Maxwell-Boltzmann distribution — full derivation: Exercises

Throughout we use the boxed results from the parent note. Let me restate the toolkit so no symbol is unearned:

Level 1 — Recognition

L1.1 — Which factor kills the high-speed tail?

In $f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- m v^{2} /2 k_{B} T}$ , name the factor that makes $f (v) \to 0$ as $v \to \infty$ , and the factor that makes $f (0) = 0$ .

Recall Solution

WHAT: Two competing factors fight for control of the shape.

As $v \to \infty$ : the Boltzmann factor $e^{- m v^{2} /2 k_{B} T}$ decays faster than any power, so it wins and $f (v) \to 0$ .
At $v = 0$ : the geometric factor $v^{2}$ (the shell volume $4 π v^{2}$ ) is zero, so $f (0) = 0$ .

WHY it looks like this: a bell that starts at zero, rises because of $v^{2}$ , then is crushed by the exponential. See Boltzmann factor and partition function for where $e^{- E / k_{B} T}$ comes from.

L1.2 — Order the three speeds

Without computing anything, write $v_{p}, ⟨ v ⟩, v_{r m s}$ in increasing order and give the numeric ratio.

Recall Solution

$v_{p} : ⟨ v ⟩ : v_{r m s} = 2 : 8/ π : 3 \approx 1.414 : 1.596 : 1.732.$ So $== v_{p} < ⟨ v ⟩ < v_{r m s} ==$ . The long high-speed tail drags the mean and rms to the right of the peak. Mnemonic: Please Make Room (Peak, Mean, RMS).

Level 2 — Application

L2.1 — Most probable speed of helium at 300 K

Helium atom mass $m = 4 u$ . Compute $v_{p}$ at $T = 300$ K.

Recall Solution

WHAT: plug into $v_{p} = 2 k_{B} T / m$ . $m = 4 \times 1.66 \times 1 0^{- 27} = 6.64 \times 1 0^{- 27} kg .$ $v_{p} = \frac{2 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{6.64 \times 1 0 ^{- 27}} = 1.247 \times 1 0^{6} \approx 1117 m/s .$ WHY: light atoms move fast — He is $28/4 = 7 \approx 2.6 \times$ faster than N₂ at the same $T$ .

L2.2 — Ratio of mean to most-probable speed

Show $⟨ v ⟩ / v_{p}$ is a pure number and evaluate it.

Recall Solution

$\frac{⟨ v ⟩}{v _{p}} = \frac{8 k _{B} T / π m}{2 k _{B} T / m} = \frac{8}{2 π} = \frac{4}{π} = \frac{2}{π} \approx 1.128.$ WHY it's universal: $k_{B} T$ and $m$ cancel completely, so this ratio is the same for every ideal gas at every temperature — a fingerprint of the MB shape itself.

L2.3 — Temperature needed to double $v_{p}$

A gas has $v_{p} = 400$ m/s at $T_{1} = 300$ K. At what temperature is $v_{p} = 800$ m/s?

Recall Solution

WHAT: use $v_{p} \propto T$ (mass fixed). $\frac{v _{p, 2}}{v _{p, 1}} = \frac{T _{2}}{T _{1}} \Rightarrow \frac{800}{400} = 2 = \frac{T _{2}}{300} .$ Square: $T_{2} = 4 \times 300 = 1200$ K. WHY: doubling speed needs $4 \times$ the temperature because kinetic energy $\propto v^{2} \propto T$ . Links to Equipartition theorem.

Level 3 — Analysis

L3.1 — Why does the peak height fall when $T$ rises?

The total area under $f (v)$ is always $1$ . Explain, using $v_{p} \propto T$ , why heating the gas lowers the peak of $f (v)$ . See figure.

Figure — Maxwell-Boltzmann distribution — full derivation

Recall Solution

WHAT the figure shows: two MB curves, cooler (lavender) and hotter (coral). The hotter one peaks further right and lower. WHY: the area $\int_{0}^{\infty} f (v) d v = 1$ is conserved (it's a probability). When the peak slides right to a larger $v_{p} \propto T$ , the curve also gets wider (its spread scales with $T$ too). A fixed area spread over a wider base must be shorter. So: hotter = faster and flatter, never taller. This is exactly the Kinetic theory of gases picture of more energetic, more spread-out molecules.

L3.2 — Fraction of molecules moving in the $+ x$ hemisphere

What fraction of molecules have $v_{x} > 0$ ? And what fraction have both $v_{x} > 0$ and $v_{y} > 0$ ?

Recall Solution

WHAT: each component $g (v_{x})$ is an even Gaussian centred at $0$ , so $\int_{0}^{\infty} g (v_{x}) d v_{x} = \frac{1}{2} .$ Fraction with $v_{x} > 0$ is $1/2$ . Components are independent, so joint probability multiplies: $P (v_{x} > 0 and v_{y} > 0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} .$ WHY: isotropy forbids any preferred direction, forcing $⟨ v_{x} ⟩ = 0$ and perfect symmetry between $+ x$ and $- x$ .

L3.3 — Where is the energy distribution peaked?

Convert $f (v)$ to an energy distribution and find the most probable kinetic energy $E_{p}$ . Does $E_{p} = \frac{1}{2} m v_{p}^{2}$ ?

Recall Solution

WHAT we do: change variable $E = \frac{1}{2} m v^{2}$ , so $v = 2 E / m$ and $d v = \frac{d E}{2 m E}$ . Substitute into $f (v) d v$ : $f (v) d v \propto v^{2} e^{- m v^{2} /2 k_{B} T} d v \propto \frac{2 E}{m} e^{- E / k_{B} T} \cdot \frac{d E}{2 m E} \propto E e^{- E / k_{B} T} d E .$ So $F (E) \propto E^{1/2} e^{- E / k_{B} T}$ . Maximize: $\frac{d}{d E} (E^{1/2} e^{- E / k_{B} T}) = 0$ gives $\frac{1}{2} E^{- 1/2} e^{- E / k_{B} T} - \frac{1}{k _{B} T} E^{1/2} e^{- E / k_{B} T} = 0 \Rightarrow \frac{1}{2} = \frac{E}{k _{B} T} \Rightarrow E_{p} = \frac{1}{2} k_{B} T .$ The twist: compare $\frac{1}{2} m v_{p}^{2} = \frac{1}{2} m \cdot \frac{2 k _{B} T}{m} = k_{B} T$ . So $E_{p} = \frac{1}{2} k_{B} T \neq = k_{B} T$ ! WHY they differ: the Jacobian $d v / d E \propto E^{- 1/2}$ warps the peak. The most-probable speed and the most-probable energy live in different coordinate systems — nonlinear change of variable moves peaks. A beautiful, sneaky fact.

Level 4 — Synthesis

L4.1 — Effusion favours fast molecules

The rate at which molecules escape through a tiny hole is proportional to $v f (v)$ (faster molecules hit the hole more often). Find the most probable speed of the escaping molecules, and compare to $v_{p}$ inside.

Recall Solution

WHAT: the effusion beam density is $\propto v \cdot f (v) \propto v \cdot v^{2} e^{- α v^{2}} = v^{3} e^{- α v^{2}}$ , with $α = m /2 k_{B} T$ . Maximize: $\frac{d}{d v} (v^{3} e^{- α v^{2}}) = (3 v^{2} - 2 α v^{4}) e^{- α v^{2}} = 0 \Rightarrow 3 = 2 α v^{2}$ , so $v_{p, eff} = \frac{3}{2 α} = \frac{3 k _{B} T}{m} = v_{r m s} .$ WHY: the extra factor of $v$ from "faster = escapes more often" pushes the peak of the escaping beam up to the interior $v_{r m s}$ — the beam is hotter than the gas left behind. This is the physics behind Effusion and Graham's law and isotope enrichment.

L4.2 — Graham's law from the effusion flux

Two gases at the same $T$ effuse through the same hole. Show carefully that their effusion rates obey $\frac{R _{1}}{R _{2}} = \frac{m _{2}}{m _{1}}$ , then find the rate ratio for H₂ ( $m = 2 u$ ) vs O₂ ( $m = 32 u$ ).

Recall Solution

WHAT governs the rate — which moment? The number of molecules crossing a hole of area $A$ per unit time is the flux: count molecules with a positive velocity component $v_{z}$ toward the hole, weighted by how fast they approach: $R = n A \int_{v_{z} > 0} v_{z} g (v_{z}) d v_{z},$ where $n$ is number density and $g (v_{z})$ is the one-component Gaussian. WHY $v_{z}$ and not $v_{r m s}$ ? Only motion along the hole's normal carries molecules through it; sideways motion ( $v_{x}, v_{y}$ ) doesn't matter. So the governing quantity is $⟨ v_{z} ⟩_{+}$ , the mean of the one-directional speed over the escaping half — not the rms speed and not the full mean speed $⟨ v ⟩$ . Evaluate the moment: with $α = m /2 k_{B} T$ and $g (v_{z}) = α / π e^{- α v_{z}^{2}}$ , $\int_{0}^{\infty} v_{z} g (v_{z}) d v_{z} = \frac{α}{π} \cdot \frac{1}{2 α} = \frac{1}{2} \frac{1}{π α} = \frac{1}{2} \frac{2 k _{B} T}{π m} .$ The mass dependence is all that matters here: at fixed $T$ and equal $n$ , $A$ , $R \propto \frac{1}{m} \Rightarrow \frac{R _{1}}{R _{2}} = \frac{m _{2}}{m _{1}} .$ WHY the moment integral scales as $1/ m$ : the prefactor $α \cdot α^{- 1} = α^{- 1/2} = 2 k_{B} T / m$ carries the only mass dependence — so the whole flux $\propto m^{- 1/2}$ , even though the specific moment is the directional mean, not $v_{r m s}$ . For H₂ vs O₂: $\frac{R _{H_{2}}}{R _{O_{2}}} = \frac{32}{2} = 16 = 4.$ WHY: hydrogen effuses $4 \times$ faster because it's $16 \times$ lighter and the flux scales as $1/ m$ . That is Graham's law.

L4.3 — Recover $b$ purely from the Gaussian integral

Starting from $g (v_{x}) = A e^{- b v_{x}^{2}}$ and only the constraint $⟨ v_{x}^{2} ⟩ = k_{B} T / m$ , rederive $b$ using Gaussian integrals. Do not quote the parent's answer.

Recall Solution

WHAT we need: two Gaussian integrals. $\int_{- \infty}^{\infty} e^{- b x^{2}} d x = \frac{π}{b}, \int_{- \infty}^{\infty} x^{2} e^{- b x^{2}} d x = \frac{1}{2} \frac{π}{b ^{3}} .$ The variance is the ratio (the $A$ and $π$ cancel): $⟨ v_{x}^{2} ⟩ = \frac{\frac{1}{2} π / b ^{3}}{π / b} = \frac{1}{2 b} .$ Set equal to the physics input $\frac{k _{B} T}{m}$ : $\frac{1}{2 b} = \frac{k _{B} T}{m} \Rightarrow b = \frac{m}{2 k _{B} T} .$ WHY the second integral? It's obtained by differentiating $\int e^{- b x^{2}} d x = π / b$ with respect to $b$ — a slick trick that converts "multiply by $x^{2}$ " into " $- \partial_{b}$ ". See Gaussian integrals.

Level 5 — Mastery

L5.1 — Full normalization check of $⟨ v ⟩$

Prove $⟨ v ⟩ = \frac{8 k _{B} T}{π m}$ by evaluating $\int_{0}^{\infty} v f (v) d v$ from scratch.

Recall Solution

WHAT: with $α = m /2 k_{B} T$ and $C = 4 π (α / π)^{3/2}$ (so $f (v) = C v^{2} e^{- α v^{2}}$ ), $⟨ v ⟩ = \int_{0}^{\infty} v \cdot C v^{2} e^{- α v^{2}} d v = C \int_{0}^{\infty} v^{3} e^{- α v^{2}} d v .$ Key integral: $\int_{0}^{\infty} v^{3} e^{- α v^{2}} d v = \frac{1}{2 α ^{2}}$ (substitute $u = v^{2}$ ). So $⟨ v ⟩ = 4 π (\frac{α}{π})^{3/2} \cdot \frac{1}{2 α ^{2}} = \frac{2}{π} α^{- 1/2} = \frac{2}{π} \frac{2 k _{B} T}{m} = \frac{8 k _{B} T}{π m} . ✓$ WHY the odd power was easy: integrals of $v^{odd} e^{- α v^{2}}$ over $[0, \infty)$ are elementary (sub $u = v^{2}$ ); only the even powers need $π$ .

L5.2 — Fraction faster than $v_{p}$

Show that the fraction of molecules with $v > v_{p}$ is $\frac{2}{π} e^{- 1} + erfc (1)$ , and evaluate numerically. (Here $erfc (x) = \frac{2}{π} \int_{x}^{\infty} e^{- t^{2}} d t$ is the "complementary error function" — the tail area of a Gaussian, the exact tool for "fraction beyond a threshold".)

Recall Solution

WHAT: use the dimensionless speed $s = v / v_{p}$ , so $α v^{2} = s^{2}$ (since $v_{p}^{2} = 1/ α$ ) and $f (v) d v = \frac{4}{π} s^{2} e^{- s^{2}} d s$ . We need $P (v > v_{p}) = \frac{4}{π} \int_{1}^{\infty} s^{2} e^{- s^{2}} d s .$ Integrate by parts with $u = s, d w = s e^{- s^{2}} d s$ (so $w = - \frac{1}{2} e^{- s^{2}}$ ): $\int_{1}^{\infty} s^{2} e^{- s^{2}} d s = [- \frac{1}{2} s e^{- s^{2}}]_{1}^{\infty} + \frac{1}{2} \int_{1}^{\infty} e^{- s^{2}} d s = \frac{1}{2} e^{- 1} + \frac{1}{2} \cdot \frac{π}{2} erfc (1) .$ Multiply by $4/ π$ : $P (v > v_{p}) = \frac{4}{π} \cdot \frac{1}{2} e^{- 1} + \frac{4}{π} \cdot \frac{π}{4} erfc (1) = \frac{2}{π} e^{- 1} + erfc (1) .$ Numbers: $\frac{2}{π} e^{- 1} = 1.1284 \times 0.3679 = 0.4151$ ; $erfc (1) = 0.1573$ . Sum $\approx 0.572$ . WHY erfc? "Fraction beyond a threshold" of a Gaussian is exactly the definition of the complementary error function; no elementary closed form exists for the $s^{2}$ integral without it. About 57% of molecules move faster than the most-probable speed — because the peak is not the middle of an asymmetric hump.

L5.3 — 2D gas: rederive the speed distribution

In a two-dimensional gas, velocity space is a plane, so the shell is a ring of circumference $2 π v$ instead of a sphere. Derive $f_{2 D} (v)$ and its most probable speed. See figure.

Recall Solution

WHAT changes: in 2D there are only two components $v_{x}, v_{y}$ , each still Gaussian $e^{- α v_{i}^{2}}$ with $α = m /2 k_{B} T$ . The geometric factor becomes the ring area $2 π v d v$ (circumference $\times$ thickness), not $4 π v^{2} d v$ . The joint density is the product of two identical Gaussians, each normalized so that the 2D prefactor is $α / π$ : $f_{2 D} (v) d v = 2D norm \frac{α}{π} e^{- α (v_{x}^{2} + v_{y}^{2})} \cdot ring 2 π v d v = 2 α v e^{- α v^{2}} d v .$ Check normalization: substitute $u = α v^{2}$ , $d u = 2 α v d v$ : $\int_{0}^{\infty} 2 α v e^{- α v^{2}} d v = \int_{0}^{\infty} e^{- u} d u = 1. ✓$ So $f_{2 D} (v) = 2 α v e^{- α v^{2}}$ is a genuine probability density. Most probable speed: maximize $v e^{- α v^{2}}$ : $\frac{d}{d v} (v e^{- α v^{2}}) = (1 - 2 α v^{2}) e^{- α v^{2}} = 0 \Rightarrow 1 = 2 α v^{2} \Rightarrow v_{p}^{2 D} = \frac{1}{2 α} = \frac{k _{B} T}{m} .$ WHY it is smaller than 3D ( $v_{p}^{3 D} = 2 k_{B} T / m$ ): the geometric factor grows only as $v^{1}$ (a ring's circumference) instead of $v^{2}$ (a sphere's surface area). Fewer dimensions means less "room to be fast", so the falling exponential wins sooner and the peak sits at a lower speed. The visual: in the figure the 2D ring (left) has far less area added per unit $v$ than the 3D shell (right), so the balance point between "more room" and "energy cost" moves inward. Physically, a molecule confined to a plane has one fewer axis to store kinetic energy, so its typical speed is lower at the same temperature — consistent with Equipartition theorem ( $\frac{1}{2} k_{B} T$ per degree of freedom, now only two of them).

Active Recall

Recall Universal ratio

$⟨ v ⟩ / v_{p}$ ? $2/ π \approx 1.128$ — cancels $T$ and $m$ , same for every ideal gas.

Recall Why does heating lower the peak of

$f (v)$ ? Fixed area $= 1$ ; peak slides right ( $v_{p} \propto T$ ) and curve widens, so it must get shorter.

Recall Most probable energy vs energy of most probable speed?

$E_{p} = \frac{1}{2} k_{B} T$ , but $\frac{1}{2} m v_{p}^{2} = k_{B} T$ — they differ because of the Jacobian in the change of variable.

Recall Which moment governs effusion rate, and how does it scale with mass?

The directional flux $\int_{0}^{\infty} v_{z} g (v_{z}) d v_{z} = \frac{1}{2} 2 k_{B} T / π m$ — the mean of the inward component, not $v_{r m s}$ ; it scales as $m^{- 1/2}$ , giving Graham's law.

Recall 2D most-probable speed?

$k_{B} T / m$ , from $f_{2 D} (v) = 2 α v e^{- α v^{2}}$ (ring factor $2 π v$ ).