Maxwell-Boltzmann distribution of speeds — most probable, mean, rms: Worked examples

The scenario matrix

Every question about MB speeds is one (or a mix) of these cells. The examples below each carry a tag like [Cell A] so you can see the coverage is complete.

Cell	Case class	What makes it tricky	Example
A	Plain "find a speed"	pick right formula, SI units	Ex 1
B	Convert between the three speeds	use the fixed ratio, don't re-integrate	Ex 2
C	Two gases, same $T$	speeds scale as $1/ M$	Ex 3
D	Same gas, change $T$	speeds scale as $T$	Ex 4
E	Solve for the unknown ( $T$ or $M$ )	invert the formula	Ex 5
F	Degenerate / limiting ( $T \to 0$ , $T \to \infty$ , $M \to \infty$ )	what the curve does at the edges	Ex 6
G	Real-world word problem (escape from atmosphere)	connect speed to a physical threshold	Ex 7
H	Curve-shape / geometry read-off	which speed is where, area = 1	Ex 8
I	Exam twist (energy vs speed trap)	KE uses $v_{r m s}$ , not $\overset{v}{ˉ}$	Ex 9

Example 1 — Plain find-a-speed [Cell A]

Worked example

$v_{r m s}$ of nitrogen at 300 K Nitrogen $N_{2}$ : $M = 28 g/mol$ . Find $v_{r m s}$ at $T = 300 K$ .

Forecast: guess — will it be nearer 100, 500, or 5000 m/s?

Convert mass to SI. $M = 28 g/mol = 0.028 kg/mol$ . Why this step? $R$ is in joules, and a joule is $kg m^{2} s^{- 2}$ . Feeding g/mol makes the answer $1000 \approx 31.6 \times$ too small.
Pick the energy speed. "rms" $\Rightarrow v_{r m s} = \frac{3 R T}{M}$ . Why this step? The word rms names the energy-weighted average; only this formula has the factor 3.
Plug in. $v_{r m s} = \frac{3 ( 8.314 ) ( 300 )}{0.028} = 2.673 \times 1 0^{5} \approx 517 m/s$ . Why this step? Straight substitution once units are fixed.

Verify: $517 m/s \approx 1860 km/h$ — roughly 1.5× the speed of sound in air ( $\sim 340$ m/s), which is physically right (sound travels at a typical molecular speed). Units: $J/(kg/mol) \cdot mol = m^{2} / s^{2} = m/s$ . ✓

Example 2 — Convert between the three [Cell B]

Example 3 — Two gases, same temperature [Cell C]

Example 4 — Same gas, change temperature [Cell D]

Example 5 — Solve for the unknown [Cell E]

Worked example At what

$T$ does $H_{2}$ match $O_{2}$ 's speed? At what temperature does $H_{2}$ ( $M = 2$ g/mol) have the same $v_{r m s}$ that $O_{2}$ ( $M = 32$ g/mol) has at $300 K$ ?

Forecast: hydrogen is light and fast, so it should need a lower $T$ to be that slow.

Equal $v_{r m s}$ means equal $T / M$ . Set $\frac{3 R T _{H_{2}}}{M _{H_{2}}} = \frac{3 R ( 300 )}{M _{O_{2}}} \Rightarrow \frac{T _{H_{2}}}{M _{H_{2}}} = \frac{300}{M _{O_{2}}}$ . Why this step? $v_{r m s}$ depends only on the combination $T / M$ , so matching speed = matching $T / M$ .
Solve for $T_{H_{2}}$ . $T_{H_{2}} = 300 \times \frac{M _{H_{2}}}{M _{O_{2}}} = 300 \times \frac{2}{32} = 18.75 K$ . Why this step? Isolate the unknown; the mass ratio (not its square root — because we matched $T / M$ not $T / M$ … wait: we matched the whole speed, so $T / M$ equal, giving the plain ratio).

Verify: plug back — $3 R (18.75) /0.002 = 3 R (300) /0.032$ ? Both give $T / M = 9375$ , so yes, identical $v_{r m s}$ ✓. Lower $T$ for the lighter gas, as forecast.

Example 6 — Degenerate & limiting cases [Cell F]

Worked example The edges:

$T \to 0$ , $T \to \infty$ , $M \to \infty$ Describe what happens to $f (v)$ and to $v_{m p}$ in three extreme limits. This is the "corner case" cell — the reader must never be surprised by an edge.

Forecast: at absolute zero, what speed do molecules have?

$T \to 0$ . $v_{m p} = 2 R T / M \to 0$ . Every speed collapses to zero; the curve becomes an infinitely tall, infinitely thin spike at $v = 0$ (all molecules frozen). Why this step? $T$ sits in the numerator under the root; zero $T$ zeroes the speed. Physically: no thermal energy, no motion.
$T \to \infty$ . $v_{m p} \to \infty$ ; the peak marches right without bound and the curve flattens into an ever-wider spread. See the widening in the figure. Why this step? Unbounded $T$ means unbounded typical speed — but the area under $f (v)$ stays exactly 1 (still 100 % of molecules).
$M \to \infty$ (very heavy molecule). $v_{m p} = 2 R T / M \to 0$ . Heavy molecules are sluggish — same $T$ , but the curve pulls left and grows tall & narrow. Why this step? $M$ is in the denominator under the root; large $M$ crushes the speed.

Verify: in all three cases the total area = 1 is preserved (all molecules always counted); only the shape moves. Limits are consistent: $v_{m p} \to 0$ for $T \to 0$ and $M \to \infty$ , since both send $T / M \to 0$ . ✓

Example 7 — Real-world word problem [Cell G]

Worked example Why Earth keeps

$N_{2}$ but loses $H_{2}$ A molecule escapes Earth if its speed reaches the escape velocity $v_{esc} \approx 11200 m/s$ . Compare $v_{r m s}$ of $H_{2}$ ( $M = 2$ g/mol) and $N_{2}$ ( $M = 28$ g/mol) at an upper-atmosphere temperature $T = 1000 K$ . Which is closer to escaping?

Forecast: neither $v_{r m s}$ reaches 11 200 m/s — so how does escape happen at all?

$v_{r m s}$ of $H_{2}$ . $3 (8.314) (1000) /0.002 = 1.247 \times 1 0^{7} \approx 3532 m/s$ . Why this step? Direct formula; use the hottest relevant $T$ since escape happens high up.
$v_{r m s}$ of $N_{2}$ . $3 (8.314) (1000) /0.028 = 8.908 \times 1 0^{5} \approx 944 m/s$ . Why this step? Same $T$ , heavier $\Rightarrow$ slower by $28/2 = 14 = 3.74 \times$ .
Compare to $v_{esc}$ . $H_{2}$ reaches $\sim 31%$ of $v_{esc}$ ; $N_{2}$ only $\sim 8%$ . Why this step? Escape isn't about the average molecule — it's the fast tail. A gas whose $v_{r m s}$ is a larger fraction of $v_{esc}$ has vastly more molecules in the tail above $v_{esc}$ (the exponential punishes the ratio harshly).

Verify: ratio $v_{r m s}^{H_{2}} / v_{r m s}^{N_{2}} = 3532/944 = 3.74 = 14$ ✓, matching the pure mass-ratio prediction. Light $H_{2}$ sits far higher toward escape, so it drains from the atmosphere over geological time while $N_{2}$ stays — exactly what the parent note claimed.

Example 8 — Read the curve (geometry) [Cell H]

Worked example Locating the three speeds on the graph

On a single MB curve for one gas, mark where $v_{m p}$ , $\overset{v}{ˉ}$ , $v_{r m s}$ sit and state which region of area each cuts.

Forecast: which one is at the tallest point of the curve?

Peak = $v_{m p}$ . The maximum of $f (v)$ is by definition the most probable speed — the tallest point. Why this step? We found $v_{m p}$ by setting $df / d v = 0$ ; that's literally the peak.
Order along the axis. Moving right from the peak: $\overset{v}{ˉ}$ (at $1.128 v_{m p}$ ), then $v_{r m s}$ (at $1.225 v_{m p}$ ). All three are to the left of the tail's far end because the tail runs to infinity. Why this step? The fixed ratio $1 : 1.128 : 1.225$ places them; the right-skew guarantees mean and rms sit right of the peak.
Area meaning. The area from $0$ to any $v_{0}$ equals the fraction of molecules slower than $v_{0}$ ; total area $= 1$ . Why this step? $f (v) d v$ is a fraction, so integrating adds fractions.

Verify: the three vertical lines in the figure obey $v_{m p} < \overset{v}{ˉ} < v_{r m s}$ and the peak sits exactly on $v_{m p}$ ✓. Numerically $1.128/1 = 1.128$ and $1.225/1 = 1.225$ match the ratio.

Example 9 — Exam twist: energy vs speed trap [Cell I]

Worked example Which speed for kinetic energy?

A student computes the average kinetic energy of $O_{2}$ at $300 K$ as $\frac{1}{2} m \overset{v}{ˉ}^{2}$ using $\overset{v}{ˉ}$ . Find the correct $\overline{K E}$ per molecule, and show the student's number is wrong.

Forecast: will the two answers differ by a lot or a little?

Correct energy uses $v_{r m s}$ . $\overline{K E} = \frac{1}{2} m \overline{v^{2}} = \frac{1}{2} m v_{r m s}^{2} = \frac{3}{2} k T$ . Why this step? The average of $v^{2}$ is $v_{r m s}^{2}$ by definition — not $\overset{v}{ˉ}^{2}$ . Energy is built from $\overline{v^{2}}$ , so only $v_{r m s}$ is legal here.
Numeric value. $\frac{3}{2} k T = \frac{3}{2} (1.381 \times 1 0^{- 23}) (300) = 6.21 \times 1 0^{- 21} J$ per molecule. Why this step? Use $k$ (per-molecule) since we asked per molecule.
Student's error size. Using $\overset{v}{ˉ}$ gives $\frac{1}{2} m \overset{v}{ˉ}^{2} = \frac{1}{2} m (0.921 v_{r m s})^{2} = 0.92 1^{2} \times \overline{K E} = 0.848 \overline{K E}$ — about 15 % too low. Why this step? $\overset{v}{ˉ}^{2} \neq = \overline{v^{2}}$ ; the gap is exactly $(\overset{v}{ˉ} / v_{r m s})^{2} = 0.848$ .

Verify: $0.92 1^{2} = 0.848$ ✓, so the student underestimates by $1 - 0.848 = 0.152$ , i.e. $\approx 15%$ — not negligible. Correct $\overline{K E} = 6.21 \times 1 0^{- 21}$ J, independent of the gas (depends only on $T$ ) ✓.

Recap

Recall Which cell needs which trick?

Same $T$ , two gases ::: speed ratio $= M_{2} / M_{1}$ Same gas, two temperatures ::: speed ratio $= T_{2} / T_{1}$ Convert among $v_{m p}, \overset{v}{ˉ}, v_{r m s}$ ::: multiply by the fixed ratio $1 : 1.128 : 1.225$ Kinetic energy question ::: always $v_{r m s}$ , never $\overset{v}{ˉ}$ $T \to 0$ or $M \to \infty$ ::: every speed $\to 0$ , area stays 1 g/mol given ::: convert to kg/mol before plugging into $3 R T / M$

Related vault topics: Kinetic Theory of Gases, Average Kinetic Energy and Temperature, Graham's Law of Effusion, Mean Free Path and Collision Frequency, Boltzmann Distribution, Real Gases and van der Waals Equation.