Maxwell-Boltzmann distribution — full derivation: Worked examples

This page is the exhaustive drill sheet for the Maxwell-Boltzmann distribution. The parent note derived the curve; here we hit every kind of question it can throw at you — every limiting case, every degenerate input, every sign, plus a word problem and an exam twist.

Before we start, one reminder of the two objects we will use over and over.

Here $m$ = mass of one molecule (kg), $k_{B} = 1.38 \times 1 0^{- 23} J/K$ (Boltzmann's constant, the "energy per kelvin per molecule"), $T$ = temperature (K). The symbol N₂ below just means a nitrogen molecule (two nitrogen atoms bonded, molar mass 28 g/mol); O₂ is an oxygen molecule (32 g/mol). The three landmark speeds are

$v_{p} = \frac{2 k _{B} T}{m}, ⟨ v ⟩ = \frac{8 k _{B} T}{π m}, v_{r m s} = \frac{3 k _{B} T}{m} .$

The scenario matrix

Every question about this distribution falls into exactly one of these case classes. The examples below are labelled by which cell they fill.

Cell	Case class	What is "extreme" about it	Example
A	Plug-and-chug landmark speed	ordinary numbers	Ex 1
B	Sign / direction of a component	$v_{x} < 0$ vs $v_{x} > 0$ , both halves	Ex 2
C	Degenerate input $v = 0$	speed exactly zero	Ex 3
D	Limiting behaviour $v \to \infty$	the far tail	Ex 4
E	Limit $T \to 0$ and $T \to \infty$	freezing / infinitely hot	Ex 5
F	Heavy vs light molecule at same $T$	mass extremes	Ex 6
G	Real-world word problem	escape from atmosphere	Ex 7
H	Exam twist — dimensionless ratio	temperature/mass cancel	Ex 8
I	Fraction in a finite band	an integral you must set up	Ex 9

We now walk cells A→I.

Example 1 — Cell A: plug-and-chug landmark speed

Worked example Mean speed of oxygen (O₂) at

$T = 300 K$ One O₂ molecule has mass $m = 32 \times 1.66 \times 1 0^{- 27} = 5.31 \times 1 0^{- 26} kg$ . Find the mean speed $⟨ v ⟩$ .

Forecast: A nitrogen molecule (N₂, molar mass 28 g/mol) at 300 K has a most-probable speed near 420 m/s. O₂ is heavier than N₂, so guess a bit slower — a few hundred m/s. Write your guess down.

Step 1. Pick the right formula. We want the mean, so use $⟨ v ⟩ = 8 k_{B} T / (π m)$ . Why this step? Three landmark speeds exist; "mean" specifically carries the factor $8/ π$ , not 2 or 3.

Step 2. Assemble the numerator inside the root: $8 k_{B} T = 8 (1.38 \times 1 0^{- 23}) (300) = 3.312 \times 1 0^{- 20}$ . Why this step? Group all constants before dividing, so units stay clean (J = kg·m²/s²).

Step 3. Divide and take the root: $⟨ v ⟩ = \frac{3.312 \times 1 0 ^{- 20}}{π ( 5.31 \times 1 0 ^{- 26} )} = 1.986 \times 1 0^{5} \approx 446 m/s .$ Why this step? The root converts (m²/s²) back to (m/s).

Verify: Units: $J / kg = m^{2} / s^{2} = m/s$ ✓. Value ~446 m/s, in the "few hundred" range we forecast ✓.

Example 2 — Cell B: sign of a component (both halves)

Worked example Fraction of molecules moving

leftward faster than $v_{p}$ along $x$ Using the single-component Gaussian $g (v_{x})$ , what fraction have $v_{x} < - v_{p}$ ? (Here $- v_{p}$ means moving in the $- x$ direction with a component speed exceeding the most-probable speed magnitude.)

Forecast: $g$ is symmetric about 0, so leftward and rightward tails are equal. Guess: "same as the fraction with $v_{x} > + v_{p}$ ."

Step 1. Recognise symmetry: $g (- v_{x}) = g (v_{x})$ because it depends only on $v_{x}^{2}$ . Why this step? It converts the leftward question into an equivalent rightward one — no new integral. $P (v_{x} < - v_{p}) = P (v_{x} > + v_{p}) .$

Step 2. Rescale by substituting $u = v_{x} / v_{p}$ . Two things change together:

The exponent: since $v_{p}^{2} = 2 k_{B} T / m$ , we get $\frac{m v _{x}^{2}}{2 k _{B} T} = \frac{v _{x}^{2}}{v _{p}^{2}} = u^{2}$ .
The measure (Jacobian): $v_{x} = u v_{p}$ so $d v_{x} = v_{p} d u$ — every slice of width $d v_{x}$ becomes a slice of width $v_{p} d u$ .

Now track the prefactor. Writing $g (v_{x}) d v_{x}$ and substituting both pieces: $g (v_{x}) d v_{x} = (\frac{m}{2 π k _{B} T})^{1/2} e^{- u^{2}} (v_{p} d u) .$ The prefactor times the Jacobian is $(\frac{m}{2 π k _{B} T})^{1/2} v_{p}$ . Since $v_{p} = 2 k_{B} T / m$ , that product is $\frac{m}{2 π k _{B} T} \cdot \frac{2 k _{B} T}{m} = \frac{1}{π} = \frac{1}{π}$ . So $P (v_{x} > v_{p}) = \frac{1}{π} \int_{1}^{\infty} e^{- u^{2}} d u = \frac{1}{2} erfc (1) .$ Why this step? We must carry the Jacobian $d v_{x} = v_{p} d u$ , otherwise the integral would be off by a factor $v_{p}$ . The Jacobian is exactly what cancels the dimensional part of the prefactor, leaving the clean dimensionless number $1/ π$ — the standard normalised Gaussian tail (which is the $erfc$ integral defined above).

Step 3. Evaluate: $erfc (1) = 0.1573$ , so $P = \frac{1}{2} (0.1573) = 0.0786 \approx 7.9%$ . Why this step? $erfc (z)$ is our "area past $z$ " function; here $z = 1$ because $u = v_{p}$ maps to 1.

Verify: In figure s01 the two red shaded tails (the leftward tail $u < - 1$ and the rightward tail $u > 1$ of the symmetric bell $g$ ) are mirror images, each carrying ~7.9% of the area, with dashed vertical lines marking $\pm v_{p}$ . Total in both tails $= 15.7%$ , leaving $84%$ with $∣ v_{x} ∣ < v_{p}$ , which matches a Gaussian's "most of the mass near the centre" ✓.

Figure s01: The symmetric single-component Gaussian $g (v_{x})$ plotted against $v_{x} / v_{p}$ ; the two red-shaded regions beyond $\pm 1$ are the equal tails, each ~7.9% of the total area.

Example 3 — Cell C: the degenerate input $v = 0$

Worked example What is

$f (0)$ , and how many molecules are exactly at rest?

Forecast: The Boltzmann factor $e^{0} = 1$ is maximal at $v = 0$ . Naïve guess: "most molecules are near rest." Is that right?

Step 1. Plug $v = 0$ into $f (v) = 4 π (\dots) v^{2} e^{- \dots}$ . The $v^{2}$ factor gives $0$ . Why this step? $f$ is a product; if any factor is 0 the whole thing is 0, no matter how large the exponential. $f (0) = 4 π (\dots) \cdot 0 \cdot 1 = 0.$

Step 2. Interpret. "Exactly at rest" needs $v_{x} = v_{y} = v_{z} = 0$ simultaneously — a single point in 3D velocity space, which has zero volume. Why this step? The shell $4 π v^{2} d v$ shrinks to nothing as its radius $\to 0$ .

Verify: In figure s02 the red speed curve visibly touches zero at the origin then climbs to its hump. Compare with the dashed single-component Gaussian, which is maximal at 0. The difference is purely the geometric $v^{2}$ shell factor ✓.

Figure s02: Red = speed distribution $f (v)$ , which starts at $f (0) = 0$ ; dashed black = a velocity-component Gaussian, which peaks at the origin. The gap between them is the $v^{2}$ shell factor.

Example 4 — Cell D: the far tail $v \to \infty$

Worked example Does

$f (v)$ ever stop decreasing, and how fast does the tail die?

Forecast: Guess whether the falling exponential eventually always wins over the rising $v^{2}$ .

Step 1. Compare the two factors for large $v$ . $v^{2}$ grows polynomially; $e^{- α v^{2}}$ (with $α = m /2 k_{B} T > 0$ ) decays super-exponentially. Why this step? A key limit: $lim_{v \to \infty} v^{2} e^{- α v^{2}} = 0$ , because an exponential of $v^{2}$ beats any power of $v$ .

Step 2. Find where the curve turns over, right here on this page. Set the derivative of $v^{2} e^{- α v^{2}}$ to zero: $\frac{d}{d v} (v^{2} e^{- α v^{2}}) = (2 v - 2 α v^{3}) e^{- α v^{2}} = 0 \Rightarrow 1 = α v^{2} \Rightarrow v = \frac{1}{α} = \frac{2 k _{B} T}{m} = v_{p} .$ So the single peak sits at $v_{p}$ ; the curve rises for $v < v_{p}$ and falls monotonically for $v > v_{p}$ . Why this step? This locates the one hump without borrowing from any other page — the tail we care about is everything to the right of $v_{p}$ .

Step 3. Estimate the tail at, say, $v = 3 v_{p}$ : exponent $= α v^{2} = α (9/ α) = 9$ , so factor $e^{- 9} \approx 1.23 \times 1 0^{- 4}$ . Relative to the peak, the height ratio $f (3 v_{p}) / f (v_{p}) = 9 e^{- 9} / (1 \cdot e^{- 1}) = 9 e^{- 8} \approx 3.0 \times 1 0^{- 3}$ . Why this step? Numbers make "the tail is tiny but nonzero" concrete — 0.3% of the peak height.

Verify: $9 e^{- 8} = 9 (3.355 \times 1 0^{- 4}) = 3.02 \times 1 0^{- 3}$ ✓. The tail never reaches zero for finite $v$ but is negligible — this is why some fast molecules can always escape (see Ex 7).

Example 5 — Cell E: temperature limits $T \to 0$ and $T \to \infty$

Worked example What shape does

$f (v)$ approach as the gas freezes ( $T \to 0$ ) or becomes infinitely hot ( $T \to \infty$ )?

Forecast: Guess where the peak $v_{p} = 2 k_{B} T / m$ goes in each limit.

Step 1 ( $T \to 0$ ). $v_{p} \to 0$ and the exponent $- m v^{2} /2 k_{B} T \to - \infty$ for any $v > 0$ , so the curve is pushed to zero everywhere except right at the origin. But the height prefactor $(\frac{m}{2 π k _{B} T})^{3/2} \to \infty$ as $T \to 0$ . These two effects are not in conflict — they cooperate:

the width of the surviving region shrinks like $v_{p} \propto T \to 0$ ,
the prefactor blows up like $T^{- 3/2} \to \infty$ ,

and they blow up/shrink at exactly the rate that keeps $\int_{0}^{\infty} f (v) d v = 1$ at every temperature. The limiting object is therefore an infinitely tall, infinitely thin spike of unit area at $v = 0$ — a Dirac delta $δ (v)$ (a "spike of area 1"). Why this step? Without the diverging prefactor the total area would collapse to 0 and $f$ would stop being a probability distribution. Normalisation ( $\int f = 1$ ) is forced at all $T$ , which is precisely why the height must run to infinity as the width runs to zero. Physically: a frozen gas has every molecule essentially at rest.

Step 2 ( $T \to \infty$ ). $v_{p} \to \infty$ ; the peak slides right without bound and the curve flattens (its area must stay 1, so a wider curve must be lower). Why this step? Hotter = faster and broader, as the parent's Forecast-then-Verify predicted.

Step 3. Sanity number: doubling $T$ multiplies every landmark speed by $2 \approx 1.414$ . Why this step? $v \propto T$ means a $4 \times$ temperature jump only doubles speeds — heating is an inefficient way to speed molecules up.

Verify: $2 = 1.41421$ ✓. In figure s03 the red curve ( $2 T$ ) sits to the right of and lower than the black ( $T$ ) curve; both enclose the same area 1, confirming the "wider means lower" trade-off ✓.

Figure s03: Two speed distributions at temperature $T$ (black) and $2 T$ (red). Heating pushes the peak right and lowers it; both curves keep unit area.

Example 6 — Cell F: heavy vs light at the same $T$

Example 7 — Cell G: real-world word problem

Worked example Can hydrogen escape Earth's atmosphere, but not oxygen?

Earth's escape speed is $v_{esc} = 1.12 \times 1 0^{4} m/s$ . Upper-atmosphere temperature $\approx 1000 K$ . Compare $v_{r m s}$ for H₂ ( $m = 2$ u) and O₂ ( $m = 32$ u). A gas leaks away over geological time once $v_{esc} ≲ 6 v_{r m s}$ .

Forecast: Light H₂ is fast — guess it can leak. Heavy O₂ is slow — guess it stays.

Step 1. H₂: $m = 2 \times 1.66 \times 1 0^{- 27} = 3.32 \times 1 0^{- 27}$ kg. $v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 1000 )}{3.32 \times 1 0 ^{- 27}} = 1.247 \times 1 0^{7} \approx 3530 m/s .$ Why this step? $v_{r m s}$ (factor 3) is the fairest "energy speed" for an escape argument.

Step 2. O₂: $m = 16 \times$ heavier, so $v_{r m s}$ is $16 = 4$ × smaller: $3530/4 \approx 883 m/s$ . Why this step? Reuse the $1/ m$ scaling instead of recomputing from scratch.

Step 3. Compare to threshold $v_{esc} /6 = 1.12 \times 1 0^{4} /6 = 1867$ m/s.

H₂: $3530 > 1867$ ⇒ its fast tail escapes. Leaks away. ✓ forecast
O₂: $883 < 1867$ ⇒ tail negligible at $v_{esc}$ . Retained. ✓ forecast Why this step? Escape is governed by the tail (Ex 4), and the tail sits at a fixed multiple of $v_{r m s}$ .

Verify: $1.247 \times 1 0^{7} = 3531$ m/s ✓; $3531/4 = 883$ m/s ✓. This is exactly why Earth kept its O₂ and N₂ but lost nearly all its free hydrogen.

Example 8 — Cell H: exam twist, dimensionless ratio

Example 9 — Cell I: fraction in a finite speed band

Worked example What fraction of molecules have speeds between

$v_{p}$ and $2 v_{p}$ ?

Forecast: This band straddles the hump's peak and part of the falling side — guess a sizeable chunk, maybe half.

Step 1. Non-dimensionalise with $u = v / v_{p}$ , so $v = u v_{p}$ and $d v = v_{p} d u$ , and the exponent $m v^{2} /2 k_{B} T = u^{2}$ . Then $f (v) d v \propto u^{2} e^{- u^{2}} d u$ and, using $\int_{0}^{\infty} u^{2} e^{- u^{2}} d u = π /4$ for normalisation, $fraction = \frac{\int _{1}^{2} u ^{2} e ^{- u^{2}} d u}{\int _{0}^{\infty} u ^{2} e ^{- u^{2}} d u} = \frac{4}{π} \int_{1}^{2} u^{2} e^{- u^{2}} d u .$ Why this step? Scaling removes $m, T$ ; the answer is universal (holds for any gas at any $T$ ). See Gaussian integrals.

Step 2. Evaluate the finite integral using the antiderivative $\int u^{2} e^{- u^{2}} d u = \frac{π}{4} erf (u) - \frac{u}{2} e^{- u^{2}}$ (with $erf$ as defined at the top): $\int_{1}^{2} u^{2} e^{- u^{2}} d u = [\frac{π}{4} erf (u) - \frac{u}{2} e^{- u^{2}}]_{1}^{2} = 0.2332.$ Why this step? A closed antiderivative avoids numerical guessing; $erf$ handles the Gaussian part.

Step 3. Multiply by the normalisation prefactor to get the final fraction: $fraction = \frac{4}{π} (0.2332) = \frac{4}{1.7725} (0.2332) = 0.526 \approx 53%.$ Why this step? The prefactor $4/ π$ is the reciprocal of the normalising integral $π /4$ ; multiplying converts the raw integral into a fraction of all molecules.

Verify: In figure s04 the red shaded band from $v_{p}$ to $2 v_{p}$ visibly covers about half the total area under the curve, matching $0.53$ ✓. (Our "maybe half" forecast was right on the money — the band brackets the peak, so it captures the bulk of the molecules.)

Figure s04: Speed distribution vs $v / v_{p}$ ; the red-shaded region between $v_{p}$ and $2 v_{p}$ is ~53% of the total area, marked by dashed lines at $u = 1$ and $u = 2$ .

Recall Which cell is each example?

A→Ex1 (plug-in) · B→Ex2 (signs, both tails) · C→Ex3 ( $v = 0$ ) · D→Ex4 ( $v \to \infty$ ) · E→Ex5 ( $T \to 0, \infty$ ) · F→Ex6 (mass extremes) · G→Ex7 (word problem) · H→Ex8 (ratio) · I→Ex9 (finite band).

Active Recall

Why is

f (0) = 0

even though the Boltzmann factor is maximal there?

The geometric shell factor

v^{2} \to 0

: a sphere of radius 0 has no volume, so no velocity vectors have exactly zero magnitude.

For two gases at the same

T

, how do their mean speeds compare?

⟨ v ⟩ \propto 1/ m

— the mass ratio's inverse square root;

T

cancels.

⟨ v ⟩ / v_{p}

gas-dependent?

No — it equals

2/ π \approx 1.128

for every ideal gas at every temperature.

T \to 0

, what happens to

f (v)

It collapses to a Dirac delta spike at

v = 0

; the diverging prefactor

\propto T^{- 3/2}

keeps the area equal to 1 as the width

\propto T

shrinks.

What does

erfc (z)

measure?

The fraction of a Gaussian's area lying in the tail beyond

z

, i.e.

erfc (z) = 1 - erf (z)