Maxwell-Boltzmann speed distribution — derivation (key for propulsion): Worked examples

Everything below uses only two constants and one formula, so let us pin them once.

The scenario matrix

Every MB problem is one (or a blend) of these case classes. The last column says which worked example covers it.

#	Case class	What makes it tricky	Covered by
C1	Plain numeric speed	Get $m$ from molar mass right	Ex 1
C2	Ratio / universal	$k_{B} T / m$ cancels — answer is pure number	Ex 2
C3	Temperature scaling	All speeds $\propto T$	Ex 3
C4	Mass scaling (propulsion)	All speeds $\propto 1/ m$	Ex 4
C5	Degenerate input $v = 0$	$f (0) = 0$ , not a peak	Ex 5
C6	Limiting $T \to 0$ and $T \to \infty$	Curve collapses / spreads	Ex 5
C7	Peak height vs peak position	Height falls as $T$ rises	Ex 6
C8	Fraction in a speed window (integral)	Needs the $f (v) d v$ meaning	Ex 7
C9	Real-world word problem (nozzle)	Turn physics into $v_{r m s}$	Ex 8
C10	Exam twist: mixture / two gases	Each gas keeps its own $m$	Ex 9

Ex 1 — C1: plain numeric speed

Worked example RMS and mean speed of oxygen O₂ at 300 K

Molar mass of O₂ is $M = 32 g/mol = 0.032 kg/mol$ . Avogadro $N_{A} = 6.022 \times 1 0^{23}$ . Find $v_{r m s}$ and $\overset{v}{ˉ}$ at $T = 300 K$ .

Forecast: N₂ was $\approx 517$ m/s in the parent. O₂ is slightly heavier (32 vs 28). Guess: a bit below 500 m/s.

Step 1 — Get the mass of one molecule. $m = \frac{M}{N _{A}} = \frac{0.032}{6.022 \times 1 0 ^{23}} = 5.31 \times 1 0^{- 26} kg .$ Why this step? The formula needs mass per molecule, not per mole — this is the #1 place people slip.

Step 2 — Plug into $v_{r m s} = 3 k_{B} T / m$ . $v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{5.31 \times 1 0 ^{- 26}} = 2.339 \times 1 0^{5} \approx 484 m/s .$ Why this step? "rms" weights the fast molecules — the speed relevant to pressure and energy.

Step 3 — Mean speed via the fixed ratio $\overset{v}{ˉ} / v_{r m s} = 8/ π / 3 = 8/ (3 π)$ . $\overset{v}{ˉ} = v_{r m s} \frac{8}{3 π} = 484 \times 0.9213 \approx 446 m/s .$ Why this step? Once you have one speed, the others are just constant multiples — no need to re-integrate.

Verify: $484$ m/s is just under $500$ and below N₂'s $517$ (O₂ heavier ⟹ slower) ✓. Units: $J / kg = m^{2} / s^{2} =$ m/s ✓.

Ex 2 — C2: universal ratio

Ex 3 — C3: temperature scaling

Ex 4 — C4: mass scaling (propulsion punchline)

Worked example Helium vs xenon as thruster propellant

Compare $v_{r m s}$ of He ( $M = 4 g/mol$ ) and Xe ( $M = 131 g/mol$ ) at the same temperature. Which makes a better cold-gas thruster (higher exhaust speed = more specific impulse)?

Forecast: Xe is $\approx 33 \times$ heavier. Speed $\propto 1/ m$ , so He should be roughly $33 \approx 5.7 \times$ faster.

Step 1 — Same $T$ , so $v_{r m s} \propto 1/ m$ , and molar mass ratio equals molecular mass ratio. $\frac{v _{He}}{v _{Xe}} = \frac{M _{Xe}}{M _{He}} = \frac{131}{4} = 32.75 .$ Why this step? The heavier mass sits under a square root in the denominator, so it enters as $1/ m$ ; the ratio wipes out $k_{B} T$ .

Step 2 — Evaluate. $\frac{v _{He}}{v _{Xe}} = 5.72.$ Why this step? This is the whole reason light gases give more thrust per kilogram — the exhaust leaves far faster.

Verify: He faster than Xe ✓, and magnitude near the forecast $5.7$ ✓. (Consistent with Graham's law, rate $\propto 1/ m$ .)

Ex 5 — C5 & C6: degenerate and limiting inputs

Worked example What happens at

$v = 0$ , and as $T \to 0$ or $T \to \infty$ ? Evaluate $f (0)$ , and describe the curve in the two temperature limits.

Forecast: At $v = 0$ nothing is moving — should be rare, so $f (0)$ small. At $T \to 0$ everything freezes; at $T \to \infty$ everything spreads out.

Step 1 — Plug $v = 0$ into the formula. $f (0) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} \cdot 0^{2} \cdot e^{0} = 0.$ Why this step? The $v^{2}$ shell factor kills the value: there is zero surface area on a sphere of radius $0$ , so no velocity vectors have exactly zero speed. The curve starts at the origin, not at a peak.

Step 2 — Limit $T \to 0$ . Then $v_{p} = 2 k_{B} T / m \to 0$ and the Gaussian width shrinks — the whole curve collapses into a spike at $v = 0$ . Why this step? No thermal energy means no motion; a distribution of a single value $0$ is a spike.

Step 3 — Limit $T \to \infty$ . Then $v_{p} \to \infty$ , the peak marches right, the curve flattens and widens (area stays $1$ ). Why this step? Infinite heat spreads molecules over all speeds; probability must stay normalized, so height drops as width grows.

Verify (figure): the three curves below show exactly this — cold = tall narrow spike near zero, hot = short wide curve far right, and all three touch $0$ at $v = 0$ ✓.

Ex 6 — C7: peak height vs peak position

Worked example Does a hotter gas have a taller peak?

As $T$ increases, does the height of $f (v)$ at its peak go up or down? Find how the peak height scales with $T$ .

Forecast: The peak moves right (faster). Since total area is fixed at $1$ , a curve that spreads wider must get shorter. Guess: height falls.

Step 1 — The peak sits at $v_{p} = 2 k_{B} T / m$ . Substitute $v = v_{p}$ into $f (v)$ . Note $m v_{p}^{2} / (2 k_{B} T) = 1$ , so the exponential becomes $e^{- 1}$ . $f (v_{p}) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v_{p}^{2} e^{- 1} .$ Why this step? Evaluating at the peak turns the messy exponent into the constant $e^{- 1}$ , isolating the $T$ -dependence.

Step 2 — Track the powers of $T$ . The prefactor gives $T^{- 3/2}$ ; $v_{p}^{2} = 2 k_{B} T / m$ gives $T^{+ 1}$ . $f (v_{p}) \propto T^{- 3/2} \cdot T^{+ 1} = T^{- 1/2} .$ Why this step? Collecting all $T$ 's tells us the height scales as $1/ T$ — it falls as the gas heats.

Verify: From 300 K to 1200 K ( $\times 4$ ), peak height $\to (4)^{- 1/2} = 1/2$ of its value. In the figure from Ex 5 the hottest curve is indeed the shortest ✓. Area under each curve stays $1$ (spread up, height down) ✓.

Ex 7 — C8: fraction of molecules in a speed window

Worked example Fraction with speed within

$\pm 1%$ of $v_{p}$ Estimate the fraction of molecules with speed between $v_{p}$ and $1.01 v_{p}$ (a thin window right at the peak) using $f (v) d v$ .

Forecast: A 1% window is thin. Roughly (fraction) $\approx f (v_{p}) \times$ (window width). Guess: about 1%.

Step 1 — Use the definition: fraction $\approx f (v_{p}) Δ v$ with $Δ v = 0.01 v_{p}$ . Why this step? $f (v) d v$ is the fraction in a narrow slice; near the peak $f$ is almost flat, so a rectangle is a good approximation.

Step 2 — Compute $f (v_{p})$ in units of $v_{p}$ . With $β = 1/ v_{p}^{2}$ the formula gives $f (v_{p}) = \frac{4}{π} \frac{1}{v _{p}} e^{- 1} = \frac{4}{π} e^{- 1} \frac{1}{v _{p}} = 0.8302 \frac{1}{v _{p}} .$ Why this step? Writing everything in terms of $v_{p}$ makes the $v_{p}$ 's cancel with $Δ v$ , leaving a pure number.

Step 3 — Multiply by the window. $fraction \approx 0.8302 \frac{1}{v _{p}} \times 0.01 v_{p} = 8.30 \times 1 0^{- 3} \approx 0.83%.$ Why this step? This is the concrete meaning of "fraction of molecules" — an actual small number you could count.

Verify: Result is dimensionless ✓ (units of $v_{p}$ cancelled), positive, and less than 1% as forecast ✓.

Ex 8 — C9: real-world word problem (nozzle)

Worked example Exhaust speed of a hot-hydrogen thruster

A nuclear-thermal rocket heats hydrogen gas (H₂, $M = 2 g/mol$ ) to $2800 K$ before it leaves the nozzle. Estimate the characteristic thermal speed $v_{r m s}$ of the H₂ molecules. Why does this hint at a high specific impulse?

Forecast: Very light gas, very hot — expect several thousand m/s.

Step 1 — Mass of one H₂ molecule. $m = \frac{0.002}{6.022 \times 1 0 ^{23}} = 3.32 \times 1 0^{- 27} kg .$ Why this step? Again convert molar → per-molecule before touching $k_{B}$ .

Step 2 — Compute $v_{r m s} = 3 k_{B} T / m$ . $v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 2800 )}{3.32 \times 1 0 ^{- 27}} = 3.491 \times 1 0^{7} \approx 5908 m/s .$ Why this step? $v_{r m s}$ sets the energy scale of the exhaust; nearly 6 km/s is the thermal speed feeding the directed jet.

Step 3 — Interpret. Specific impulse $\propto$ exhaust velocity $\propto v_{r m s} \propto 1/ m$ . Hydrogen's tiny $m$ is what makes its molecules so fast at a given $T$ . Why this step? Ties the number back to the propulsion payoff from the parent note.

Verify: $\sim 5900$ m/s is in the expected thousands ✓. Compare to O₂ at 300 K ( $484$ m/s): H₂ is lighter ( $\times 16$ ) and hotter ( $\times 9.3$ ), factor $16 \times 9.3 = 12.2$ , and $484 \times 12.2 = 5905$ ✓.

Ex 9 — C10: exam twist — a two-gas mixture

Worked example Same box, two gases — do they share a speed?

A sealed box holds He and Ar in thermal equilibrium at one temperature $T$ . A student claims "in equilibrium both gases have the same $v_{r m s}$ ." True or false? Find the ratio $v_{r m s, He} / v_{r m s, Ar}$ ( $M_{He} = 4$ , $M_{Ar} = 40$ g/mol).

Forecast: Equilibrium means equal temperature (equal mean energy), not equal speed. Lighter He must move faster. Guess ratio $\approx 10 \approx 3.2$ .

Step 1 — Equilibrium equalises average kinetic energy per molecule (via equipartition): $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ for each gas. Why this step? Temperature — not speed — is what two gases share when they mix. The student confused "same $T$ " with "same $v$ ."

Step 2 — Equal energy but different mass ⟹ different speed. Take the ratio. $\frac{v _{r m s, He}}{v _{r m s, Ar}} = \frac{M _{Ar}}{M _{He}} = \frac{40}{4} = 10 = 3.162.$ Why this step? With $k_{B} T$ fixed (same box), only mass differs, entering as $1/ m$ .

Step 3 — Verdict: the claim is false; only $v_{r m s, He} = 10 v_{r m s, Ar}$ holds. Why this step? Names the specific misconception so it never recurs.

Verify: He faster ✓, ratio $3.16 > 1$ ✓; equal energies check: $\frac{1}{2} m_{He} v_{He}^{2}$ vs $\frac{1}{2} m_{Ar} v_{Ar}^{2}$ have ratio $(1/10) (10) = 1$ ✓.

Recall Which cell was hardest? Self-test

Peak height of $f (v)$ scales as ::: $T^{- 1/2}$ (falls as gas heats). $f (0) = ?$ and why ::: $0$ , because the $v^{2}$ shell factor vanishes at zero speed. Two gases mixed at equilibrium share ::: temperature (mean energy), not speed. $v_{p} / v_{r m s}$ for any gas ::: $2/3 = 0.8165$ , universal.

Connections

Parent: MB derivation
Kinetic Theory of Gases — pressure links to $v_{r m s}$
Equipartition Theorem — equal energy in a mixture (Ex 9)
Boltzmann Distribution — the energy-space parent law
Mean Free Path — uses $\overset{v}{ˉ}$
Rocket Propulsion & Specific Impulse — Ex 4 and Ex 8 payoff
Effusion and Graham's Law — $1/ m$ scaling reused