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

Before we start, one shared reminder that every problem below leans on:

Level 1 — Recognition

L1.1 — Name the law

A balloon (free to expand, so pressure stays at atmospheric) is carried from a cold room into a warm car. Which single gas law governs its change in size?

Recall Solution

WHAT is held fixed? A free balloon means the inside pressure always matches the outside → pressure constant. The amount of gas $n$ is fixed (nothing leaks). WHICH law? Constant $P$ , changing $T$ and $V$ → Charles's Law, $V / T =$ const. Answer: Charles's Law. The warm car heats the gas, so $V$ grows.

L1.2 — Read the keyword

A rigid steel cylinder of gas is heated. Which law predicts its pressure change?

Recall Solution

Keyword "rigid" → the walls cannot move → volume constant. Constant $V$ , changing $T$ and $P$ → Gay-Lussac's Law, $P / T =$ const. Answer: Gay-Lussac's Law.

L1.3 — Spot the isothermal

A diver's air bubble rises and is squeezed at the same water temperature. Which law?

Recall Solution

Keyword "same temperature" (isothermal) → $T$ constant. Constant $T$ , changing $P$ and $V$ → Boyle's Law, $P V =$ const. Answer: Boyle's Law.

Level 2 — Application

L2.1 — Straight Boyle

A gas occupies $4.0 L$ at $2.0 atm$ . At constant $T$ , it is expanded to $8.0 L$ . Find the new pressure.

Recall Solution

Why Boyle? $T$ constant → $P V =$ const. $P_{2} = \frac{P _{1} V _{1}}{V _{2}} = \frac{( 2.0 ) ( 4.0 )}{8.0} = 1.0 atm .$ Sanity: volume doubled → pressure halved. ✔

L2.2 — Straight Charles

A gas has volume $250 mL$ at $300 K$ , constant $P$ . Find its volume at $450 K$ .

Recall Solution

Why Charles? $P$ constant → $V / T =$ const, $T$ already in Kelvin. $V_{2} = V_{1} \frac{T _{2}}{T _{1}} = 250 \times \frac{450}{300} = 375 mL .$ Sanity: $T \times 1.5 \Rightarrow V \times 1.5$ . ✔

L2.3 — Straight Gay-Lussac

A sealed rigid tank reads $1.5 atm$ at $250 K$ . It is cooled to $200 K$ . Find the pressure.

Recall Solution

Why Gay-Lussac? Rigid → $V$ constant → $P / T =$ const. $P_{2} = P_{1} \frac{T _{2}}{T _{1}} = 1.5 \times \frac{200}{250} = 1.2 atm .$ Sanity: cooling a locked tank lowers pressure. ✔

Level 3 — Analysis

L3.1 — The Celsius trap

A gas at $2 7^{\circ} C$ occupies $600 mL$ at constant $P$ . To what Celsius temperature must it be cooled so its volume becomes $400 mL$ ?

Recall Solution

Step 1 — convert to Kelvin (why? Charles is proportional to absolute $T$ ; $^{\circ}$ C ratios are meaningless). $T_{1} = 27 + 273 = 300 K$ . Step 2 — Charles: $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$ $T_{2} = T_{1} \frac{V _{2}}{V _{1}} = 300 \times \frac{400}{600} = 200 K .$ Step 3 — back to Celsius: $t_{2} = 200 - 273 = - 7 3^{\circ} C$ . Answer: $- 7 3^{\circ} C$ .

L3.2 — Two-step, changing units

A gas at $760 mmHg$ and $2.0 L$ is compressed isothermally to $1520 mmHg$ . Give the final volume in litres and confirm it in atmospheres ( $760 mmHg = 1 atm$ ).

Recall Solution

Why Boyle? Isothermal → $P V =$ const. Units cancel as long as both pressures share the same unit. In mmHg: $V_{2} = \frac{( 760 ) ( 2.0 )}{1520} = 1.0 L$ . Cross-check in atm: $P_{1} = 1 atm$ , $P_{2} = 2 atm$ → $V_{2} = (1) (2.0) /2 = 1.0 L$ . ✔ Answer: $1.0 L$ .

L3.3 — Absolute-zero reasoning

Look at the figure below. A constant-pressure gas has $V = 300 mL$ at $200 K$ and $V = 450 mL$ at $300 K$ . Extrapolate the straight $V$ – $T$ line to find the temperature where $V = 0$ .

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

Recall Solution

WHY a straight line? Charles says $V = (n R / P) T$ — a line through the origin in Kelvin, so $V = 0$ exactly at $T = 0 K$ . Check the two points give the same slope through origin: $\frac{V}{T} = \frac{300}{200} = 1.5, \frac{450}{300} = 1.5 mL/K . ✓$ A single constant $V / T$ ⇒ the line passes through the origin ⇒ $V = 0$ at $T = 0 K$ , which is $- 27 3^{\circ} C$ — absolute zero. Answer: $0 K (- 27 3^{\circ} C)$ .

Level 4 — Synthesis

L4.1 — Combined gas law

A gas occupies $1.0 L$ at $300 K$ and $1.0 atm$ . It is heated to $600 K$ and compressed to $0.5 L$ . Find the final pressure.

Recall Solution

Why combined? $P$ , $V$ , and $T$ all change → no single law; use $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ . $P_{2} = P_{1} \frac{V _{1}}{V _{2}} \frac{T _{2}}{T _{1}} = 1.0 \times \frac{1.0}{0.5} \times \frac{600}{300} = 1.0 \times 2 \times 2 = 4.0 atm .$ Sanity: halving $V$ pushes $P$ up ( $\times 2$ , Boyle-like); doubling $T$ pushes $P$ up again ( $\times 2$ , Gay-Lussac-like). Both raise $P$ → $\times 4$ . ✔

L4.2 — Collapse the combined law

Starting from $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ , show algebraically that holding $V$ constant recovers Gay-Lussac's law.

Recall Solution

WHAT we do: set $V_{1} = V_{2} = V$ (rigid container). $\frac{P _{1} V}{T _{1}} = \frac{P _{2} V}{T _{2}} .$ WHY it works: the identical factor $V$ appears on both sides, so it cancels: $\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}} .$ That is exactly Gay-Lussac's law. (Set $P$ equal instead → Charles; set $T$ equal → Boyle. See Combined Gas Law.)

L4.3 — Two changes, find the missing temperature

A $2.0 L$ gas at $1.0 atm$ , $250 K$ is changed to $3.0 L$ at $2.0 atm$ . Find the new temperature.

Recall Solution

Why combined? All three vary. Solve $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ for $T_{2}$ : $T_{2} = T_{1} \frac{P _{2} V _{2}}{P _{1} V _{1}} = 250 \times \frac{( 2.0 ) ( 3.0 )}{( 1.0 ) ( 2.0 )} = 250 \times 3 = 750 K .$ Sanity: both $P$ and $V$ rose, and $P V \propto T$ , so $T$ must rise a lot. ✔

Level 5 — Mastery

L5.1 — Reason from the molecular picture

Without any numbers: a sealed rigid can is left in a fire. Using the kinetic picture (molecules bouncing off walls), explain in cause-and-effect steps why the pressure rises, and name the law.

Recall Solution

Heat raises the average kinetic energy of the molecules → they move faster (temperature is average kinetic energy).
Volume is locked (rigid), so molecules cannot spread out; the density of hits stays but each molecule now travels between walls faster → it strikes more often.
Each faster molecule also carries more momentum → each strike delivers a harder push.
More hits + harder hits = greater total push per area = higher pressure.
Constant $V$ , $P ↑$ with $T ↑$ ⇒ this is Gay-Lussac's Law, $P / T$ = const.

L5.2 — Quantitative safety limit (Synthesis of L5)

An aerosol can is safe up to $3.0 atm$ . It is filled at $1.0 atm$ , $300 K$ , and the can is rigid. At what temperature (in $^{\circ}$ C) does it reach the danger pressure?

Recall Solution

Why Gay-Lussac? Rigid can → $V$ constant → $P / T =$ const. $T_{2} = T_{1} \frac{P _{2}}{P _{1}} = 300 \times \frac{3.0}{1.0} = 900 K .$ Convert: $t_{2} = 900 - 273 = 62 7^{\circ} C$ . Answer: $\approx 62 7^{\circ} C$ — a fire easily exceeds this, which is why the can can burst. ✔

L5.3 — Trap-detection challenge

A student writes: "A gas doubles its Celsius temperature from $2 0^{\circ} C$ to $4 0^{\circ} C$ at constant pressure, so by Charles's law its volume doubles." Diagnose the error and give the correct volume ratio.

Recall Solution

The error: Charles's law is proportional to absolute temperature, not Celsius. Doubling $^{\circ}$ C is not doubling $T$ . Correct: $T_{1} = 20 + 273 = 293 K$ , $T_{2} = 40 + 273 = 313 K$ . $\frac{V _{2}}{V _{1}} = \frac{T _{2}}{T _{1}} = \frac{313}{293} \approx 1.068.$ Answer: the volume grows by only about $6.8%$ ( $\approx 1.07 \times$ ), not double.

Recall One-line self-test recap

Which law when? ::: One knob locked → single law (T→Boyle, P→Charles, V→Gay-Lussac); two or more move → Combined law $P_{1} V_{1} / T_{1} = P_{2} V_{2} / T_{2}$ ; always Kelvin for any $T$ ratio.

Connections

Ideal Gas Equation PV=nRT — every solution above is one frozen-variable slice of this.
Combined Gas Law — the L4 workhorse.
Kinetic Theory of Gases — the L5 molecular reasoning.
Absolute Zero and Kelvin Scale — why L3.3 and all Kelvin conversions matter.
Dalton's Law of Partial Pressures · Avogadro's Law — next knobs (mixtures, amount).