Gas laws — Boyle (PV const at T), Charles (V - T const at P), Gay-Lussac (P - T const at V): Worked examples

Before anything, the one machine we use everywhere:

The scenario matrix

Every cell below is a distinct case-class this topic can throw at you. The examples that follow are tagged with the cell(s) they cover.

#	Case class	What is held fixed	Trap / feature to watch
A	Boyle — squeeze	$T$ , $n$	inverse trade-off; no Kelvin needed
B	Charles — heat & expand	$P$ , $n$	MUST convert $^{\circ}$ C → K
C	Gay-Lussac — sealed & heat	$V$ , $n$	keyword "rigid/sealed"
D	Combined — everything changes	only $n$	do NOT use Boyle alone
E	Degenerate — nothing changes	one variable identical	answer = unchanged (sanity anchor)
F	Limiting — $T \to 0$ K	$P$ or $V$	volume/pressure → 0, absolute zero
G	Sign / negative-Celsius trap	—	never divide by $^{\circ}$ C
H	Real-world word problem	to be decided	translate words → which law
I	Exam twist — two-step / find $n$	mixed	reason in stages

We now clear every cell.

Reading a $P$ – $V$ trade-off visually (Cells A, F)

Figure — Gas laws — Boyle (PV const at T), Charles (V - T const at P), Gay-Lussac (P - T const at V)

Look at the burnt-orange curve: it is Boyle's hyperbola, $P = const / V$ . Pick a point, halve its $V$ , and the curve forces $P$ to double — that is what "inverse" looks like. The teal dashed line shows Charles/Gay-Lussac's straight proportionality ( $V$ or $P$ rising in step with $T$ ), and the plum arrow shows where that line, extended, would hit zero — absolute zero. Keep this picture in mind for every example.

Cell A — Boyle, the squeeze

Cell B — Charles, heat & expand

Worked example Cell B: the free piston

A gas under a freely-moving piston (so pressure stays atmospheric) occupies $250 mL$ at $2 0^{\circ} C$ . It is warmed to $8 0^{\circ} C$ . Find the new volume.

Forecast: temperature rose only $6 0^{\circ}$ C — does the volume rise by a lot or a little? Guess.

Convert to Kelvin. Why this step? Charles is a proportionality to absolute temperature; $^{\circ}$ C is not proportional to molecular energy. $T_{1} = 20 + 273 = 293 K$ , $T_{2} = 80 + 273 = 353 K$ .
Pick the law. Why? Free piston ⇒ constant $P$ ⇒ Charles: $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$ .
Solve & plug in. $V_{2} = V_{1} \frac{T _{2}}{T _{1}} = 250 \times \frac{353}{293} = 301.2 mL$ .

Verify: absolute $T$ rose by factor $353/293 \approx 1.205$ , so $V$ rose by the same tiny factor — a $6 0^{\circ}$ jump is only a $\sim 20%$ volume change because $293$ K is a big baseline. ✔ Warning: if you'd wrongly used $80/20 = 4$ , you'd get $1000$ mL — the Kelvin trap.

Cell C — Gay-Lussac, sealed can

Cell D — Combined, everything moves

Worked example Cell D: weather balloon rising

A balloon has $2.0 L$ at ground level ( $1.0 atm$ , $2 7^{\circ} C$ ). High up, pressure drops to $0.40 atm$ and temperature to $- 2 3^{\circ} C$ . Find the new volume.

Forecast: lower pressure inflates it, but colder shrinks it — which wins? Guess.

Convert to Kelvin. $T_{1} = 27 + 273 = 300 K$ , $T_{2} = - 23 + 273 = 250 K$ .
Choose the combined law. Why this step? Both $P$ AND $T$ change, so no single simple law applies — using Boyle alone would be the classic error. Use $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ .
Solve for $V_{2}$ . $V_{2} = V_{1} \cdot \frac{P _{1}}{P _{2}} \cdot \frac{T _{2}}{T _{1}}$ .
Plug in. $V_{2} = 2.0 \times \frac{1.0}{0.40} \times \frac{250}{300} = 2.0 \times 2.5 \times 0.8333 = 4.167 L$ .

Verify: pressure fell to $0.4 \times$ (expands, $\times 2.5$ ), temperature fell to $0.833 \times$ (shrinks). Net $2.5 \times 0.833 = 2.083$ , times $2.0 L = 4.17 L$ . The pressure drop wins → balloon grows. ✔

Cell E — Degenerate: nothing actually changes

Cell F — Limiting behaviour toward absolute zero

Worked example Cell F: Charles extrapolated

A gas has $300 mL$ at $300 K$ under constant pressure. (a) What volume does the formula predict at $150 K$ ? (b) What does it predict at $0 K$ , and why is that only a limit?

Forecast: at half the Kelvin temperature, half the volume? And at $0$ K? Guess.

Half temperature. $V_{2} = 300 \times \frac{150}{300} = 150 mL$ . Why? Charles: $V \propto T$ , so half $T$ ⇒ half $V$ .
Zero temperature. $V = 300 \times \frac{0}{300} = 0 mL$ .
Interpret the limit. Why this step? The straight $V$ – $T$ line hits $V = 0$ at $T = 0 K = - 273.1 5^{\circ} C$ = absolute zero. Real gases liquefy long before this, so $V = 0$ is an idealised limit, not a physical outcome — see the plum arrow in the figure above.

Verify: (a) $150 mL$ ✔; (b) $0 mL$ as a mathematical limit ✔. The line can never give a negative volume, which is why temperature can't drop below $0$ K.

Cell G — The sign / negative-Celsius trap

Cell H — Real-world word problem

Cell I — Exam twist: two-step, find $n$

Worked example Cell I: reconcile with the ideal gas law

A rigid $10 L$ tank holds gas at $2.0 atm$ , $300 K$ . (a) It is heated to $600 K$ — find the new pressure. (b) Using $R = 0.0821 L\cdotatm\cdotmol^{- 1} K^{- 1}$ , how many moles are in the tank?

Forecast: doubling Kelvin at fixed volume — double the pressure? And is $n$ big or small?

Part (a): pick the law. Why? Rigid ⇒ constant $V$ ⇒ Gay-Lussac: $P_{2} = P_{1} \frac{T _{2}}{T _{1}} = 2.0 \times \frac{600}{300} = 4.0 atm$ .
Part (b): switch tools. Why this step? To get the actual amount $n$ we need the full ideal gas equation, not a two-state ratio — ratios cancel $n$ out. Use $P V = n R T$ at the original state.
Solve for $n$ . $n = \frac{P V}{R T} = \frac{( 2.0 ) ( 10 )}{( 0.0821 ) ( 300 )} = \frac{20}{24.63} = 0.812 mol$ .

Verify: (a) $T$ doubled ⇒ $P$ doubled: $4.0 atm$ ✔. (b) plug back: $n R T = 0.812 \times 0.0821 \times 300 = 20.0 = P V = 2.0 \times 10$ ✔. Units of $n$ : $\frac{atm \cdot L}{L\cdotatm\cdotmol ^{- 1} K ^{- 1} \cdot K} = mol$ ✔.

Recall check

Recall Which cell is which law?

Rigid / sealed / constant volume ::: Gay-Lussac, $P / T$ constant (Cells C, I, E). Free piston / balloon / constant pressure ::: Charles, $V / T$ constant (Cells B, F, G). Isothermal / same temperature ::: Boyle, $P V$ constant (Cells A, H). Both $P$ and $T$ change ::: Combined law $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ (Cell D). You need the actual amount of gas $n$ ::: full $P V = n R T$ , ratios won't give it (Cell I).

Connections

Parent topic
Combined Gas Law — the unifier used in Cell D.
Ideal Gas Equation PV=nRT — needed in Cell I to find $n$ .
Absolute Zero and Kelvin Scale — the limit in Cell F.
Kinetic Theory of Gases — why pressure rises with temperature.
Dalton's Law of Partial Pressures · Avogadro's Law — next knobs.