Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms: Worked examples

Every symbol used here was defined in the parent:

$v_{r m s} = \overline{v^{2}}$ — root-mean-square speed (the "typical" speed that pressure feels).
$ρ = N m / V$ — density (mass per volume).
$R = 8.314 J mol^{- 1} K^{- 1}$ , $k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}$ , $N_{A} = 6.022 \times 1 0^{23}$ .
$M$ — molar mass in kg/mol (always convert grams!).
$T$ — absolute temperature in kelvin.

The scenario matrix

Every question this topic throws is one of these cells. Each worked example below is tagged with the cell it covers.

Cell	What varies / the trap	Given → Find	Example
A	Straight plug-in	$T, M \to v_{r m s}$	Ex 1
B	Bulk route (no $T$ , no $M$ )	$P, ρ \to v_{r m s}$	Ex 2
C	Ratio of two gases	$M_{1}, M_{2}$ same $T \to$ ratio	Ex 3
D	Ratio, same gas	$T_{1}, T_{2} \to$ ratio	Ex 4
E	Reverse solve	$v_{r m s}, M \to T$	Ex 5
F	Degenerate input	$T = 0$ , and $N = 1$	Ex 6
G	Limiting behaviour	$T \to \infty$ , $M \to 0$	Ex 7
H	Real-world word problem	balloon leak (with Graham's Law of Diffusion)	Ex 8
I	Exam twist ( $v_{r m s}$ vs $\overline{v}$ )	mixed distribution	Ex 9

Cell B — the bulk (density) route

Worked example Ex 2 — helium from pressure and density (Cell B)

A tank of helium reads $P = 2.0 \times 1 0^{5} Pa$ and $ρ = 0.32 kg/m^{3}$ . Find $v_{r m s}$ . No temperature given.

Forecast: Can you even do it without $T$ or $M$ ? (Yes — that's the whole point of this cell.)

Start from the bulk form $P = \frac{1}{3} ρ v_{r m s}^{2}$ and solve: $v_{r m s} = \frac{3 P}{ρ}$ Why this step? $P = \frac{1}{3} ρ v_{r m s}^{2}$ links three measurable bulk quantities. We chose this form (not $3 R T / M$ ) precisely because $T$ and $M$ are missing — pick the tool whose inputs you actually have.
Plug in: $v_{r m s} = \frac{3 ( 2.0 \times 1 0 ^{5} )}{0.32} = 1.875 \times 1 0^{6} \approx 1369 m/s$ Why this step? Helium is very light, so a large speed is expected.

Verify: Cross-check via Ideal Gas Equation PV = nRT. If this is He ( $M = 0.004$ ), density $ρ = P M / R T$ gives $T = P M / (R ρ) = (2 \times 1 0^{5}) (0.004) / (8.314 \times 0.32) \approx 300.6 K$ ; then $3 R T / M = 3 (8.314) (300.6) /0.004 \approx 1369$ m/s ✓ — the two routes agree.

Cell C — ratio of two different gases

Worked example Ex 3 — CO₂ vs H₂ at the same temperature (Cell C)

At the same $T$ , find $v_{r m s} (H_{2}) / v_{r m s} (CO_{2})$ . $M (H_{2}) = 2$ , $M (CO_{2}) = 44$ .

Forecast: Same $T$ ⇒ same average kinetic energy for both. Which is faster — and by roughly what factor?

Write the ratio; $R$ and $T$ cancel because they're identical: $\frac{v _{r m s} ( H _{2} )}{v _{r m s} ( CO _{2} )} = \frac{3 R T / M _{H_{2}}}{3 R T / M _{C O_{2}}} = \frac{M _{C O_{2}}}{M _{H_{2}}}$ Why this step? In a ratio at equal $T$ , the only thing that survives is mass — this is the deep meaning of "equal KE." Heavier molecules crawl.
Compute: $44/2 = 22 \approx 4.69$ . Why this step? The mass ratio is inverted under the root because $v \propto 1/ M$ .

Verify: Hydrogen is much lighter, so must be much faster — factor $\approx 4.69 > 1$ ✓. This is exactly the reasoning behind Graham's Law of Diffusion.

Cell D — same gas, two temperatures

Cell E — reverse solve for temperature

Cell F — degenerate inputs

Worked example Ex 6 — the edge cases:

$T = 0$ and $N = 1$ (Cell F) (a) What is $v_{r m s}$ at absolute zero, $T = 0 K$ ? (b) Does the formula even make sense for a single molecule ( $N = 1$ )?

Forecast: Does the pressure vanish at $T = 0$ ? Is "root-mean-square of one number" a sensible thing?

(a) $v_{r m s} = 3 R (0) / M = 0$ . Why this step? At absolute zero all thermal motion stops (in this classical model), so no molecule moves, no collisions, $P = \frac{1}{3} ρ (0)^{2} = 0$ . Pressure genuinely goes to zero — the degenerate case is not a bug, it's the definition of the temperature floor.
(b) For one molecule, $\overline{v^{2}} = v_{1}^{2}$ , so $v_{r m s} = ∣ v_{1} ∣$ — the rms of a single value is just its magnitude. Why this step? The average over one item is that item. Physically pressure would then be a train of separate spikes, not steady; the smooth formula is a statistical limit valid only for huge $N$ (postulate 1). It's mathematically defined at $N = 1$ but physically meaningless.

Verify: (a) $\frac{1}{3} ρ \cdot 0^{2} = 0$ ✓. (b) rms of ${v}$ is $v^{2} = ∣ v ∣$ ✓. This is why the parent's [!mistake] insists on "huge number of particles."

Cell G — limiting behaviour

Worked example Ex 7 — pushing to the extremes (Cell G)

Describe $v_{r m s}$ as (a) $T \to \infty$ and (b) $M \to 0$ .

Forecast: Do these blow up, saturate, or stay finite?

(a) $T \to \infty$ : $v_{r m s} = 3 R T / M \to \infty$ , but only as $T$ — it grows without bound yet ever more slowly. Why this step? The square root tames the growth. Doubling speed needs quadrupling $T$ (Cell D generalised). Classically nothing caps it; relativity eventually would, but that's outside KMT.
(b) $M \to 0$ : $v_{r m s} = 3 R T / M \to \infty$ as $M$ shrinks. Why this step? Lighter molecules carry the same fixed energy $\frac{3}{2} k_{B} T$ with less mass, so they must move faster; a massless particle would need infinite speed — signalling KMT breaks down there (real ultralight species like electrons need quantum treatment).

Verify (numeric, from Cell A): doubling $T$ from 300 to 600 K for $N_{2}$ : $v_{r m s} = 3 (8.314) (600) /0.028 \approx 731$ m/s, and $731/517 \approx 1.414 \approx 2$ ✓ — confirms the slow $T$ climb.

Cell H — real-world word problem

Worked example Ex 8 — the two-hole leaking balloon (Cell H)

A balloon holds a 50/50 mole mixture of $H_{2}$ ( $M = 2$ ) and $O_{2}$ ( $M = 32$ ) at $300 K$ . Through a tiny hole, escape rate $\propto v_{r m s}$ . (a) Find $v_{r m s}$ of each. (b) Which leaks faster and by what factor?

Forecast: Both share the same $T$ . Guess the speed of each and the leak-rate ratio.

$v_{r m s} (H_{2}) = 3 (8.314) (300) /0.002 = 3.741 \times 1 0^{6} \approx 1934 m/s$ . Why this step? Cell-A plug-in for the light gas; convert 2 g/mol → 0.002 kg/mol.
$v_{r m s} (O_{2}) = 3 (8.314) (300) /0.032 \approx 484 m/s$ (matches parent Ex 1). Why this step? Same $T$ , heavier mass, so far slower — a concrete instance of Cell C.
Ratio $= 1934/484 \approx 4.0 = 32/2$ . Why this step? In a mixture each gas keeps its own speed (independent motions), so leak rates differ; the light gas escapes ~4× faster — the balloon becomes O₂-rich over time. This is Graham's Law of Diffusion in action.

Verify: $16 = 4$ ✓; and $\frac{1}{2} m \overline{v^{2}}$ is equal for both (equal $T$ ): $\frac{1}{2} (2) (1934)^{2} \approx \frac{1}{2} (32) (484)^{2}$ → $3.74 \times 1 0^{6} \approx 3.75 \times 1 0^{6}$ (per mole, × $1 0^{- 3}$ for kg) ✓.

Cell I — the exam twist: $v_{r m s}$ vs average speed

Worked example Ex 9 — three molecules, one trap (Cell I)

Three molecules have speeds $200, 400, 600 m/s$ . Find (a) the average speed $\overline{v}$ and (b) $v_{r m s}$ . Which does pressure depend on?

Forecast: Will $v_{r m s}$ be equal to, above, or below $\overline{v}$ ?

Average speed: $\overline{v} = \frac{200 + 400 + 600}{3} = 400 m/s$ . Why this step? $\overline{v}$ is the plain arithmetic mean — it treats all molecules equally.
Root-mean-square: square, average, root. $v_{r m s} = \frac{20 0 ^{2} + 40 0 ^{2} + 60 0 ^{2}}{3} = \frac{40000 + 160000 + 360000}{3} = 186667 \approx 432 m/s$ Why this step? Pressure comes from $\overline{v^{2}}$ (each kick $\propto v_{x}^{2}$ ), so we must average the squares. Squaring gives the fast molecule (600) extra weight — that's why $v_{r m s}$ ends up above $\overline{v}$ .
Pressure depends on $v_{r m s}$ , since $P = \frac{1}{3} ρ v_{r m s}^{2}$ . Why this step? The whole derivation carried $v^{2}$ ; using $\overline{v}$ instead would undercount the hard-hitting fast molecules.

Verify: $432 > 400$ ✓, consistent with the general rule $v_{r m s} \geq \overline{v}$ (equality only if all speeds identical). This matches the ordering $v_{r m s} > \overline{v} > v_{m p}$ from Maxwell-Boltzmann Speed Distribution.

Figure — Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms

Figure above: the scenario matrix as a map — each example is a route from "given" boxes to "find" boxes, showing which tool ( $3 R T / M$ vs $3 P / ρ$ vs a ratio) each cell uses.

Active recall

Recall Which formula for which given? (reveal)

Given $T$ and $M$ ::: use $v_{r m s} = 3 R T / M$ (Cell A/E/H).
Given $P$ and $ρ$ , no $T$ ::: use $v_{r m s} = 3 P / ρ$ (Cell B).
Two gases, same $T$ ::: ratio $= M_{2} / M_{1}$ (Cell C).
Same gas, two temperatures ::: ratio $= T_{2} / T_{1}$ (Cell D).
At $T = 0$ , $v_{r m s} =$ ::: $0$ (Cell F).
Pressure depends on == $v_{r m s}$ == not on $\overline{v}$ (Cell I).

Connections

Parent derivation (Hinglish)
Ideal Gas Equation PV = nRT — used to cross-check Ex 2.
Average Kinetic Energy and Temperature — the $T$ scaling of Ex 4/7.
Graham's Law of Diffusion — Ex 3 and Ex 8 are Graham's law.
Maxwell-Boltzmann Speed Distribution — the $v_{r m s} > \overline{v} > v_{m p}$ ordering (Ex 9).
Newton's Laws — Impulse and Momentum — the $2 m v_{x}$ kick underlying it all.
Real Gases and van der Waals Equation — where these ideal results start to fail.

Worked examples — Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms