In simple terms
A friendly intro before the formal notes — no formulas yet.
The Rules of Gas Behaviour
An ideal gas is a simplified model of a gas made of tiny particles in constant, random motion that don't attract each other. One equation, PV = nRT, ties together how much gas you have, how much space it fills, how hard it pushes, and how hot it is.
Imagine a bouncy castle. The pressure (P) is how hard the kids push against the inside walls. The volume (V) is the size of the castle. The number of moles (n) is the number of kids inside. The temperature (T) is how energetically they are all bouncing. The ideal gas law, PV = nRT, is the 'rulebook' connecting them: add more kids (increase n) or make them bounce faster (increase T) and the pressure on the walls rises, if the castle size stays the same.
- 1
Identify all known variables (P, V, n, T) and the one you must find.
- 2
Convert every value into SI units: pressure in pascals (Pa), volume in cubic metres (m³), temperature in kelvin (K).
- 3
Rearrange PV = nRT to make the unknown the subject.
- 4
Substitute, solve, then convert the answer to the units the question asks for and quote the correct significant figures.
Explore the concept
Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.
Step-synced diagram — highlights what to look for in the simulation above.
Step 1
Identify all known variables (P, V, n, T) and the one you must find.
2 more simulations for this topic — run them in the Simulations section below
Simulations
Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.
2 simulations
- PhETCoreIB S1.5
States of Matter
Squeeze and cool a gas of real particles under a movable lid until attractions make it condense.
Try this
- On Phase Changes choose argon and press Gas.
- Cool it and push the lid down; watch for particles starting to cluster.
- Repeat with neon, whose attractions are weaker.
Look for At low temperature and high pressure the particles are slow and close enough for intermolecular attractions and their own volume to matter — the two kinetic-theory assumptions that fail for a real gas.
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
- GeoGebraIB S1.5
Isothermal p–V curve
A pressure–volume curve for a fixed amount of gas at constant temperature: set T and the two volumes.
Try this
- Set V1 and V2 and read the pressure at each end of the curve.
- Halve the volume and check that the pressure doubles.
- Raise T and watch the whole curve move away from the axes.
Look for pV is the same at every point on an isotherm (Boyle’s law); a hotter gas sits on a higher curve.
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Key formulas
Tap any symbol to reveal exactly what it means and its units.
Full topic notes
Formal explanation with the rigour you need for the exam.
The kinetic molecular theory and the ideal gas
The kinetic molecular theory pictures a gas as a very large number of tiny particles in continuous, rapid, random motion. They collide with one another and with the container walls, and it is these wall collisions that we measure as pressure. The collisions are treated as perfectly elastic, meaning no kinetic energy is lost, and the average kinetic energy of the particles is directly proportional to the absolute (kelvin) temperature — hotter gas means faster particles.
An 'ideal' gas is an idealisation of this picture built on two assumptions that make the mathematics simple. These assumptions are the thing examiners most often ask you to state, so learn them precisely.
Assumption 1 — negligible particle volume: the volume of the particles themselves is negligible compared with the volume of the container. The gas is treated as mostly empty space.
Assumption 2 — no intermolecular forces: there are no forces of attraction or repulsion between the particles, so between collisions they travel in straight lines at constant speed.
Consequence: because average kinetic energy ∝ T(K), raising the temperature raises the speed of the particles and the force of their collisions with the walls.
Ideal gases vs. real gases
No real gas is truly ideal, because real particles do occupy space and do attract one another (through London dispersion and other intermolecular forces). Real gases behave most ideally at high temperature and low pressure: the particles are far apart and moving fast, so both their own volume and the weak forces between them are negligible. Deviations become significant under the opposite conditions.
High pressure → deviation: particles are forced close together, so their finite volume is no longer negligible relative to the container. Assumption 1 fails.
Low temperature → deviation: particles move slowly, so intermolecular attractions have time to act and pull particles together. Assumption 2 fails.
Most ideal behaviour: low pressure and high temperature — the conditions under which both assumptions hold best.
The individual gas laws and their graphs
Before the single combined equation, it helps to see the three pairwise relationships, each holding one pair of variables constant. Every temperature here is the absolute (kelvin) temperature.
Combining all three for a fixed amount of gas gives the combined gas law, which lets you follow one sample of gas between two sets of conditions without needing R.
Pressure–volume (Boyle's law) — constant n and T: P ∝ 1/V, so PV = constant. A graph of P against V is a curve (a hyperbola); a graph of P against 1/V is a straight line through the origin.
Volume–temperature (Charles's law) — constant n and P: V ∝ T, so V/T = constant. A graph of V against T(K) is a straight line through the origin.
Pressure–temperature (Gay-Lussac's law) — constant n and V: P ∝ T, so P/T = constant. A graph of P against T(K) is a straight line through the origin.
The ideal gas equation
When the amount of gas is allowed to vary, the individual laws and Avogadro's principle (equal volumes of gases at the same T and P contain equal numbers of particles) combine into a single relationship — the ideal gas equation. It is the central equation of this topic.
PV = nRT
The proportionality constant R is the ideal (universal) gas constant. Its value comes from the IB Data Booklet, so you never memorise it — but its units dictate the units of everything else, so all variables must be in SI units before substitution.
Pressure (P): pascals (Pa). Questions often give kPa or atm. .
Volume (V): cubic metres (m³). Lab data are usually cm³ or dm³. .
Amount (n): moles (mol) — you may have to find this from mass ÷ molar mass first.
Temperature (T): kelvin (K). (adding 273 is accepted in IB working).
Gas constant (R): . 'J K⁻¹ mol⁻¹' is the reminder that you need Pa, m³ and K.
Unit conversion is the most common source of lost marks in this whole topic. Before rearranging anything, write out P, V, n, T in a column and convert each to SI units (Pa, m³, K, mol). This one habit protects the answer mark on almost every gas calculation.
Molar volume of a gas
A direct consequence of PV = nRT is that, at a fixed temperature and pressure, V/n is the same for every ideal gas — one mole always occupies the same volume regardless of identity. This is the molar volume. From the Data Booklet, at STP (0 °C = 273 K and 100 kPa) the molar volume is 22.7 dm³ mol⁻¹. So for a gas at STP you can shortcut between moles and volume without R: n = V(in dm³) ÷ 22.7. Away from STP, go back to PV = nRT.
Common mistakes examiners penalise
Leaving temperature in °C — every gas law needs kelvin. This is the most heavily penalised slip in the topic; add 273 before you do anything else.
Mismatched units in PV = nRT — using kPa with cm³, or forgetting to convert dm³/cm³ to m³. R = 8.31 demands Pa, m³ and K together, or none of them.
Confusing direct and inverse relationships — P ∝ 1/V (inverse) but V ∝ T and P ∝ T (direct). Halving V doubles P; doubling T(K) doubles V.
Assuming molar volume applies at any conditions — 22.7 dm³ mol⁻¹ only holds at STP. Away from STP, use PV = nRT.
Rounding n too early in a molar-mass calculation, then quoting a molar mass that is out by several units.
Not showing the rearrangement — write V = nRT/P (or the relevant form) explicitly; the method mark depends on it, and it protects you under error-carried-forward marking.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
Calculate the pressure, in kPa, exerted by 0.250 mol of an ideal gas in a 5.00 dm³ flask at 22.0 °C. (R = 8.31 J mol⁻¹ K⁻¹)
- 1
List variables and convert to SI units:
A sample of a volatile liquid was injected into a gas syringe. When vaporised at 98.0 °C and 101 kPa, 0.150 g of the substance occupied 65.0 cm³. Determine the molar mass of the substance. (R = 8.31 J mol⁻¹ K⁻¹)
- 1
List variables and convert to SI units:
Calculate the volume, in dm³, occupied by 0.250 mol of an ideal gas at 300 K and 100 kPa. (R = 8.31 J mol⁻¹ K⁻¹) [3]
- 1
List and convert to SI units:
How it all connects
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Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Key terms for this topic — skim now; the Check step will test them.
- Pressure–volume law (Boyle's law)
At constant n and T, P ∝ 1/V, so PV = constant. A P–V graph is a curve (hyperbola); a P vs 1/V graph is a straight line through the origin.
- Volume–temperature law (Charles's law)
At constant n and P, V ∝ T (T in kelvin), so V/T = constant. A V vs T(K) graph is a straight line through the origin.
- Pressure–temperature law (Gay-Lussac's law)
At constant n and V, P ∝ T (T in kelvin), so P/T = constant. A P vs T(K) graph is a straight line through the origin.
- The combined gas law
For a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂ (T in kelvin). Useful when a fixed sample of gas changes between two sets of conditions.
- Molar volume of a gas
The volume occupied by one mole of any ideal gas at a stated temperature and pressure. At STP (0 °C, 100 kPa) it is 22.7 dm³ mol⁻¹ (IB Data Booklet). It follows from PV = nRT because at fixed T and P, V/n is the same for all ideal gases.
- Unit conversions you must have automatic
1 kPa = 1000 Pa; 1 m³ = 1000 dm³ = 1,000,000 cm³; T(K) = T(°C) + 273.15 (273 is accepted in IB working).
Name it
Read the meaning, then pick which of this lesson’s terms it describes. Miss one and you see what your choice really means.
At constant n and P, V ∝ T (T in kelvin), so V/T = constant. A V vs T(K) graph is a straight line through the origin.
Quick check
Write your answer first, then compare it with the model one — the gap is what you would have lost.
Teach it back
If you can explain it simply, you own it — gaps here are marks you’d lose.
Teach it back
Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.
Revision flashcards
Guess first, then flip — retrieval beats re-reading.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
Assumption 1 — negligible particle volume: the volume of the particles themselves is negligible compared with the volume of the container. The gas is treated as mostly empty space.
Assumption 2 — no intermolecular forces: there are no forces of attraction or repulsion between the particles, so between collisions they travel in straight lines at constant speed.
Consequence: because average kinetic energy ∝ T(K), raising the temperature raises the speed of the particles and the force of their collisions with the walls.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Get a Paper 2 gas calculation marked: use PV = nRT with correct unit handling
Get a Paper 2 gas calculation marked: use PV = nRT with correct unit handling
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Frequently asked
Checkpoint
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Reading it isn’t knowing it — prove it.
Before you move on: do Get a Paper 2 gas calculation marked: use PV = nRT with correct unit handling on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.
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