Pearson Edexcel International GCSE in Physics · 4PH1

The Gas Laws

Two equations that put numbers on everything the previous page explained.

Topic 5 · Solids, liquids and gases — one of 7 lessons in this topic, and one of 65 in Physics.

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

  • 5.21 — Use p₁ ÷ T₁ = p₂ ÷ T₂
  • 5.22 — Use p₁V₁ = p₂V₂

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

The kinetic theory told you which way things go: heat a gas at constant volume and the pressure rises; squeeze it at constant temperature and the pressure rises. These two equations say by how much. Both compare a gas before and after a change, which is why every symbol carries a 1 or a 2.

The first is for a change at CONSTANT VOLUME: p₁ ÷ T₁ = p₂ ÷ T₂. Pressure divided by kelvin temperature stays the same, which is another way of saying pressure is directly proportional to temperature. Double the kelvin temperature and the pressure doubles.

The second is for a change at CONSTANT TEMPERATURE: p₁V₁ = p₂V₂. Pressure multiplied by volume stays the same, which is to say pressure is inversely proportional to volume. Halve the volume and the pressure doubles. This is Boyle's law, established in the 1660s with a J-shaped tube and a quantity of mercury.

Think of it like a fixed budget split two ways

Imagine you have a fixed sum to spend on exactly two things, so whatever you spend on one you cannot spend on the other. If one doubles, the other must halve — their product is fixed. That is p₁V₁ = p₂V₂: pressure and volume trade against each other with the total held constant, and neither can change without the other compensating. The temperature relation is a different shape entirely. There, pressure and temperature rise and fall TOGETHER, keeping their ratio constant rather than their product — like two quantities pegged to the same exchange rate. Knowing which of the two shapes you are in tells you immediately whether the answer should be bigger or smaller than what you started with, before you calculate anything.

Both equations, with the check that catches errors

(a) A gas at 300 K exerts a pressure of 120 kPa in a rigid container. It is heated to 400 K. Find the new pressure. (b) A gas occupies 250 cm³ at 100 kPa. It is compressed to 100 cm³ at the same temperature. Find the new pressure.

  1. (a) The container is rigid, so the volume is constant: use p₁ ÷ T₁ = p₂ ÷ T₂.
  2. 120 ÷ 300 = p₂ ÷ 400, so p₂ = 120 × 400 ÷ 300 = 160 kPa.
  3. Check: the temperature went UP, so the pressure should go UP — and 160 > 120.
  4. (b) The temperature is constant: use p₁V₁ = p₂V₂.
  5. 100 × 250 = p₂ × 100, so p₂ = 25 000 ÷ 100 = 250 kPa.
  6. Check: the volume went DOWN, so the pressure should go UP — and 250 > 100.

Answer: 160 kPa, and 250 kPa.

The routine, and the three ways these questions go wrong

  1. DECIDE WHICH EQUATION. If the temperature is constant, or the gas is compressed or expanded without heating, use p₁V₁ = p₂V₂. If the volume is constant — a rigid or sealed container — use p₁ ÷ T₁ = p₂ ÷ T₂.
  2. CONVERT ANY TEMPERATURE TO KELVIN. This is not optional and not an approximation: the relationship is simply false in Celsius.
  3. Substitute, rearrange and solve for the unknown.
  4. CHECK THE DIRECTION. Ask whether the answer should be bigger or smaller than the starting value, and see whether it is.
  5. Units do NOT need converting as long as both pressures are in the same unit and both volumes are in the same unit — the equations are ratios, so kilopascals and cubic centimetres are perfectly acceptable provided you are consistent.

That last point is worth stating clearly because it runs against the habit built up everywhere else in the course. In p₁V₁ = p₂V₂ the units cancel, so a pressure in kPa on one side and kPa on the other is fine, as are cubic centimetres throughout. What must NEVER be left unconverted is the temperature, because kelvin and Celsius do not differ by a factor — they differ by an offset, and an offset does not cancel.

The direction check in step four is the most valuable habit on this page. Almost every error in these questions produces an answer that is wrong in an obvious direction — a compressed gas at a lower pressure, or a heated gas at a lower pressure — and a two-second sanity check catches it. Work out from the physics which way the answer should move before you look at what your arithmetic gave you.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

p₁ ÷ T₁ = p₂ ÷ T₂

p₁ and p₂ are the pressures before and after, T₁ and T₂ the temperatures IN KELVIN, at constant volume

Units: Both pressures in the same unit; both temperatures in KELVIN. Pressure is directly proportional to kelvin temperature.

p₁V₁ = p₂V₂

p₁ and V₁ are the pressure and volume before, p₂ and V₂ after, at constant temperature

Units: Both pressures in the same unit; both volumes in the same unit. Pressure is inversely proportional to volume.

Learn this definition · Boyle's law

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume, so the product of pressure and volume has the same value before and after any change.

Learn this definition · The pressure law

For a fixed mass of gas at constant volume, the pressure is directly proportional to the temperature in kelvin, so pressure divided by kelvin temperature has the same value before and after any change.

Answering any gas-law question

  1. Decide which quantity is CONSTANT: volume fixed means p ÷ T; temperature fixed means pV.
  2. Convert every temperature to KELVIN by adding 273.
  3. Substitute the values, keeping both pressures in the same unit and both volumes in the same unit.
  4. Rearrange and solve for the unknown.
  5. Check the DIRECTION: should the answer be larger or smaller than the starting value?

Which equation, and how to tell

A. p₁V₁ = p₂V₂ — a change at CONSTANT TEMPERATURE. Look for compressed, expanded, squeezed, or 'at the same temperature'. Pressure and volume move in OPPOSITE directions.

B. p₁ ÷ T₁ = p₂ ÷ T₂ — a change at CONSTANT VOLUME. Look for a rigid, sealed or fixed container being heated or cooled. Pressure and temperature move in the SAME direction.

The three errors that account for nearly every lost mark

  • USING CELSIUS in the temperature equation. The relationship is false in Celsius — it is an offset, not a factor, so it does not cancel
  • CHOOSING THE WRONG EQUATION. Ask what is being held constant before anything else
  • NOT CHECKING THE DIRECTION. A compressed gas must be at higher pressure; a cooled one at lower
  • Note what is NOT an error: leaving pressures in kPa or volumes in cm³ is fine, provided both sides use the same unit

Model answer [4 marks]

A gas in a rigid container is at 27 °C and a pressure of 150 kPa. It is heated to 127 °C. Calculate the new pressure. [4]

The container is rigid so the volume is constant, and both temperatures must be converted to kelvin: T₁ = 27 + 273 = 300 K and T₂ = 127 + 273 = 400 K. Using p₁ ÷ T₁ = p₂ ÷ T₂, we have 150 ÷ 300 = p₂ ÷ 400, so p₂ = 150 × 400 ÷ 300 = 200 kPa. The temperature has increased so the pressure should increase, and it has.

Model answer [3 marks]

A gas occupies 400 cm³ at a pressure of 90 kPa. It is compressed at constant temperature to 150 cm³. Calculate the new pressure. [3]

The temperature is constant, so use p₁V₁ = p₂V₂. Substituting, 90 × 400 = p₂ × 150, so p₂ = 36 000 ÷ 150 = 240 kPa. The gas has been compressed so the pressure should rise, and it has.

Not this: All the units have to be converted to SI before using the gas laws.

This: Pressures and volumes do NOT need converting, provided both sides use the same unit — the equations are ratios and the units cancel. Temperature is the exception and must always be in kelvin, because Celsius differs by an offset rather than a factor.

Mark-losing trap. ALWAYS convert temperature to kelvin. Celsius makes the equation false, not merely imprecise.

Mark-losing trap. Rigid or sealed container → p ÷ T. Compressed or expanded at constant temperature → pV.

Mark-losing trap. Check the direction of your answer before moving on. It catches nearly every slip.

Mark-losing trap. Pressures and volumes need only be CONSISTENT, not converted to SI. kPa and cm³ are fine.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

  1. Grade 6 · State [1 mark] — A gas is compressed into half its original volume at constant temperature. What happens to its pressure?
  2. Grade 7 · Calculate [3 marks] — A gas occupies 300 cm³ at a pressure of 100 kPa. It is compressed at constant temperature to 120 cm³. Calculate the new pressure.
  3. Grade 8 · Calculate [3 marks] — A gas in a sealed rigid container is at 300 K and 180 kPa. It is cooled to 200 K. Calculate the new pressure.
  4. Grade 9 · Calculate [5 marks] — A sealed rigid canister of gas is at 17 °C and a pressure of 200 kPa. It is left in sunshine and warms to 47 °C. Calculate the new pressure.
  5. 9+ · Analyse [6 marks] — A student calculates that a sealed gas canister at 20 °C and 250 kPa will reach 3750 kPa if left in a car where it warms to 300 °C. They used p₁ ÷ T₁ = p₂ ÷ T₂ with temperatures in Celsius. Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one whose law this is, found with a tube and some mercury, and the one who derived the law from the molecules rather than measuring it. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

Robert Boyle — the one whose law this is, found with a tube and some mercury

Statement 5.22 is Boyle's law, published in 1662 and obtained with equipment simple enough to describe in a sentence: a J-shaped glass tube sealed at the short end, trapping air, with mercury poured into the long open end. Adding mercury increased the pressure on the trapped air and he recorded how the volume shrank, finding that the product of the two stayed constant. He published the whole procedure and the readings so that others could repeat it, which was itself a novelty at the time.

  • “How did you set up the experiment with the J-shaped tube?”
  • “What made you look for a numerical relationship rather than just a trend?”
  • “Why did you publish all your readings rather than only the conclusion?”
  • “What did you think was happening inside the air as it was squeezed?”
  • “How accurate could a measurement like that be in your day?”

James Clerk Maxwell — the one who derived the law from the molecules rather than measuring it

Boyle found the relationship by measurement two centuries before anyone could say why it held. Maxwell's kinetic theory derived it: if a gas is a swarm of molecules in random motion, then halving the volume halves the distance between collisions with the walls, doubling their frequency and hence the pressure — and pV comes out constant without measuring anything. Deriving a known experimental law from a model is one of the strongest tests a model can pass, because the answer is already known and cannot be adjusted to fit.

  • “How can a law found by experiment be derived from moving molecules?”
  • “Why is deriving a known result a good test of a model?”
  • “What assumptions did you have to make about the molecules?”
  • “Where do the gas laws stop describing a real gas accurately?”
  • “Did you regard the molecules as real, or as a useful picture?”

Then defend it

On Incandio a lesson is not finished when the questions come out right. You teach the idea back to Ember, an AI apprentice who asks the awkward question, and then you argue it against Robert Boyle in a structured debate marked against descriptors you can read before you enter. Learn it, teach it, then defend it — all three happen on this page once the app loads.

Carry on through the course