Boyle’s Law states that for a fixed mass of gas at constant temperature, absolute pressure is inversely proportional to volume: pressure times volume stays constant, so P1V1 = P2V2. Halve the volume and the pressure doubles. It is exact for an ideal gas and accurate for real gases at ordinary pressures.
Take a breath. Right now, a sheet of muscle under your lungs has pulled downward, your chest cavity has grown by roughly half a litre, and the pressure inside it has dropped about one per cent below the pressure of the room. Air rushed in to close the gap.
You did not pull the air in. You made space, and the atmosphere did the rest. That trade — more room, less pressure — is Boyle’s Law, and you have been running it about twenty thousand times a day since the moment you were born.
What Is Boyle’s Law?
Boyle’s Law says that if you seal a fixed amount of gas in a container and keep its temperature steady, squeezing the gas into a smaller space raises its pressure by exactly the same factor that the volume shrinks. Squeeze it to one-third the volume, and the pressure triples.
Stated formally: at constant temperature, the absolute pressure of a fixed mass of gas is inversely proportional to its volume. “Inversely proportional” is the whole idea in two words. As one goes up, the other goes down, and their product refuses to move.
Three conditions are doing quiet work in that sentence, and every one of them matters:
- Fixed amount of gas — nothing leaks in or out. The number of molecules is locked.
- Constant temperature — the process is isothermal. The gas has time to shed or absorb heat and settle back to its surroundings’ temperature.
- Absolute pressure — measured from a perfect vacuum, not from atmospheric pressure. This is where most mistakes begin.
Who Was Robert Boyle?
In 1662 the Anglo-Irish natural philosopher Robert Boyle published the result that carries his name. He poured mercury into a sealed J-shaped glass tube, trapping air in the short arm, and recorded how the trapped column shortened as he added more mercury.
Boyle was building on a suggestion from Richard Towneley and Henry Power, and the apparatus was largely the work of his assistant Robert Hooke — the same Hooke of spring fame. The French physicist Edme Mariotte reported the relationship independently in the 1670s, which is why continental textbooks often call it the Boyle–Mariotte law. NASA’s Glenn Research Center still teaches it to aeronautics students in exactly the form Boyle left it.
The Boyle’s Law Formula
The law is usually written as a statement about a constant:
Because the product cannot change, the initial state and the final state must give the same answer. That gives the form you will actually use:
Rearranged for whichever quantity is missing:
Every symbol, with its SI unit:
- P1 — initial absolute pressure of the gas. SI unit: pascal (Pa). 1 Pa = 1 N/m2.
- V1 — initial volume of the gas. SI unit: cubic metre (m3).
- P2 — final absolute pressure. SI unit: pascal (Pa).
- V2 — final volume. SI unit: cubic metre (m3).
Here is the practical mercy of this equation: the units cancel. Because pressure appears on both sides and volume appears on both sides, you may work in kilopascals and litres, or atmospheres and millilitres, provided you are consistent within each pair.
What you may not do is mix. Litres on the left and cubic metres on the right will hand you an answer that is wrong by a factor of a thousand — and it will look perfectly reasonable.
Prefer to skip the algebra? Drop your three known values into our Boyle’s Law Calculator and it returns the missing pressure or volume, in whatever units you choose.

Halving the volume of a sealed gas at constant temperature doubles its absolute pressure. The molecules never speed up — they simply hit the walls twice as often.
How Boyle’s Law Works: Inside the Gas
Pressure is not a substance sitting inside a container. It is a drumroll. Billions of molecules are hammering the walls every second, and pressure is nothing more than the average force of that hammering, divided by the wall area.
So ask the obvious question: what happens to the drumroll when the room shrinks?
Two things, and they reinforce each other. First, each molecule has less distance to cross before it slams into a wall, so it returns more often. Second, the same crowd is now packed into less space, so more of them are near any given patch of wall at any moment.
Halve the volume and you double the collision rate on every square metre. Double the collision rate and you double the pressure. Nothing about the individual molecule has changed — it is still travelling at the same speed, delivering the same punch. It is just punching more often.
Why Temperature Must Stay Constant
That last point is the hinge of the whole law. The speed of the molecules is set by temperature: absolute temperature is a direct measure of the average kinetic energy of the molecules.
Hold the temperature fixed and you hold the molecular speeds fixed. Only the geometry changes, and the pressure obeys a clean inverse rule. Let the temperature climb, and the molecules start hitting harder as well as more often — a second effect, on top of Boyle’s, which the equation P1V1 = P2V2 knows nothing about. If the distinction between heat and temperature feels slippery, this is the sentence to anchor it to.
The Shape of the Graph
Plot pressure against volume for a gas at fixed temperature and you get a curve that falls steeply, then flattens, never quite touching either axis. That curve is a rectangular hyperbola, and physicists call it an isotherm.
Plot pressure against 1/V instead, and the same data snaps into a straight line through the origin, with gradient equal to the constant PV. That is the plot examiners want when they ask you to “verify Boyle’s law graphically” — a hyperbola is hard to judge by eye, a straight line is not.

The same Boyle’s Law data, plotted two ways. The hyperbola is the physics; the straight line is the proof.
See Boyle’s Law in Action
Drag the piston and watch the pressure gauge climb while the product PV sits stubbornly still. Push the volume down to a tenth and the needle goes tenfold — the hyperbola on the graph traces itself as you go.
Boyle’s Law vs the Other Gas Laws
Boyle’s Law is one of four experimental laws that were eventually stitched into a single equation. Each one freezes two variables and watches the other two dance. Learn which variable is being held and you will never confuse them again.
| Law | Held constant | Relationship | Equation | Graph shape |
|---|---|---|---|---|
| Boyle’s | Temperature, amount | P ∝ 1/V | P1V1 = P2V2 | Hyperbola on a P–V plot |
| Charles’s | Pressure, amount | V ∝ T | V1/T1 = V2/T2 | Straight line aimed at 0 K |
| Gay-Lussac’s | Volume, amount | P ∝ T | P1/T1 = P2/T2 | Straight line aimed at 0 K |
| Avogadro’s | Pressure, temperature | V ∝ n | V1/n1 = V2/n2 | Straight line through origin |
| Combined gas law | Amount only | PV/T fixed | P1V1/T1 = P2V2/T2 | Family of isotherms |
| Ideal gas law | Nothing | All four linked | PV = nRT | A surface in P–V–T space |
Every temperature in that table is an absolute temperature, in kelvin. Celsius will destroy Charles’s law and Gay-Lussac’s law instantly. Boyle’s law is the one law in the family where temperature never appears — which is precisely why it is the one students reach for by reflex, sometimes when they shouldn’t.
Real-World Examples of Boyle’s Law
1. Breathing
Your diaphragm contracts and flattens; the chest cavity expands. By Boyle’s Law the pressure inside your lungs drops roughly 1–2 kPa below atmospheric, and air floods in. Exhaling reverses it. You are a soft-walled piston, running at about twelve strokes a minute.
2. Scuba Diving and the Growing Bubble
Sea water adds about one atmosphere of pressure for every 10 metres of depth. At 10 m a diver breathes air at roughly twice atmospheric pressure; at 30 m, four times.
Release a bubble at 30 m and it swells to about four times its volume by the time it reaches the surface. So does the air in a diver’s lungs. This is why the first rule of scuba is never to hold your breath while ascending — a full breath taken at 30 m would expand to four lungfuls at the surface, and lungs do not stretch.
3. The Bicycle Pump and the Syringe
Block the outlet of a bicycle pump and push. The barrel volume falls, the pressure climbs, and the handle fights back harder with every centimetre. A syringe with its nozzle capped does the same thing with your thumb.
4. Weather Balloons
A meteorological balloon launched limp and floppy at ground level swells as it climbs, because the outside pressure falls away. By around 30 km it may be many times its launch volume — and eventually it bursts, which is exactly what the design expects.
5. Aerosols, Vacuum Packing and the Marshmallow Trick
Put a marshmallow under a bell jar and pump the air out. The tiny bubbles trapped in the sugar foam find themselves at a fraction of their original external pressure, so they expand, and the marshmallow inflates like a balloon. Let the air back in and it collapses into a sad, dense lump.
Common Misconceptions About Boyle’s Law
Trap 1: Using the Pressure Your Gauge Shows
This is the single most expensive error in gas-law questions. A tyre gauge, a pressure sensor on a lab bench, most everyday instruments — they read gauge pressure, the excess above atmospheric. Boyle’s Law demands absolute pressure.
| Reading | Measured from | Safe to put into P1V1 = P2V2? |
|---|---|---|
| Gauge pressure | Local atmospheric pressure | No — convert it first |
| Absolute pressure | A perfect vacuum | Yes |
| Conversion | Pabs = Pgauge + Patm | Patm ≈ 101 kPa at sea level |
In practice: a flat tyre reads zero on a gauge, but it is emphatically not a vacuum. It still holds air at 101 kPa absolute.
Trap 2: Assuming Any Squeeze Is a Boyle’s Law Squeeze
Pump a bicycle tyre vigorously and grip the barrel. It is hot. That heat is real work being converted into internal energy, and it means the temperature did not stay constant — so Boyle’s Law does not describe what just happened.
A fast compression is approximately adiabatic, following PVγ = constant, where γ ≈ 1.4 for air. The pressure rises faster than Boyle predicts. Boyle’s Law needs a slow squeeze, or a container that leaks heat away quickly enough for the gas to keep pace with its surroundings. How fast that happens depends on the material’s specific heat capacity.
Trap 3: Expecting a Straight Line on a P–V Graph
“Inversely proportional” is not “decreasing”. A student who sketches a downward-sloping straight line has drawn P = a − bV, which is a different physical claim entirely. The Boyle isotherm curves, hugging both axes without ever meeting them. Only the P against 1/V plot is straight.
Trap 4: Believing It Works Everywhere
Boyle’s Law is exact for an ideal gas — one whose molecules have no volume and no attraction for one another. Real molecules have both.
Squeeze hard enough and the molecules’ own volume stops being negligible, so the gas resists more than Boyle predicts. Cool it enough and intermolecular attraction pulls molecules inward, so the pressure comes in lower than predicted. Near the point where a gas is about to liquefy, PV stops being constant altogether. The van der Waals equation exists to patch exactly these two effects.
How Boyle’s Law Relates to the Ideal Gas Law and Thermodynamics
Boyle’s Law is not a separate truth. It is a special case, and it falls out of the ideal gas law in one line. Start with:
where n is the amount of gas in moles, R is the universal gas constant (8.314 J mol−1 K−1) and T is the absolute temperature in kelvin. Hold n and T fixed. The entire right-hand side is now a number that cannot change — so PV must be constant, and P1V1 = P2V2. Georgia State’s HyperPhysics sets out the same one-line reduction, alongside the Charles’s law case.
Freeze pressure instead and you recover Charles’s law. Freeze volume and you get Gay-Lussac’s. Three famous laws, one parent equation.
Boyle’s Law and the First Law of Thermodynamics
Compress a gas isothermally and you do work on it. Where does that energy go? Not into internal energy — internal energy depends only on temperature for an ideal gas, and temperature has not moved.
It leaves as heat. Every joule you put in through the piston flows straight out through the walls into the surroundings. That bookkeeping is the first law of thermodynamics in its cleanest possible costume, and the work done comes out as W = nRT ln(V1/V2). Problem 7 below puts numbers on it.
Worked Problems
Show Solution
Solution:
Step 1: Temperature and amount of gas are constant, so Boyle’s Law applies: P1V1 = P2V2
Step 2: Rearrange for the unknown: P2 = P1V1 / V2
Step 3: Substitute with units: P2 = (100 kPa × 2.00 L) / 0.500 L = 200 kPa·L / 0.500 L
Step 4: Solve: P2 = 400 kPa
Sanity check: the volume fell to one quarter, so the pressure rose four-fold. It did.
Answer: P2 = 400 kPa (absolute)
Show Solution
Solution:
Step 1: Apply Boyle’s Law: P1V1 = P2V2
Step 2: Rearrange for volume: V2 = P1V1 / P2
Step 3: Substitute with units: V2 = (1.50 atm × 3.60 L) / 4.00 atm = 5.40 atm·L / 4.00 atm
Step 4: Solve: V2 = 1.35 L
The atmospheres cancel, leaving litres. Mixed but consistent units are fine.
Answer: V2 = 1.35 L
Show Solution
Solution:
Step 1: A 20 per cent reduction means V2 = 0.800 V1
Step 2: From Boyle’s Law: P2 = P1V1 / V2 = P1V1 / (0.800 V1)
Step 3: The V1 terms cancel: P2 = P1 / 0.800 = 1.25 P1
Step 4: The pressure is 1.25 times its original value, an increase of 0.25 P1
A common student slip is to answer “20 per cent”. Inverse proportionality is not symmetric — a 20 per cent cut in volume gives a 25 per cent rise in pressure.
Answer: the absolute pressure increases by 25 per cent
Show Solution
Solution:
Step 1: Convert to absolute pressure before touching Boyle’s Law: P1 = P_gauge + P_atm = 0 + 101 = 101 kPa
Step 2: Apply P1V1 = P2V2, so P2 = P1V1 / V2
Step 3: Substitute with units: P2 = (101 kPa × 500 cm3) / 125 cm3 = 50 500 kPa·cm3 / 125 cm3
Step 4: Solve for absolute pressure: P2 = 404 kPa
Step 5: Convert back to gauge pressure: P_gauge = 404 − 101 = 303 kPa
Skipping Step 1 gives 0 × 500 / 125 = 0 kPa — an “answer” that says compressing air does nothing.
Answer: the gauge reads 303 kPa (404 kPa absolute)
Show Solution
Solution:
Step 1: Absolute pressure at depth is atmospheric plus the water column: P1 = P_atm + ρgh
Step 2: Substitute with units: ρgh = 1025 kg/m3 × 9.81 m/s2 × 30.0 m = 301 658 Pa = 301.7 kPa
So P1 = 101 + 301.7 = 402.7 kPa
Step 3: At the surface, P2 = 101 kPa
Step 4: Apply Boyle’s Law: V2 = P1V1 / P2 = (402.7 kPa × 25.0 mL) / 101 kPa
Step 5: Solve: V2 = 99.7 mL
Sanity check: 30 m of sea water is roughly three extra atmospheres, so the total is about four. The bubble should grow about four-fold, and 99.7 / 25.0 = 3.99.
Answer: V2 ≈ 99.7 mL, about four times its original volume
Show Solution
Solution:
Step 1: The amount of gas is unchanged and the final temperature equals the initial temperature, so Boyle’s Law applies between the initial and final equilibrium states.
Step 2: The gas now fills both cylinders: V2 = 2.00 L + 3.00 L = 5.00 L
Step 3: Apply P1V1 = P2V2, so P2 = P1V1 / V2
Step 4: Substitute with units: P2 = (500 kPa × 2.00 L) / 5.00 L = 1000 kPa·L / 5.00 L
Step 5: Solve: P2 = 200 kPa
Note the evacuated cylinder contributes volume but no gas, so its initial pressure of 0 kPa never enters the arithmetic.
Answer: P2 = 200 kPa (absolute)
Show Solution
Solution:
Step 1: Find the initial pressure from PV = nRT, so P1 = nRT / V1
nRT = 1.00 mol × 8.314 J/(mol·K) × 300 K = 2494.2 J
V1 = 20.0 L = 0.0200 m3
P1 = 2494.2 J / 0.0200 m3 = 124 710 Pa = 125 kPa (3 s.f.)
Step 2: Find the final pressure the same way, with V2 = 0.0100 m3
P2 = 2494.2 J / 0.0100 m3 = 249 420 Pa = 249 kPa (3 s.f.)
Step 3: Verify Boyle’s Law: P1V1 = 124 710 × 0.0200 = 2494.2 J and P2V2 = 249 420 × 0.0100 = 2494.2 J. The products match, and the pressure has doubled as the volume halved.
Step 4: The work done on the gas during a reversible isothermal compression is W = nRT ln(V1 / V2)
Step 5: Substitute with units: W = 2494.2 J × ln(20.0 / 10.0) = 2494.2 J × ln 2 = 2494.2 × 0.6931
Step 6: Solve: W = 1729 J
Because the temperature never changed, the internal energy did not change either — all 1729 J left the gas as heat.
Answer: P1 = 125 kPa, P2 = 249 kPa, and the work done on the gas is W ≈ 1.73 kJ, all of it released as heat