Educerie · IB Diploma · Chemistry
Structure 1 Models of the particulate nature of matter · S1.5 Ideal gases
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Recognise the key assumptions of the ideal gas model | SL, HL | "State two assumptions of the kinetic theory of an ideal gas" (2 marks) |
| Explain the limitations of the model: why real gases deviate at low temperature and high pressure | SL, HL | "Explain why real gases deviate from ideal behaviour at high pressure" (2 marks); no calculation required |
| Use the molar volume of an ideal gas at STP from the data booklet | SL, HL | "Calculate the volume of carbon dioxide produced at STP" (2 marks) |
| Investigate and analyse graphs relating pressure, volume and temperature for a fixed mass of gas | SL, HL | Paper 1A graph shapes; Paper 1B data to plot or extrapolate |
| Solve problems with the ideal gas equation pV = nRT, in SI units | SL, HL | Paper 2, 2 to 3 marks, working shown |
| Solve problems with the combined gas law | SL, HL | Paper 2, 2 marks |
Before you start
You need S1.1: the kinetic molecular theory, gases as fast particles far apart, and temperature in kelvin as a measure of average kinetic energy. You need S1.4: n = m ÷ M, and Avogadro's law. The data booklet gives the gas constant R = 8.31 J K⁻¹ mol⁻¹, the ideal gas equation, the combined gas law, and the molar volume of an ideal gas at STP.
1The idea in one paragraph
A gas is mostly empty space with particles flying through it. If you pretend the particles take up no room at all and ignore any pull between them, the behaviour of every gas can be predicted by one equation, pV = nRT, whatever the gas is. That pretend gas is the ideal gas. No real gas is ideal, but at room temperature and ordinary pressures most gases are close enough that the equation gives answers within a few per cent. The model breaks down when its two big pretences stop being reasonable: at high pressure, where the particles are squeezed so close that their own size matters, and at low temperature, where they move so slowly that their attractions for each other matter. Knowing why it fails is as much a part of this subtopic as using it.
2The ideal gas model
Figure 1 sets out the assumptions. You must be able to recognise them and state them.
An ideal gas consists of moving particles with negligible volume and no intermolecular forces. All collisions between particles are elastic.
Each assumption has a job.
- Constant random motion. The particles move in straight lines until they hit another particle or a wall. Their average kinetic energy is proportional to the kelvin temperature (S1.1).
- Negligible volume. The particles themselves are treated as points. All of the gas's volume is the space between them, so the volume available to the particles is the whole container.
- No intermolecular forces. Particles neither attract nor repel one another except at the instant of collision. Nothing slows a particle as it heads for a wall.
- Elastic collisions. No kinetic energy is lost in collisions, so a gas left alone does not slow down and settle.
- Pressure is the result of particles hitting the walls. More collisions per second, or harder collisions, mean higher pressure.
Every gas law in section 4 follows from these. Squeeze the same gas into half the volume at the same temperature and particles hit each area of wall twice as often, so the pressure doubles. Heat it and the particles move faster and hit harder and more often, so either the pressure rises or, if the container can expand, the volume does.
3Real gases and the limits of the model
Real particles do have a size, and they do attract one another. Under most conditions neither matters much. Figure 2 shows when each one starts to.
At high pressure, particle volume is no longer negligible. Compress a gas and the empty space shrinks while the particles stay the same size. The particles now occupy a real fraction of the container, so the space they can move in is less than the container's volume. The model's assumption of point particles fails, and the real gas takes up more volume than pV = nRT predicts.
At low temperature, intermolecular forces are no longer negligible. Slow particles spend longer near each other, and their attractions have time to act. A particle about to hit the wall is pulled back slightly by its neighbours, so it hits more softly, and some particles cluster. The real pressure is lower than the model predicts. Cool far enough and the attractions win entirely: the gas condenses to a liquid, which an ideal gas can never do.
Real gases deviate most from ideal behaviour at high pressure (particle volume matters) and low temperature (intermolecular forces matter). They behave most ideally at high temperature and low pressure.
A full answer names the condition and the assumption that fails and why. "Real gases are not ideal at high pressure" states; "at high pressure the particles are close together, so their own volume is a significant fraction of the container and can no longer be ignored" explains.
How big is the deviation? Figure 3 is a sketch, not required by the guide as mathematics, that makes both effects visible at once. For an ideal gas, pV ÷ nRT = 1 at every pressure. For a real gas it starts near 1 at low pressure, dips below 1 as attractions pull the particles together, and rises above 1 at high pressure as particle volume takes over.
Some gases deviate more than others. The guide's linking question points to Structure 2.2. Under the same conditions, a gas whose particles are larger or attract each other more strongly deviates more. Helium, with tiny atoms and very weak attractions, is one of the most nearly ideal gases. Ammonia, whose molecules attract each other strongly (by hydrogen bonding, which Structure 2.2 explains), deviates much more and is easy to liquefy.
4Pressure, volume and temperature
The guide asks you to investigate how pressure, volume and temperature are related for a fixed mass of gas, and to analyse the graphs. The names of the individual gas laws are not assessed; the relationships and the graph shapes are. In each case one variable is held constant.
Pressure and volume, at constant temperature. Pressure is inversely proportional to volume: halve the volume and the pressure doubles, so p × V stays constant. A graph of p against V is a curve that falls ever more slowly; a graph of p against 1/V is a straight line through the origin, which is the easy way to show the relationship from data. Figure 4 draws both.
Volume and temperature, at constant pressure. Heat a gas in a container that can expand, such as a syringe with a free plunger, and its volume rises in a straight line with temperature. Plotted against temperature in °C, the line does not go through the origin; extend it backwards and it reaches zero volume at about −273 °C. That temperature is absolute zero, 0 K. Plotted against temperature in kelvin, the same data give a straight line through the origin: V is directly proportional to T. Figure 5, panels (a) and (b), shows the two plots.
Pressure and temperature, at constant volume. Heat a gas in a sealed rigid container and its pressure rises in the same way: p is directly proportional to T in kelvin, panel (c).
That is why every gas calculation uses kelvin. In °C, doubling the number does not double anything: 20 °C to 40 °C is only 293 K to 313 K.
Sketched or plotted? The guide's Nature of science question asks when to sketch a graph and when to plot one. A sketch shows the shape and the key features (straight, through the origin, curving towards the axis) without values; use it to explain a relationship. A plotted graph uses measured points on a scale, with a line of best fit, and lets you read values, find a gradient or extrapolate, as in Try it Q5. The first communicates the idea; the second tests it.
5Molar volume
Avogadro's law (S1.4) says that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. Turn it round and one mole of any ideal gas occupies the same volume at a given temperature and pressure: the molar volume. It is a constant only once the temperature and pressure are fixed.
The data booklet gives the value at standard temperature and pressure (STP), which is 273.15 K (0 °C) and 100 kPa:
molar volume of an ideal gas at STP = 22.7 dm³ mol⁻¹ (2.27 × 10⁻² m³ mol⁻¹)
Worked example. Calcium carbonate decomposes on heating: CaCO₃(s) → CaO(s) + CO₂(g). Calculate the volume of carbon dioxide, at STP, released by 5.00 g of calcium carbonate.
At any other temperature or pressure, 22.7 dm³ mol⁻¹ is wrong and you must use pV = nRT instead. At room temperature (298 K and 100 kPa) the molar volume is about 24.8 dm³ mol⁻¹, which pV = nRT gives you in one line.
6The ideal gas equation
All the relationships of section 4, and the molar volume, combine into one equation:
pV = nRT p in Pa, V in m³, n in mol, T in K, R = 8.31 J K⁻¹ mol⁻¹
The guide requires SI units, and R has the value 8.31 only in SI units. So convert before you substitute:
| Given | Convert to | How |
|---|---|---|
| kPa | Pa | × 10³ |
| dm³ | m³ | × 10⁻³ (÷ 1000) |
| cm³ | m³ | × 10⁻⁶ |
| °C | K | + 273.15 |
Worked example 1: find a pressure. 2.00 g of oxygen gas is held in a 1.50 dm³ flask at 25.0 °C. Calculate the pressure.
Worked example 2: find a molar mass. The guide's linking question asks how pV = nRT gives the molar mass of a gas from experimental data. Figure 6 shows one method: inject a weighed mass of a volatile liquid into a gas syringe inside an oven hot enough to vaporise it, and read the volume of vapour.
Since n = m ÷ M, the equation rearranges to M = mRT ÷ pV. A student injects 0.500 g of liquid; the vapour occupies 333 cm³ at 100.0 °C and 101 kPa.
The value matches ethanol, C₂H₅OH (46.08 g mol⁻¹). The main errors in this experiment come from the model itself: near its boiling point the vapour is not ideal, because attractions between molecules still act, so the measured volume tends to be too small and the molar mass too large.
The combined gas law. When a fixed amount of gas moves from one set of conditions to another, n and R cancel:
p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂
Here the units of p and V need not be SI, as long as they are the same on both sides. T must still be in kelvin, because the equation relies on proportionality to absolute temperature.
Worked example 3. A weather balloon holds 2.50 dm³ of helium at 101 kPa and 20.0 °C at ground level. It rises to where the pressure is 35.0 kPa and the temperature is −40.0 °C. Calculate its new volume.
Check the direction: the pressure fell to about a third, which alone would nearly triple the volume; the cold shrinks it a little. 5.74 dm³ is a little over twice the start. Sensible.
7Where marks are lost
Using °C in any gas equation. Every gas calculation needs kelvin, including the combined gas law.
Leaving dm³ or kPa in pV = nRT. With R = 8.31, p must be in Pa and V in m³. A volume in dm³ gives an answer a thousand times too big or too small.
Using 22.7 dm³ mol⁻¹ at room temperature. That molar volume belongs to STP (273.15 K, 100 kPa) only.
Saying real gases deviate "because the model is wrong". Name the assumption that fails and why: particle volume at high pressure, intermolecular attractions at low temperature.
Reversing the conditions. Real gases are most ideal at high temperature and low pressure. Low temperature and high pressure are where they deviate most.
Drawing p against V as a straight line. At constant temperature p against V is a curve; p against 1/V is the straight line.
Stopping at n in a molar mass question. M = m ÷ n: the last step is the one most often left out.
8Draw it right
- p against V (constant T): a smooth curve falling towards the V-axis and never touching either axis. p against 1/V: a straight line through the origin.
- V against T in kelvin (constant p) and p against T in kelvin (constant V): straight lines through the origin. With temperature in °C the line cuts the temperature axis at −273 °C; draw the part below your data as a dashed extrapolation.
- Label axes with quantity and unit, and state what is held constant.
- Plotted graphs: points plotted accurately, one straight line of best fit, extrapolated with a ruler.
- Particle diagrams: for an ideal gas, small particles far apart with random motion arrows; for high pressure, particles crowded so their size is a large part of the space; for low temperature, slow particles near each other.
9Try it
Marks in brackets. Answers and marker's notes are at the end. Use the data booklet for constants.
Q1. Under which conditions does a real gas behave most like an ideal gas? 1 mark
A. High temperature and low pressure B. Low temperature and high pressure C. High temperature and high pressure D. Low temperature and low pressure
Q2. (a) State two assumptions of the ideal gas model. 2 marks
(b) Explain why ammonia deviates more from ideal behaviour than helium under the same conditions. 2 marks
Q3. A 5.00 dm³ cylinder contains nitrogen gas at 200 kPa and 27 °C. Calculate the mass of nitrogen in the cylinder. 3 marks
Q4. A sample of gas occupies 250 cm³ at 27 °C and 100 kPa. Calculate its volume at 127 °C and 150 kPa. 2 marks
Q5. A student traps air in a syringe and measures its volume at different temperatures, keeping the pressure constant.
| Temperature (°C) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Volume (cm³) | 50.0 | 53.7 | 57.3 | 61.0 | 64.6 | 68.3 |
(a) Determine, from the data, the temperature in °C at which the volume of the gas would become zero. 2 marks
(b) State the significance of this temperature. 1 mark
(c) Predict the volume of the air at 150 °C. 1 mark
Q6. 0.486 g of magnesium reacts with excess hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Calculate the volume of hydrogen produced at STP. 2 marks
10In one breath
An ideal gas is a model: particles in constant random motion, with negligible volume, no intermolecular forces and elastic collisions, and pressure comes from their collisions with the walls. Real gases deviate most at high pressure, where the particles' own volume is no longer negligible, and at low temperature, where intermolecular attractions are no longer negligible; they are most ideal at high temperature and low pressure, and larger or more strongly attracting particles deviate more. For a fixed mass of gas, p is inversely proportional to V at constant T, so p against 1/V is a straight line through the origin; V and p are each directly proportional to T in kelvin, and in °C the lines meet zero at −273 °C, absolute zero. One mole of ideal gas occupies 22.7 dm³ at STP (273.15 K, 100 kPa) only. pV = nRT needs Pa, m³, mol and K with R = 8.31 J K⁻¹ mol⁻¹, and with n = m ÷ M it gives a molar mass as M = mRT ÷ pV. For a change of conditions, p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂, in any consistent units of p and V, but always kelvin.
Answers
Q1. A. At high temperature the particles move fast enough that attractions hardly matter, and at low pressure they are far enough apart that their own volume is negligible. A only.
Q2. (a) Any two: the particles are in constant random motion; the particles have negligible volume; there are no intermolecular forces between the particles; collisions are elastic. 1 each. "The gas has no volume" scores 0: it is the particles' volume that is negligible, compared with the container. (b) Ammonia molecules are larger than helium atoms and attract each other much more strongly (hydrogen bonding), so both the particle volume and the intermolecular forces that the ideal model ignores are more significant for ammonia. 1 for stronger intermolecular forces in ammonia, 1 for larger particle size / volume, each linked to an ideal-gas assumption.
Q3.
M1 for SI conversions of p, V and T, M1 for n = pV ÷ RT, A1 for 11.2 g. Using M = 14.01 (N atoms, not N₂) gives 5.62 g and loses the A1. Leaving V in dm³ loses the first M1.
Q4.
M1 for the combined gas law with temperatures in kelvin, A1 for 222 cm³. Using 27 and 127 °C gives 784 cm³ and scores 0.
Q5. (a) The volume rises by 68.3 − 50.0 = 18.3 cm³ over 100 °C, a gradient of 0.183 cm³ °C⁻¹. Extending back from 50.0 cm³ at 0 °C, the volume reaches zero after a fall of 50.0 ÷ 0.183 = 273 °C, at −273 °C. M1 for the gradient or a correct extrapolated graph, A1 for −273 °C (accept −270 to −276 °C from a graph). (b) It is absolute zero, 0 K: the lowest possible temperature, at which the particles would have minimum kinetic energy. [1] (c) 50.0 + 0.183 × 150 = 77.5 cm³ (or 50.0 × 423.15 ÷ 273.15). accept 77 to 78 cm³.
Q6.
M1 for moles of Mg equal to moles of H₂, A1 for 0.454 dm³ (454 cm³). Using 24.0 dm³ mol⁻¹ or 24.8 dm³ mol⁻¹ loses the A1: the question fixes STP.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section S1.5 Ideal gases. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.