75\text{K}$ The lighter gas needs a proportionally lower temperature to be as slow (on rms) as the heavy gas.

WHAT: invert $v_{mp}=\sqrt{2RT/M}$ for $M$. WHY: square, then isolate $M$. $v_{mp}^2=\frac{2RT}{M}\Rightarrow M=\frac{2RT}{v_{mp}^2}=\frac{2(8.314)(300)}{395^2}$ $M=\frac{4988.4}{156025}\approx 0.03197\text{kg/mol}=32.0\text{g/mol}$ That is O₂ (matches Example 2 in the parent note, where O₂ at 300 K had $v_{mp}\approx 395\text{m/s}$).

WHAT: use $v_{rms}\propto \sqrt{T}$ (same gas ⇒ $M$ cancels). WHY: the speed grows as the square root of $T$, so a speed factor $1.5$ needs a temperature factor $1.5^2$. $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}=1.5\Rightarrow \frac{T_2}{T_1}=1.5^2=2.25$

WHY two ways: $v_{rms}$ is defined so that $\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}kT$ — this is the whole reason $v_{rms}$ is the "energy" speed. See Average Kinetic Energy and Temperature. Mass of one O₂: $m=\frac{M}{N_A}=\frac{0.032}{6.022\times 10^{23}}=5.314\times 10^{-26}\text{kg}$. (i) $\frac{1}{2}m{v}_{rms}^2=\frac{1}{2}(5.314\times 10^{-26})(484{)}^2=\frac{1}{2}(5.314\times 10^{-26})(2.343\times 10^5)$ $=6.22\times 10^{-21}\text{J}$ (ii) $\frac{3}{2}kT=\frac{3}{2}(1.381\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}$ They agree (the tiny gap is rounding in $v_{rms}=484$). Both $\approx 6.2\times 10^{-21}\text{J}$.

(a) WHAT: $v_{mp}\propto \sqrt{T}$. WHY: direct from the formula. Going $T\to 4T$ gives $v_{mp}\to \sqrt{4}{v}_{mp}=2v_{mp}$. The peak moves to twice the speed. (b) WHY the height must drop: substitute $v=v_{mp}=\sqrt{2RT/M}$ into the explicit $f(v)$ from the top of this page. The messy $T$-dependence collapses because at the peak the exponent is fixed ($e^{-1}$) and the $v^2$ and $(M/2\pi RT{)}^{3/2}$ factors combine to give $f(v_{mp})=4\pi {\left(\frac{M}{2\pi RT}\right)}^{3/2}\left(\frac{2RT}{M}\right)e^{-1}=\frac{4}{\sqrt{\pi }}{e}^{-1}{\left(\frac{M}{2RT}\right)}^{1/2}\propto \frac{1}{\sqrt{T}}$ So the peak height $\propto 1/\sqrt{T}$ (the curve stretches horizontally by $\sqrt{T}$, so it must shrink vertically by the same factor to keep area 1). Going $T\to 4T$: $\frac{f(v_{mp}){|}_{4T}}{f(v_{mp}){|}_T}=\frac{1}{\sqrt{4}}=\frac{1}{2}$ Peak drops to half height and the curve is twice as wide — area preserved. This is exactly what the two curves in the figure show: the blue ($T$) curve is tall and narrow, the red ($4T$) curve is half as tall and twice as wide.

(a) WHAT: ratio of $\bar{v}$. WHY: at equal $T$, $\bar{v}\propto 1/\sqrt{M}$. $\frac{{\bar{v}}_{He}}{{\bar{v}}_{C{H}_4}}=\sqrt{\frac{M_{C{H}_4}}{M_{He}}}=\sqrt{\frac{16}{4}}=2$ (b) WHY effusion rate $\propto \bar{v}\propto 1/\sqrt{M}$: the rate is set by how fast molecules stream through the hole. So the rate ratio equals the speed ratio — this is Graham's Law of Effusion: $\frac{r_{He}}{r_{C{H}_4}}=\sqrt{\frac{M_{C{H}_4}}{M_{He}}}=2$ He effuses twice as fast.

WHAT: invert $\bar{v}=\sqrt{8RT/\pi M}$ for $T$. WHY: square both sides, isolate $T$. ${\bar{v}}^2=\frac{8RT}{\pi M}\Rightarrow T=\frac{\pi M{\bar{v}}^2}{8R}=\frac{\pi (0.020)(500{)}^2}{8(8.314)}$ $T=\frac{\pi (0.020)(250000)}{66.512}=\frac{15707.96}{66.512}\approx 236.2\text{K}$ Then $v_{rms}$ at the same $T$ uses the fixed ratio (no re-integration): $v_{rms}=\frac{1.225}{1.128}\bar{v}=1.086\times 500\approx 543\text{m/s}$.

WHY $1000\text{K}$: the outer atmosphere (thermosphere) is far hotter than the ground. Threshold $=11200/6\approx 1867\text{m/s}$. H₂: $v_{rms}=\sqrt{\frac{3(8.314)(1000)}{0.002}}=\sqrt{1.247\times 10^7}\approx 3532\text{m/s}$. N₂: $v_{rms}=\sqrt{\frac{3(8.314)(1000)}{0.028}}=\sqrt{8.908\times 10^5}\approx 944\text{m/s}$. Decision: H₂ ($3532>1867$) escapes; N₂ ($944<1867$) is retained. WHY: the light gas is faster at the same $T$ (same KE, smaller mass), so its rms speed clears the escape threshold while the heavy gas stays bound. This is why Earth keeps N₂ but has almost no free H₂.

$v_{rms}$ the largest of the three speeds? Squaring in $\overline{v^2}$ over-weights the fast tail (a molecule twice as fast counts $4\times$), so the root-mean-square is dragged highest.

$v\propto \sqrt{T}$, so you need $2^2=4\times$ the absolute temperature.

$f(v)$ is a fraction with total area fixed at 1; heating spreads molecules to higher speeds, so the curve gets wider and shorter, not taller.