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

Figure — Maxwell-Boltzmann speed distribution — derivation (key for propulsion)

The figure above is your visual cheat-sheet: the same curve with the three speeds marked. The peak $v_{p}$ sits at the maximum, $\overset{v}{ˉ}$ a little right of it, and $v_{r m s}$ further right still, out toward the long tail.

Level 1 — Recognition

L1.1

Which formula gives the fraction of molecules with speed between $v$ and $v + d v$ ? Write it and state its units.

Recall Solution (L1.1)

The distribution function itself: $f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- \frac{m v ^{2}}{2 k _{B} T}} .$ The fraction is $f (v) d v$ . Since $f (v) d v$ must be a pure number (a fraction) and $d v$ has units of speed, $f (v)$ has units of $(speed)^{- 1}$ , i.e. $s/m$ .

L1.2

Order the three characteristic speeds from smallest to largest, and state $f (0)$ .

Recall Solution (L1.2)

Always $v_{p} < \overset{v}{ˉ} < v_{r m s}$ (ratios $1.000 : 1.128 : 1.225$ ). And $f (0) = 0$ , because the $v^{2}$ factor kills the function at $v = 0$ — no molecule sits at exactly zero speed as a "most likely" value.

L1.3

In the product $v^{2} e^{- m v^{2} /2 k_{B} T}$ , which factor rises with $v$ and which falls? What shape does their product make?

Recall Solution (L1.3)

$v^{2}$ rises (a growing parabola), $e^{- m v^{2} /2 k_{B} T}$ falls (a decaying exponential). Their product is a skewed peak: it climbs from $0$ , reaches a maximum at $v_{p}$ , then decays with a long high-speed tail.

Level 2 — Application

L2.1

Compute $v_{r m s}$ for N₂ at $T = 300$ K.

Recall Solution (L2.1)

$v_{r m s} = \frac{3 k _{B} T}{m} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{4.65 \times 1 0 ^{- 26}} = 2.671 \times 1 0^{5} \approx 517 m/s .$ Why $v_{r m s}$ and not $\overset{v}{ˉ}$ ? RMS weights fast molecules more heavily (it comes from $⟨ v^{2} ⟩$ ), which is exactly what energy and pressure depend on.

L2.2

For the same N₂ at 300 K, compute $v_{p}$ and $\overset{v}{ˉ}$ .

Recall Solution (L2.2)

$v_{p} = \frac{2 k _{B} T}{m} = \frac{2 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{4.65 \times 1 0 ^{- 26}} = 1.781 \times 1 0^{5} \approx 422 m/s .$ $\overset{v}{ˉ} = \frac{8 k _{B} T}{π m} = \frac{8 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{π ( 4.65 \times 1 0 ^{- 26} )} = 2.267 \times 1 0^{5} \approx 476 m/s .$ Check ordering: $422 < 476 < 517$ ✓.

L2.3

If N₂ is heated from 300 K to 600 K, by what factor does $v_{p}$ change? What is the new $v_{p}$ ?

Recall Solution (L2.3)

$v_{p} \propto T$ , so doubling $T$ multiplies $v_{p}$ by $2 \approx 1.414$ . New $v_{p} = 422 \times 1.414 \approx 597 m/s$ . Why the square root? $T$ enters $v_{p}$ inside a square root ( $v_{p} = 2 k_{B} T / m$ ), so a factor 2 inside becomes $2$ outside.

Level 3 — Analysis

L3.1

Compare H₂ ( $m = 3.3 \times 1 0^{- 27}$ kg) and CO₂ ( $m = 7.3 \times 1 0^{- 26}$ kg) at the same temperature. Find $v_{r m s, H_{2}} / v_{r m s, CO_{2}}$ .

Recall Solution (L3.1)

Since $v_{r m s} \propto 1/ m$ at fixed $T$ : $\frac{v _{H_{2}}}{v _{CO_{2}}} = \frac{m _{CO_{2}}}{m _{H_{2}}} = \frac{7.3 \times 1 0 ^{- 26}}{3.3 \times 1 0 ^{- 27}} = 22.12 \approx 4.70.$ Propulsion punchline: lighter exhaust ⟹ faster molecules ⟹ higher exhaust velocity ⟹ higher specific impulse. See Rocket Propulsion & Specific Impulse.

L3.2

A gas is at temperature $T$ . At what temperature $T^{'}$ would a twice-as-heavy gas have the same $v_{r m s}$ ?

Recall Solution (L3.2)

Set $v_{r m s}$ equal: $\frac{3 k _{B} T}{m} = \frac{3 k _{B} T ^{'}}{2 m} \Rightarrow T^{'} = 2 T$ . Why: to keep the same RMS speed after doubling mass, you must double the temperature — mass and temperature trade off one-for-one inside $3 k_{B} T / m$ .

L3.3

Two speeds of interest satisfy $\overset{v}{ˉ} / v_{p} = 4/ π$ . Verify this ratio and state its numeric value.

Recall Solution (L3.3)

$\frac{v ˉ}{v _{p}} = \frac{8 k _{B} T / π m}{2 k _{B} T / m} = \frac{8}{π \cdot 2} = \frac{4}{π} \approx 1.128.$ This is universal — the $k_{B} T / m$ cancels, so the ratio is identical for every gas at every temperature.

Level 4 — Synthesis

L4.1

Starting from $g (v_{x}) = \frac{m}{2 π k _{B} T} e^{- m v_{x}^{2} /2 k_{B} T}$ , derive $⟨ v_{x}^{2} ⟩ = \frac{k _{B} T}{m}$ and use it to obtain $v_{r m s}$ .

Recall Solution (L4.1)

With $β = \frac{m}{2 k _{B} T}$ , the standard Gaussian moment is $⟨ v_{x}^{2} ⟩ = \frac{\int _{- \infty}^{\infty} v _{x}^{2} e ^{- β v_{x}^{2}} d v _{x}}{\int _{- \infty}^{\infty} e ^{- β v_{x}^{2}} d v _{x}} = \frac{1}{2 β} = \frac{2 k _{B} T}{2 m} = \frac{k _{B} T}{m} .$ By isotropy $⟨ v^{2} ⟩ = ⟨ v_{x}^{2} ⟩ + ⟨ v_{y}^{2} ⟩ + ⟨ v_{z}^{2} ⟩ = 3 \cdot \frac{k _{B} T}{m} = \frac{3 k _{B} T}{m}$ , so $v_{r m s} = ⟨ v^{2} ⟩ = \frac{3 k _{B} T}{m} . ✓$ This uses the Equipartition Theorem result $⟨ \frac{1}{2} m v_{x}^{2} ⟩ = \frac{1}{2} k_{B} T$ per component.

L4.2

Show explicitly that $\overset{v}{ˉ} = \frac{8 k _{B} T}{π m}$ by evaluating $\overset{v}{ˉ} = \int_{0}^{\infty} v f (v) d v$ .

Recall Solution (L4.2)

Write $β = \frac{m}{2 k _{B} T}$ and $C = 4 π (\frac{β}{π})^{3/2}$ so $f (v) = C v^{2} e^{- β v^{2}}$ . Then $\overset{v}{ˉ} = C \int_{0}^{\infty} v^{3} e^{- β v^{2}} d v = C \cdot \frac{1}{2 β ^{2}} .$ (Using $\int_{0}^{\infty} v^{3} e^{- β v^{2}} d v = \frac{1}{2 β ^{2}}$ .) Substituting $C$ : $\overset{v}{ˉ} = 4 π (\frac{β}{π})^{3/2} \frac{1}{2 β ^{2}} = \frac{2}{π β} = \frac{4}{π β} = \frac{4 \cdot 2 k _{B} T}{π m} = \frac{8 k _{B} T}{π m} . ✓$

L4.3

Find the ratio of the peak height $f (v_{p})$ at temperature $2 T$ to that at $T$ . (Argue via scaling, don't grind constants.)

Recall Solution (L4.3)

At the peak, $v_{p} \propto T$ and $f (v_{p}) \propto v_{p}^{2} e^{- 1} \cdot (\frac{m}{2 π k _{B} T})^{3/2}$ . Since the $e^{- m v_{p}^{2} /2 k_{B} T} = e^{- 1}$ factor is temperature-independent (it's evaluated at the peak), $f (v_{p}) \propto (T)^{2} \cdot T^{- 3/2} = T \cdot T^{- 3/2} = T^{- 1/2} .$ So $\frac{f ( v _{p} ) at 2 T}{f ( v _{p} ) at T} = (2)^{- 1/2} = \frac{1}{2} \approx 0.707$ . Interpretation: heating lowers the peak (curve spreads out) while total area stays $1$ — matching the parent note's Forecast-then-Verify.

Level 5 — Mastery

L5.1

Effusion through a tiny hole samples molecules with rate $\propto v f (v)$ (faster molecules hit the hole more often). Find the most probable speed of effusing molecules, i.e. the peak of $v f (v) \propto v^{3} e^{- β v^{2}}$ .

Recall Solution (L5.1)

Maximize $h (v) = v^{3} e^{- β v^{2}}$ . Set $h^{'} (v) = 0$ : $h^{'} (v) = (3 v^{2} - 2 β v^{4}) e^{- β v^{2}} = v^{2} (3 - 2 β v^{2}) e^{- β v^{2}} = 0.$ The nonzero root: $3 - 2 β v^{2} = 0 \Rightarrow v = \frac{3}{2 β} = \frac{3 k _{B} T}{m} = v_{r m s}$ . Beautiful result: the typical effusing molecule moves at $v_{r m s}$ — faster than the bulk peak $v_{p}$ , because the hole preferentially samples the fast tail. This underlies Effusion and Graham's Law.

L5.2

Rocket exhaust "energy speed" is best captured by $v_{r m s} = 3 k_{B} T / m$ . A chamber runs at $T = 3000$ K burning H₂ ( $m = 3.3 \times 1 0^{- 27}$ kg). Estimate $v_{r m s}$ and comment on why real exhaust velocities are lower.

Recall Solution (L5.2)

$v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 3000 )}{3.3 \times 1 0 ^{- 27}} = 3.764 \times 1 0^{7} \approx 6135 m/s .$ So $\sim 6.1$ km/s of random thermal speed. Real exhaust is lower because the nozzle converts this isotropic random motion into directed flow imperfectly (finite expansion, recombination, heat loss). Still, this shows why light hot gas is prized: $v_{r m s} \propto T / m$ — see Rocket Propulsion & Specific Impulse.

L5.3

Show that the fraction of molecules with speed above $v_{p}$ is temperature- and mass-independent (a universal number), and estimate it. (Set up the dimensionless integral; you may quote the numeric value.)

Recall Solution (L5.3)

Substitute $u = v / v_{p}$ , so $β v^{2} = v^{2} / v_{p}^{2} = u^{2}$ (since $v_{p} = 1/ β$ ). Then $f (v) d v$ becomes a pure function of $u$ : $P (v > v_{p}) = \int_{1}^{\infty} \frac{4}{π} u^{2} e^{- u^{2}} d u .$ All $k_{B} T / m$ has cancelled — the answer is a universal constant. Evaluating numerically: $P (v > v_{p}) \approx 0.572.$ So about 57% of molecules move faster than the most probable speed — direct evidence of the right-skew (the peak is not the middle of the population).

Connections

Kinetic Theory of Gases — the collision picture behind the distribution
Equipartition Theorem — supplies $\frac{1}{2} k_{B} T$ per DOF (used in L4.1)
Boltzmann Distribution — the energy-space parent of MB
Mean Free Path — uses $\overset{v}{ˉ}$
Rocket Propulsion & Specific Impulse — L5.2 payoff
Effusion and Graham's Law — L5.1 payoff