ChemistryICSE Class 9 · Study of Gas Laws

Boyle's Law: P₁V₁ = P₂V₂, Graphs, Units and Solved Numericals (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Boyle's law: at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, so PV = constant and P₁V₁ = P₂V₂. The P–V graph is a curve, the P–1/V graph a straight line through the origin and the PV–P graph a horizontal line. Pressure units must be the same on both sides (1 atm = 760 mm Hg = 76 cm Hg).

Pushing the piston to halve the volume squeezes the same molecules into less space, so pressure doubles and the P–V graph is a falling curve.

Illustration drawn for QuestionBuddy.

Understand Boyle's Law: Pressure–Volume Relation, Graphs and Calculations

Kinetic picture. A gas is made of tiny particles, far apart, moving randomly at high speed and attracting each other very weakly. Pressure is caused by the molecules hitting the walls; more or harder hits mean more pressure. Heating makes the molecules faster.

Units. $1 atm = 76 cm Hg = 760 mm Hg = 1.013 \times 1 0^{5}$ Pa; $1 dm^{3} = 1 L = 1000 cm^{3}$ .

Boyle’s law. At constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure:

$V \propto \frac{1}{P} \Rightarrow V = \frac{k}{P} \Rightarrow P V = k,$

so for two states of the same gas at the same temperature

$P_{1} V_{1} = P_{2} V_{2} .$

Explanation. Halving the volume puts the same molecules in half the space, so each unit area of wall is struck twice as often: the pressure doubles.

Graphs (T constant).

P against V: a smooth curve (rectangular hyperbola), falling as V rises.
P against $\frac{1}{V}$ (or V against $\frac{1}{P}$ ): a straight line through the origin.
PV against P (or against V): a horizontal straight line, because PV is constant.

Method. (1) List $P_{1}, V_{1}, P_{2}, V_{2}$ with units; (2) convert pressures to one unit (all in atm, or all in mm Hg) and volumes to one unit; (3) use $P_{2} = \frac{P _{1} V _{1}}{V _{2}}$ or $V_{2} = \frac{P _{1} V _{1}}{P _{2}}$ ; (4) check direction: more pressure, less volume.

Testing data. Multiply P and V in each row; if the products are equal within error, Boyle’s law holds.

Mixing and sharing. When gas at pressure $P_{1}$ in volume $V_{1}$ is allowed to share a bigger volume $V_{1} + V_{2}$ (same temperature), the new pressure follows from $P_{1} V_{1} = P_{2} (V_{1} + V_{2})$ if the other volume was empty. When extra gas is pumped into a fixed volume, add the volumes at the same pressure first.

Boyle's Law: Pressure–Volume Relation, Graphs and Calculations: formulas and key facts

Boyle's law: $P V = k; P_{1} V_{1} = P_{2} V_{2}$; Constant T, fixed mass.
Pressure equivalents: $1 atm = 76 cm Hg = 760 mm Hg = 1.013 \times 1 0^{5} Pa$; Use one unit on both sides.
Volume equivalents: $1 dm^{3} = 1 L = 1000 cm^{3}$; Convert before substituting.
Graph shapes: $P - V : curve ∣ P - (1/ V) : straight line through origin ∣ P V - P : horizontal line$; T constant.

How to solve Boyle's Law: Pressure–Volume Relation, Graphs and Calculations questions

Check that the temperature really is constant and the mass fixed.

1Write $P_{1}$ , $V_{1}$ , $P_{2}$ , $V_{2}$ ; mark the unknown.
2Convert so that P units match and V units match on both sides.
3Use $P_{1} V_{1} = P_{2} V_{2}$ and solve for the unknown.
4Check the direction of change (P up means V down).
5For data tables, compute PV for each row.

Boyle's Law: Pressure–Volume Relation, Graphs and Calculations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Pumping a tyre

Suppose a tyre has an inner volume of 800 cm³ filled with air at 1 atm, and a hand pump adds 100 cm³ of air at 1 atm with each stroke. Assume the temperature stays the same and the tyre volume does not change.

What is the pressure after 6 strokes?

Try it first, then show the solutionSolution

Total air at 1 atm $= 800 + 6 \times 100 = 1400$ cm³.
Compressed into 800 cm³: $P_{2} = \frac{1 \times 1400}{800} = 1.75$ atm.

Answer: 1.75 atm.

Memory hook: Count all the air, then compress.

Tricks to remember Boyle's Law: Pressure–Volume Relation, Graphs and Calculations

Short memory aids. Say each one aloud once, then use it on a question the same day.

PV stays put
At constant T the product PV is constant.
Use it when: Setting up the equation.
Straight line needs 1/V
Plot P against 1/V to get a straight line through the origin.
Use it when: Graph questions.
Same units both sides
atm with atm, mm Hg with mm Hg.
Use it when: Avoiding unit mistakes.

Boyle's Law: Pressure–Volume Relation, Graphs and Calculations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Doubling the pressure · 1 mark · MCQBoard level

A fixed mass of gas occupies 8 L at 2 atm. At constant temperature, its volume at 4 atm is:
(a) 2 L
(b) 4 L
(c) 8 L
(d) 16 L

Try it first, then show the solutionSolution

$V_{2} = \frac{2 \times 8}{4} = 4$ L.

Answer: (b) 4 L

Example 2: Compressing a gas · 2 marks · Short answerBoard level

500 cm³ of a gas at 750 mm Hg is compressed to 300 cm³ at constant temperature. Find the final pressure.

Try it first, then show the solutionSolution

$P_{1} V_{1} = P_{2} V_{2}$ (½)
$P_{2} = \frac{750 \times 500}{300}$ (1)
$= 1250$ mm Hg. (½)

Answer: 1250 mm Hg

Example 3: Mixed units · 3 marks · Short answerHarder board level

2.5 dm³ of a gas at 1.2 atm expands at constant temperature to 4 dm³. Find the new pressure in atm and in mm Hg.

Try it first, then show the solutionSolution

$P_{2} = \frac{1.2 \times 2.5}{4} = 0.75$ atm. (1½)
In mm Hg: $0.75 \times 760 = 570$ mm Hg. (1½)

Answer: 0.75 atm = 570 mm Hg

Example 4: Testing Boyle’s law · 3 marks · Short answerHarder board level

A gas at constant temperature gives P (kPa): 80, 120, 160, 200 and V (cm³): 90, 60, 45, 36. (a) Find PV for each row. (b) Does Boyle’s law hold? (c) Find V at 150 kPa. (d) Describe the graph of P against 1/V.

Try it first, then show the solutionSolution

(a) $P V = 7200$ in every row. (1)
(b) Yes, PV is constant. (½)
(c) $V = \frac{7200}{150} = 48$ cm³. (1)
(d) A straight line through the origin. (½)

Answer: PV = 7200; yes; 48 cm³; straight line

Example 5: A diver’s cylinder · 4 marks · Case-basedHOTS / competency

A diving cylinder of capacity 12 L holds air at 150 atm. Assume constant temperature and that air can be drawn from the cylinder only until its pressure falls to the pressure around the diver. A diver breathes 24 L of air per minute (measured at the pressure around the diver).
(a) What volume of air at 1 atm does the cylinder hold?
(b) How long does it last at the surface (1 atm)?
(c) How long does it last at a depth where the pressure is 4 atm?
(d) Explain the difference using molecules.

Try it first, then show the solutionSolution

(a) $V = \frac{150 \times 12}{1} = 1800$ L. (1)
(b) At the surface 12 L of air at 1 atm stays in the cylinder, so usable air = 1800 − 12 = 1788 L; $\frac{1788}{24} = 74.5$ minutes. (1)
(c) At 4 atm the cylinder keeps 12 L at 4 atm, which is $12 \times 4 = 48$ L at 1 atm, so usable air = 1800 − 48 = 1752 L (at 1 atm). 24 L at 4 atm equals $24 \times 4 = 96$ L at 1 atm each minute: $\frac{1752}{96} = 18.25$ minutes. (1)
(d) At 4 atm the air is compressed, so each litre breathed holds four times as many molecules; the cylinder’s fixed number of molecules lasts less time. (1)

Answer: (a) 1800 L (b) 74.5 min (c) 18.25 min (d) more molecules per breath

Where students lose marks in Boyle's Law: Pressure–Volume Relation, Graphs and Calculations

The slips examiners see most often, and what to write instead.

Mistake: Pressure units mixed
Both pressures must use the same unit.
Wrong working
P₂ = 1.2 × 2.5/4 = 0.75 and then compared with 760 mm Hg.
Right working
0.75 atm = 570 mm Hg.
Do this: Convert first.
Usually costs: The answer.
Mistake: Direction wrong
More pressure means less volume.
Wrong working
V₂ = 600 × 2.5 = 1500 cm³
Right working
V₂ = 600/2.5 = 240 cm³
Do this: Check the sense of the change.
Usually costs: The answer.
Mistake: P–V graph drawn as a straight line
It is a curve; P against 1/V is straight.
Wrong working
Straight sloping line.
Right working
Curve (hyperbola).
Do this: Remember PV = k.
Usually costs: Graph marks.

Exam tips

State "temperature and mass constant" when you quote the law.
Write both states in a small table before substituting.
Convert units and state the final unit.
Draw the three graphs with labelled axes.

Practice questions on Boyle's Law: Pressure–Volume Relation, Graphs and Calculations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For a fixed mass of gas at constant temperature, the product of pressure and volume is:
(a) increasing
(b) decreasing
(c) constant
(d) zero
Hint
Boyle’s law.
Answer
(c) constant
Q2 · 1 mark · Very short answerBoard level
How many mm Hg are there in 1 atm?
Hint
Standard pressure.
Answer
760 mm Hg.
Q3 · 2 marks · Short answerHarder board level
600 cm³ of a gas at 1.0 atm is compressed to 2.5 atm at constant temperature. Find the new volume.
Hint
V₂ = P₁V₁/P₂.
Answer
$V_{2} = \frac{1.0 \times 600}{2.5} = 240$ cm³.
Q4 · 3 marks · Short answerHarder board level
20 cm³ of a gas at 90 cm Hg is compressed at constant temperature until its pressure is 120 cm Hg. Find the volume and express the final pressure in atm.
Hint
Same unit for P, then convert.
Answer
$V_{2} = \frac{90 \times 20}{120} = 15$ cm³; $120 cm Hg = \frac{120}{76} \approx 1.58$ atm.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): When the volume of a gas is halved at constant temperature, its pressure doubles.
Reason (R): The molecules strike the walls twice as often when the same number are in half the space.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Kinetic picture.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A rigid tyre of volume 800 cm³ contains air at 1 atm. A pump delivers 100 cm³ of air at 1 atm per stroke. After how many strokes is the pressure 2 atm, assuming constant temperature?
Hint
Total air at 1 atm = 1600 cm³ needed.
Answer
For 2 atm in 800 cm³ the air would occupy $2 \times 800 = 1600$ cm³ at 1 atm. It already holds 800 cm³ so $800$ cm³ more is needed: $\frac{800}{100} = 8$ strokes.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Boyle's law?Show answer
At constant T, V ∝ 1/P for a fixed mass of gas.
2.Which graph is a straight line through the origin?Show answer
P against 1/V.
3.1 atm in cm Hg?Show answer
76 cm Hg.
4.What causes gas pressure?Show answer
Molecules hitting the walls.
5.Is PV constant?Show answer
Yes, at constant temperature for a fixed mass.

Boyle's Law: Pressure–Volume Relation, Graphs and Calculations: frequently asked questions

What is Boyle's law?

At constant temperature the volume of a fixed mass of a gas is inversely proportional to its pressure: P₁V₁ = P₂V₂.

What does the graph of Boyle's law look like?

P against V is a curve; P against 1/V is a straight line through the origin; PV against P is a horizontal line.

How does kinetic theory explain Boyle's law?

Reducing the volume packs the molecules closer, so they hit the walls more often and the pressure rises.