Ideal gas law PV = nRT — derivation from kinetic theory: Worked examples

The scenario matrix

Every problem this topic can throw is one of these cells. We tick each off with an example.

#	Cell (kind of problem)	What is asked	Worked in
A	Solve for $P$ (all others known)	direct plug-in	Ex 1
B	Solve for $T$ , convert °C↔K	unit trap	Ex 2
C	Solve for $n$ , then count molecules	bridge to $N = n N_{A}$	Ex 3
D	Two-state, one variable fixed	ratio law	Ex 4
E	Microscopic reach-through: $v_{r m s}$	kinetic core	Ex 5
F	Degenerate / limiting input ( $T \to 0$ , $V \to \infty$ )	what breaks	Ex 6
G	Real-world word problem (a tyre)	modelling	Ex 7
H	Exam twist: mixture of two gases	additivity	Ex 8
I	Two-state, two variables change	combined gas law	Ex 9

Cell A — Solve for $P$

Worked example Example 1 — pressure of a known sample

$n = 3 mol$ of ideal gas fills $V = 0.020 m^{3}$ at $T = 400 K$ . Find the pressure $P$ .

Forecast: guess the order of magnitude before computing — is $P$ closer to $1 0^{3}$ , $1 0^{5}$ , or $1 0^{7}$ Pa? (Room air is $\sim 1 0^{5}$ ; a few moles squeezed into 20 litres should be higher.)

Rearrange $P V = n R T$ for the unknown: $P = \frac{n R T}{V}$ . Why this step? We isolate the single unknown so every other symbol is a known number.
Substitute with SI units only: $P = \frac{3 \times 8.314 \times 400}{0.020}$ . Why this step? Mixing litres with $R$ in $J mol^{- 1} K^{- 1}$ would silently break the units; $V$ must be in $m^{3}$ .
Compute: numerator $= 9976.8 J$ , so $P = 498840 \approx 4.99 \times 1 0^{5} Pa$ . Why this step? We are turning the exact symbolic answer into a number so we can sanity-check its size against everyday pressures.

Verify: units $\frac{J}{m ^{3}} = \frac{N\cdotm}{m ^{3}} = \frac{N}{m ^{2}} = Pa$ ✓. Using $1 atm = 1.013 \times 1 0^{5} Pa$ , our $4.99 \times 1 0^{5} Pa$ is $4.99/1.013 \approx 4.9$ atm — higher than room air, matching the forecast. ✓

Cell B — Solve for $T$ with a Celsius trap

Worked example Example 2 — the temperature nobody in Kelvin

A sealed rigid tank holds $n = 1.5 mol$ in $V = 0.010 m^{3}$ . A gauge reads $P = 4.0 \times 1 0^{5} Pa$ . What is the temperature, in °C?

Forecast: the final answer must be in Celsius, but can we plug Celsius into $P V = n R T$ ? (No — think about why $0^{\circ}$ C would give zero energy, which is false.)

Solve for absolute $T$ : $T = \frac{P V}{n R} = \frac{4.0 \times 1 0 ^{5} \times 0.010}{1.5 \times 8.314}$ . Why this step? The law only holds for absolute (Kelvin) temperature, because $T$ is proportional to average kinetic energy — a quantity that can never be negative.
Compute: numerator $= 4000 J$ , denominator $= 12.471$ , so $T = 320.7 K$ . Why this step? We are evaluating the number to see what absolute temperature the tank sits at before any unit cosmetics.
Convert to Celsius at the very end: $T (^{\circ} C) = 320.7 - 273.15 = 47.6^{\circ} C$ . Why this step? Conversion is a final cosmetic step; it must never enter the equation itself.

Verify: back-substitute $T = 320.7 K$ : $P = \frac{1.5 \times 8.314 \times 320.7}{0.010} = 4.0 \times 1 0^{5} Pa$ ✓.

Cell C — Solve for $n$ , then count molecules

Worked example Example 3 — how many molecules?

A balloon has $V = 0.005 m^{3}$ at $P = 1.01 \times 1 0^{5} Pa$ and $T = 293 K$ . Find (a) the moles $n$ and (b) the number of molecules $N$ .

Forecast: 5 litres of air at room conditions — roughly how many moles? (A mole of gas at room conditions occupies $\sim 24$ litres, so expect a fraction of a mole.)

Solve for $n$ : $n = \frac{P V}{R T} = \frac{1.01 \times 1 0 ^{5} \times 0.005}{8.314 \times 293}$ . Why this step? $n$ is the only unknown; everything else is measured.
Compute: numerator $= 505 J$ , denominator $= 2436.0$ , so $n = 0.2073 mol$ . Why this step? We evaluate the number to compare it against the " $\sim 24$ litres per mole" mental estimate from the forecast.
Bridge to molecule count: $N = n N_{A} = 0.2073 \times 6.022 \times 1 0^{23} = 1.25 \times 1 0^{23}$ . Why this step? $N_{A}$ is the exchange rate between the human unit (moles) and the microscopic count (molecules) used in the kinetic derivation.

Verify: $N k_{B} T = 1.25 \times 1 0^{23} \times 1.381 \times 1 0^{- 23} \times 293 = 505 J = P V$ ✓ (both routes, $n R T$ and $N k_{B} T$ , agree).

Cell D — Two-state comparison, one variable fixed (ratio law)

Worked example Example 4 — heat it at constant volume

Gas in a rigid sealed can starts at $P_{1} = 2.0 \times 1 0^{5} Pa$ , $T_{1} = 300 K$ . It is heated to $T_{2} = 450 K$ . Find $P_{2}$ . ( $n$ , $V$ fixed.)

Forecast: temperature rose by a factor $450/300 = 1.5$ . If $P \propto T$ , what factor should $P$ grow by?

Write the law for both states: $P_{1} V = n R T_{1}$ and $P_{2} V = n R T_{2}$ . Why this step? Anything that stays constant ( $n$ , $R$ , $V$ ) will cancel when we divide.
Divide state 2 by state 1: $\frac{P _{2}}{P _{1}} = \frac{T _{2}}{T _{1}}$ (Gay-Lussac's law falls out). Why this step? Dividing removes the constants so we never need their numeric values.
Solve: $P_{2} = P_{1} \frac{T _{2}}{T _{1}} = 2.0 \times 1 0^{5} \times \frac{450}{300} = 3.0 \times 1 0^{5} Pa$ . Why this step? We turn the ratio into the final pressure and check it grew by the forecast factor of $1.5$ .

Verify: ratio $P_{2} / P_{1} = 1.5 = T_{2} / T_{1}$ ✓, matching the forecast. Microscopically: hotter → faster molecules → harder and more frequent kicks → more pressure. ✓ This is the constant- $V$ line traced on the $P$ – $T$ plane in the figure below.

Cell E — Reach through to molecular speed

Worked example Example 5 — how fast is an oxygen molecule?

Find the root-mean-square speed $v_{r m s} = ⟨ v^{2} ⟩$ of $O_{2}$ at $T = 300 K$ . Molar mass $M = 0.032 kg/mol$ .

Forecast: nitrogen at 300 K came out near 517 m/s. Oxygen is heavier ( $0.032$ vs $0.028$ kg/mol). Should $v_{r m s}$ be bigger or smaller?

Start from the bridge (per molecule): $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ . Why this step? This is the single equation that connects temperature to microscopic speed.
Scale to per-mole by multiplying both sides by $N_{A}$ : $\frac{1}{2} M ⟨ v^{2} ⟩ = \frac{3}{2} R T$ , using $m N_{A} = M$ and $k_{B} N_{A} = R$ . Why this step? We know $M$ (per mole) and $R$ , not $m$ and $k_{B}$ individually — this converts the equation into knowns.
Solve for the speed: $v_{r m s} = \frac{3 R T}{M} = \frac{3 \times 8.314 \times 300}{0.032}$ .
Compute: inside $= \frac{7482.6}{0.032} = 233831$ , so $v_{r m s} = 483.6 m/s$ . Why this step? We evaluate the square root to get an actual speed we can compare with the nitrogen benchmark from the forecast.

Verify: heavier gas ⇒ smaller speed, and $483 < 517$ ✓. Check the mass ratio: $\frac{v _{O_{2}}}{v _{N_{2}}} = \frac{0.028}{0.032} = 0.935$ , and $517 \times 0.935 = 483$ ✓. See Root Mean Square Speed.

Cell F — Degenerate & limiting inputs

Worked example Example 6 — what happens at the edges?

For a fixed amount of gas, examine three limits: (a) $T \to 0 K$ at fixed $V$ ; (b) $V \to \infty$ at fixed $T$ ; (c) $n \to 0$ . Say what $P$ does and whether the ideal description still makes sense.

Forecast: at absolute zero, is the pressure large, small, or exactly zero in the ideal model?

(a) $T \to 0$ : $P = \frac{n R T}{V} \to \frac{n R \cdot 0}{V} = 0$ . Why this step? Directly reading the law at the limit. Microscopically $⟨ v^{2} ⟩ \propto T \to 0$ : molecules stop, deliver no momentum, so no push. This is the definition of absolute zero — the coldest the ideal model allows, and pressure cannot go negative.
(b) $V \to \infty$ : $P = \frac{n R T}{V} \to 0$ as $V \to \infty$ . Why this step? Molecules hit the (now distant) walls ever more rarely — with $L \to \infty$ the collision rate $v_{x} /2 L \to 0$ . A gas that fills unlimited space exerts vanishing pressure.
(c) $n \to 0$ : $P \to 0$ . No molecules, no collisions, no pressure — the vacuum. Why this step? Confirms the model degrades gracefully: emptiness gives zero pressure, as it must.

Verify: all three limits give $P \to 0$ from a single common cause — fewer/slower/rarer collisions. Numerically, halving $T$ from 300 to 150 K at fixed $V$ , $n = 1$ , $V = 0.024 m^{3}$ : $P$ drops from $103925$ to $51962 Pa$ , exactly a factor $2$ ✓.

Cell G — Real-world word problem

Worked example Example 7 — the car tyre on a hot day

A car tyre is inflated to a gauge pressure of $2.0 \times 1 0^{5} Pa$ (above atmospheric $1.0 \times 1 0^{5} Pa$ ) at $T_{1} = 285 K$ (a cold morning). After highway driving the air heats to $T_{2} = 320 K$ . The tyre volume is essentially fixed and no air leaks. Find the new gauge pressure.

Forecast: will the gauge reading rise by the same factor as the absolute temperature, or by less? (Careful — the law works on absolute pressure, not gauge.)

Convert gauge to absolute pressure: $P_{1}^{ab s} = 2.0 \times 1 0^{5} + 1.0 \times 1 0^{5} = 3.0 \times 1 0^{5} Pa$ . Why this step? $P V = n R T$ uses total pressure of the gas, not the excess over atmosphere. Gauge is a reading convenience.
Fixed $V, n$ → ratio law: $P_{2}^{ab s} = P_{1}^{ab s} \frac{T _{2}}{T _{1}} = 3.0 \times 1 0^{5} \times \frac{320}{285}$ . Why this step? Same reasoning as Cell D — constants cancel.
Compute absolute: $P_{2}^{ab s} = 3.0 \times 1 0^{5} \times 1.1228 = 3.368 \times 1 0^{5} Pa$ . Why this step? We turn the ratio into an actual absolute pressure so we can subtract atmosphere next.
Convert back to gauge: $P_{2}^{g a ug e} = 3.368 \times 1 0^{5} - 1.0 \times 1 0^{5} = 2.37 \times 1 0^{5} Pa$ . Why this step? The dashboard/gauge subtracts atmosphere; we report in the same units the driver sees.

Verify: absolute rose by factor $1.123$ , but gauge rose from $2.0$ to $2.37$ — a factor $1.18$ , larger than $1.123$ . That answers the forecast: gauge rises by more, because the constant atmospheric offset is not scaled by temperature. ✓

Cell H — Exam twist: a mixture of two gases

Worked example Example 8 — two gases share one box

A rigid box of $V = 0.010 m^{3}$ at $T = 300 K$ contains $n_{1} = 0.40 mol$ of helium and $n_{2} = 0.60 mol$ of nitrogen. Find the total pressure.

Forecast: do the two gases' pressures add, or does the heavier gas dominate? (In the ideal model, molecules ignore each other's identity except during collisions with the walls.)

Each gas obeys the law independently (Dalton's law of partial pressures): $P_{i} = \frac{n _{i} R T}{V}$ . Why this step? Ideal molecules exert no forces on each other, so each species pushes on the wall as if alone — the derivation in the parent never used the molecules' chemical identity.
Total pressure is the sum: $P = P_{1} + P_{2} = \frac{( n _{1} + n _{2} ) R T}{V}$ . Why this step? Total force on the wall is the sum of independent contributions (Step 4 of the derivation), so pressures add.
Compute with $n_{t o t} = 1.00 mol$ : $P = \frac{1.00 \times 8.314 \times 300}{0.010} = 2.494 \times 1 0^{5} Pa$ . Why this step? We evaluate the total so we can split it back into partial pressures for the check.

Verify: the answer depends only on total moles, not on which gas — helium and nitrogen at the same $T$ carry the same average kinetic energy per molecule (equipartition, $\frac{3}{2} k_{B} T$ ), so molar amount is all that counts. See Equipartition Theorem. Partial pressures: $P_{H e} = 9.98 \times 1 0^{4}$ , $P_{N_{2}} = 1.50 \times 1 0^{5}$ , summing to $2.49 \times 1 0^{5} Pa$ ✓.

Cell I — Two-state, TWO variables change (combined gas law)

Worked example Example 9 — a gas that changes pressure

and volume and temperature A fixed amount of gas starts at $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.030 m^{3}$ , $T_{1} = 300 K$ . It is compressed and heated to $V_{2} = 0.020 m^{3}$ and $T_{2} = 400 K$ . Find the new pressure $P_{2}$ .

Forecast: two effects push $P$ up — squeezing (smaller $V$ ) and heating (larger $T$ ). So expect $P_{2} > P_{1}$ , but by how much? Guess before computing.

Write the law for both states: $P_{1} V_{1} = n R T_{1}$ and $P_{2} V_{2} = n R T_{2}$ . Why this step? Only $n$ and $R$ are shared and constant now — $P$ , $V$ , $T$ all move, so no single ratio law (Cell D) applies.
Divide state 2 by state 1 so $n$ and $R$ cancel: $\frac{P _{2} V _{2}}{P _{1} V _{1}} = \frac{T _{2}}{T _{1}}$ , i.e. the combined gas law $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ . Why this step? Dividing kills the only constants, leaving a relation between just the six measured quantities.
Solve for $P_{2}$ : $P_{2} = P_{1} \cdot \frac{V _{1}}{V _{2}} \cdot \frac{T _{2}}{T _{1}}$ . Why this step? We isolate the one unknown; note the two visible factors — a volume-squeeze factor $V_{1} / V_{2}$ and a heating factor $T_{2} / T_{1}$ , both $> 1$ .
Compute: $P_{2} = 1.0 \times 1 0^{5} \times \frac{0.030}{0.020} \times \frac{400}{300} = 1.0 \times 1 0^{5} \times 1.5 \times 1.333 = 2.0 \times 1 0^{5} Pa$ . Why this step? Multiplying the two enhancement factors ( $1.5$ and $1.333$ ) shows the combined effect is their product, exactly doubling the pressure.

Verify: check $\frac{P _{1} V _{1}}{T _{1}} = \frac{( 1.0 \times 1 0 ^{5} ) ( 0.030 )}{300} = 10$ and $\frac{P _{2} V _{2}}{T _{2}} = \frac{( 2.0 \times 1 0 ^{5} ) ( 0.020 )}{400} = 10$ — equal ✓. Both push pressure up, and $P_{2} = 2 P_{1}$ matches the forecast. ✓

Active Recall

Recall Which cell is each phrase testing?

"Convert to Kelvin before plugging in" ::: Cell B — the Celsius trap. "Divide state 2 by state 1 to cancel constants" ::: Cell D / G — one-variable-fixed ratio law. " $P_{1} V_{1} / T_{1} = P_{2} V_{2} / T_{2}$ " ::: Cell I — combined gas law, two variables change. " $v_{r m s} = 3 R T / M$ " ::: Cell E — microscopic reach-through. "Pressures of components add" ::: Cell H — Dalton's law for a mixture. "At $T \to 0$ , $P \to 0$ but real gas liquefies" ::: Cell F — degenerate/limiting case. " $1 atm = ?$ Pa" ::: $1.013 \times 1 0^{5}$ Pa — the unit bridge from the toolkit.