Mean free path, mean speed, RMS speed — derivations: Worked examples

This page is the drill floor for the parent derivations. The formulas are already built there; here we hit every kind of question an exam or the real world can throw. First we lay out the full map of cases, then work each one.

If a symbol here feels unfamiliar, it was defined in the parent note — but we re-state each one the moment it appears, so you never have to flip back.

The scenario matrix

Think of every possible question as a cell in this grid. Our goal: no empty cells. Each worked example below is tagged with the cell it fills.

#	Cell class	What makes it tricky	Example
A	Direct plug — $v_{r m s}$	SI unit trap ( $M$ in kg/mol)	Ex 1
B	Direct plug — $\overset{v}{ˉ}$ and $v_{p}$	three speeds, one gas	Ex 2
C	Ratio / scaling (T changes)	speeds $\propto T$	Ex 3
D	Ratio / scaling (different gas, same T)	speeds $\propto 1/ M$	Ex 4
E	Mean free path — direct	$n = P / k_{B} T$ form, the $2$	Ex 5
F	Limiting / degenerate ( $P \to 0$ , $T \to 0$ )	what blows up, what vanishes	Ex 6
G	Collision frequency (combine $\overset{v}{ˉ}$ and $λ$ )	rate $= \overset{v}{ˉ} / λ$	Ex 7
H	Real-world word problem	vacuum chamber design	Ex 8
I	Exam-style twist (reverse: find $T$ or $d$ )	invert the formula	Ex 9

Constants used throughout (memorise these):

$R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant)
$k_{B} = 1.38 \times 1 0^{- 23} J K^{- 1}$ (Boltzmann constant, $= R / N_{A}$ )
$N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$

Cell A — Direct RMS speed

Worked example

Ex 1. Find $v_{r m s}$ of oxygen ( $O_{2}$ ) at $T = 300 K$ . Molar mass $M = 32 g/mol$ .

Forecast: guess before computing — will it be tens, hundreds, or thousands of m/s? (Room-temperature gas molecules are hundreds of m/s.)

Convert $M$ to SI. $M = 32 g/mol = 0.032 kg/mol$ . Why this step? The formula $v_{r m s} = 3 R T / M$ uses $R$ in joules, which are $kg m^{2} s^{- 2}$ . If you leave $M$ in grams you divide by a number 1000× too small, so the speed comes out $1000 \approx 32 \times$ too big. This is the single most common exam error.
Plug in. $v_{r m s} = \frac{3 R T}{M} = \frac{3 ( 8.314 ) ( 300 )}{0.032} = 2.338 \times 1 0^{5}$ Why this step? $v_{r m s}$ is the "energy-carrying" speed because kinetic energy $\propto v^{2}$ ; it comes straight from $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$ .
Root it. $v_{r m s} \approx 483 m/s$ .

Verify: Units — $J/(kg/mol) = (kg m^{2} s^{- 2} / mol) / (kg/mol) = m^{2} s^{- 2} = m/s$ . ✓ Hundreds of m/s, as forecast, and lighter than… well, we'll compare in Ex 4.

Cell B — All three speeds of one gas

Worked example

Ex 2. For the same $O_{2}$ at $300 K$ , also find the mean speed $\overset{v}{ˉ}$ and most probable speed $v_{p}$ .

Forecast: the parent note gives the ordering $v_{p} < \overset{v}{ˉ} < v_{r m s}$ . So both new numbers should sit below $483 m/s$ , with $v_{p}$ lowest.

Mean speed — the ordinary arithmetic average of speed, which controls how often things happen (collisions, effusion): $\overset{v}{ˉ} = \frac{8 R T}{π M} = \frac{8 ( 8.314 ) ( 300 )}{π ( 0.032 )} \approx 445 m/s$ Why this step? The extra $8/ π$ instead of $3$ comes from integrating $v f (v)$ rather than $v^{2} f (v)$ over the Maxwell-Boltzmann distribution.
Most probable speed — where the distribution $f (v)$ peaks, the single most common speed: $v_{p} = \frac{2 R T}{M} = \frac{2 ( 8.314 ) ( 300 )}{0.032} \approx 395 m/s$ Why this step? Setting $df / d v = 0$ gives the factor $2$ under the root.

Verify: Ratios should be $v_{p} : \overset{v}{ˉ} : v_{r m s} = 2 : 8/ π : 3 \approx 1.414 : 1.596 : 1.732$ . Check: $445/395 = 1.127$ and $8/ π / 2 = 1.128$ . ✓ Ordering $395 < 445 < 483$ holds, as forecast.

Figure — Mean free path, mean speed, RMS speed — derivations

Cell C — Temperature scaling (same gas)

Cell D — Different gas, same temperature

Worked example

Ex 4. At the same temperature, compare $v_{r m s}$ of helium ( $M = 4 g/mol$ ) and oxygen ( $M = 32 g/mol$ ).

Forecast: helium is 8× lighter. Speed goes as $1/ M$ , and $8 \approx 2.83$ — so helium should be about 2.8× faster.

Ratio at fixed $T$ . $\frac{v _{He}}{v _{O_{2}}} = \frac{M _{O_{2}}}{M _{He}} = \frac{32}{4} = 8 \approx 2.83$ Why this step? At the same $T$ every molecule has the same average kinetic energy $\frac{3}{2} k_{B} T$ (this is the Equipartition theorem). Same energy $\frac{1}{2} m v^{2}$ with smaller mass means bigger speed — lighter gas flies faster.
Use Ex 1's oxygen value $v_{O_{2}} = 483 m/s$ at $300 K$ : $v_{He} = 2.83 \times 483 \approx 1367 m/s$

Verify: Direct plug: $3 (8.314) (300) /0.004 = 1.871 \times 1 0^{6} \approx 1368 m/s$ . ✓ (rounding). This is why helium leaks and escapes Earth's gravity faster than heavier gases — link to Diffusion and effusion.

Cell E — Mean free path, direct

Worked example

Ex 5. Find the mean free path $λ$ of nitrogen at STP: $T = 273 K$ , $P = 1.01 \times 1 0^{5} Pa$ , molecular diameter $d = 3.7 \times 1 0^{- 10} m$ .

Forecast: the parent note quoted ~68 nm for air — nitrogen is basically air, so expect the same ballpark.

Use the pressure form directly. $λ = \frac{k _{B} T}{2 π d ^{2} P}$ Why this step? Substituting number density $n = P / k_{B} T$ (from the Ideal gas law) lets us skip computing $n$ as a separate messy number.
Plug the numbers. $λ = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 273 )}{2 π ( 3.7 \times 1 0 ^{- 10} ) ^{2} ( 1.01 \times 1 0 ^{5} )}$ Numerator $= 3.767 \times 1 0^{- 21}$ . Denominator $= 2 π (1.369 \times 1 0^{- 19}) (1.01 \times 1 0^{5}) = 6.14 \times 1 0^{- 14}$ . $λ \approx 6.1 \times 1 0^{- 8} m = 61 nm$ Why the $2$ ? Targets are moving too. The average relative speed between two molecules is $2 \overset{v}{ˉ}$ , which raises the collision rate by $2$ and shortens the flight by $2$ .

Verify: Units — $\frac{J}{m ^{2} \cdot Pa} = \frac{J}{m ^{2} \cdot N/m ^{2}} = \frac{J}{N} = m$ . ✓ 61 nm ≈ 165× the molecular diameter — gas is mostly empty space. ✓ (Matches parent's ~68 nm for the slightly different $d$ of air.)

Cell F — Limiting & degenerate cases

Worked example

Ex 6. What happens to $v_{r m s}$ and $λ$ in these extreme limits? (a) $T \to 0$ ; (b) $P \to 0$ (perfect vacuum) at fixed $T$ ; (c) $P \to \infty$ .

Forecast: cold gas freezes → speeds die. Empty space → nothing to hit → flights get huge. Crush the gas → nowhere to fly.

(a) $T \to 0$ . $v_{r m s} = 3 R T / M \to 0 = 0$ . Why? All three speeds share the $T$ factor, so $v_{p}, \overset{v}{ˉ}, v_{r m s} \to 0$ together. Zero thermal energy means (classically) no motion. Meanwhile $λ = \frac{k _{B} T}{2 π d ^{2} P} \to 0$ if $P$ is held fixed — but real gas would liquefy first, so this is an idealised limit.
(b) $P \to 0$ at fixed $T$ . $λ = \frac{k _{B} T}{2 π d ^{2} P} \to \infty$ . Why? Fewer molecules per volume ( $n = P / k_{B} T \to 0$ ) means almost nothing to collide with — a molecule can cross the whole chamber uninterrupted. This is exactly the "molecular flow" regime used in vacuum technology (Ex 8). Note $v_{r m s}$ is unchanged — it depends only on $T$ , not on $P$ .
(c) $P \to \infty$ . $λ \to 0$ . Why? Squeeze infinitely → infinite crowding → zero flight between hits.

Verify: Limits are consistent with $λ \propto 1/ P$ and $v_{r m s} \propto T$ (independent of $P$ ). Sanity: doubling $P$ halves $λ$ but leaves $v_{r m s}$ alone. ✓

Cell G — Collision frequency (combining $\overset{v}{ˉ}$ and $λ$ )

Worked example

Ex 7. How many collisions per second does one nitrogen molecule make at STP? Use $\overset{v}{ˉ} = 445 m/s$ (from Ex 2-style, close enough for $N_{2}$ ) and $λ = 6.1 \times 1 0^{- 8} m$ (Ex 5).

Forecast: it flies 445 m each second and each flight is only ~61 nm long — so the number of flights (collisions) will be enormous, billions per second.

Collision frequency = distance travelled per second ÷ distance per flight. $Z = \frac{v ˉ}{λ} = \frac{445}{6.1 \times 1 0 ^{- 8}}$ Why this step? In one second a molecule covers $\overset{v}{ˉ}$ metres; each collision ends a flight of length $λ$ , so the count is simply their ratio. We use $\overset{v}{ˉ}$ (not $v_{r m s}$ ) because collision rate is governed by the ordinary average speed.
Compute. $Z \approx 7.3 \times 1 0^{9} s^{- 1}$ .

Verify: $7.3 \times 1 0^{9}$ collisions per second — about 7 billion. ✓ Enormous, as forecast. Time between collisions $\approx 1/ Z \approx 1.4 \times 1 0^{- 10} s$ . Units: $\frac{m/s}{m} = s^{- 1}$ . ✓

Cell H — Real-world word problem

Worked example

Ex 8. A vacuum chamber is 0.5 m across. Engineers want the mean free path to exceed the chamber size (so molecules hit walls, not each other — "molecular flow"). At $T = 300 K$ , $d = 3.7 \times 1 0^{- 10} m$ , what pressure is needed?

Forecast: normal air has $λ \sim 60$ nm. We need $λ \geq 0.5$ m — about $1 0^{7}$ × bigger. Since $λ \propto 1/ P$ , we need pressure about $1 0^{7}$ × smaller than atmospheric, i.e. deep vacuum.

Set the condition $λ = 0.5 m$ and solve for $P$ . $λ = \frac{k _{B} T}{2 π d ^{2} P} \Rightarrow P = \frac{k _{B} T}{2 π d ^{2} λ}$ Why this step? We invert the mean-free-path formula because the unknown is now $P$ , not $λ$ .
Plug in. $P = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{2 π ( 3.7 \times 1 0 ^{- 10} ) ^{2} ( 0.5 )}$ Numerator $= 4.14 \times 1 0^{- 21}$ . Denominator $= 2 π (1.369 \times 1 0^{- 19}) (0.5) = 3.04 \times 1 0^{- 19}$ . $P \approx 1.36 \times 1 0^{- 2} Pa$

Verify: Ratio to atmospheric: $1.36 \times 1 0^{- 2} /1.01 \times 1 0^{5} \approx 1.3 \times 1 0^{- 7}$ — about ten-million-fold lower, as forecast. ✓ This is why high-vacuum equipment (~ $1 0^{- 2}$ Pa and below) enters the molecular-flow regime.

Cell I — Exam twist (reverse the formula)

Worked example

Ex 9. At what temperature does hydrogen ( $M = 2 g/mol$ ) have $v_{r m s}$ equal to Earth's escape speed, $11.2 km/s$ ? (This is why Earth loses hydrogen to space.)

Forecast: escape speed is ~11 200 m/s — over 20× a typical room-temperature $v_{r m s}$ . Since $v_{r m s} \propto T$ , we'll need $T$ far above room temperature (thousands of K), but not absurdly so, because even a fraction of molecules in the high-speed tail escape at lower $T$ .

Invert $v_{r m s} = 3 R T / M$ for $T$ . $T = \frac{M v _{r m s}^{2}}{3 R}$ Why this step? The unknown is now $T$ ; squaring both sides removes the root, then rearrange.
Plug in ( $M = 0.002 kg/mol$ , $v_{r m s} = 11200 m/s$ ): $T = \frac{( 0.002 ) ( 11200 ) ^{2}}{3 ( 8.314 )} = \frac{( 0.002 ) ( 1.2544 \times 1 0 ^{8} )}{24.94}$ $T \approx 1.006 \times 1 0^{4} K \approx 10, 000 K$

Verify: Reverse-check: $3 (8.314) (10060) /0.002 = 1.254 \times 1 0^{8} \approx 11, 200 m/s$ . ✓ Thousands of K, as forecast. (In reality hydrogen escapes at much lower $T$ because the fast tail of the Maxwell-Boltzmann distribution leaks out continuously.)

Recall

Recall Which speed for which job?

Energy / KE calculations ::: use $v_{r m s}$ Collision rate, effusion, "how often" ::: use $\overset{v}{ˉ}$ Peak of the distribution ::: $v_{p}$ Distance between collisions ::: $λ = 1/ (2 nπ d^{2})$ Collisions per second ::: $Z = \overset{v}{ˉ} / λ$

Recall Scaling shortcuts (avoid re-plugging)

Same gas, $T$ changes ::: $v \propto T$ (ratio kills all constants) Same $T$ , different gas ::: $v \propto 1/ M$ Fixed $T$ , $P$ changes ::: $λ \propto 1/ P$ ; $v_{r m s}$ unchanged