IB DP Chemistry Structure 1.5: Ideal gases

Here you will learn Structure 1.5, “Ideal gases”, from scratch. The page is written against the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023) and works through understandings Structure 1.5.1, 1.5.2, 1.5.3 and 1.5.4 with worked examples. Ideal gases carry no additional higher level statements, so SL and HL candidates study identical content. It is for the course examined from 2025 onwards.

After this page, consolidate with the ideal gases revision notes and test yourself on the ideal gases question set. Avogadro’s law and the mole are taught in the guide to Structure 1.4. The Structure 1 overview, the DP Chemistry hub, a topic checklist and the free diagnostic quizzes help you plan.

What this unit covers

Guide statement	What you need to do	Level
Structure 1.5.1	Recognise the key assumptions of the ideal gas model	SL and HL
Structure 1.5.2	Explain the limitations of the model: real gases deviate most at low temperature and high pressure (no mathematics needed)	SL and HL
Structure 1.5.3	Use the molar volume of an ideal gas; investigate how temperature, pressure and volume are related for a fixed mass of gas; analyse the graphs	SL and HL
Structure 1.5.4	Solve problems using PV = nRT and P₁V₁/T₁ = P₂V₂/T₂, in SI units	SL and HL

Structure 1.5 is short: the guide allows about 3 teaching hours, the same at SL and HL.

Structure 1.5.1 The ideal gas model

An ideal gas is a model, not a real substance. It describes a gas with four simple assumptions:

1. The particles are in constant, random, straight-line motion between collisions.
2. The particles themselves take up a negligible volume; almost all of the container is empty space.
3. There are no intermolecular forces between the particles, attractive or repulsive.
4. All collisions, between particles and with the walls, are perfectly elastic: the total kinetic energy before a collision equals the total after it.

Pressure comes from the particles striking the walls: it is the total force of collisions per unit area. Two ideas from Structure 1.1 fit with this: the average kinetic energy of the particles rises in step with the Kelvin temperature, so heating a gas makes the particles move faster, and faster particles hit the walls harder and more often.

A gas with no forces between particles could never condense, so the model must fail close to condensation.

Structure 1.5.2 Why real gases deviate

Real gases behave almost ideally when the particles are far apart and moving fast. The assumptions that break down are 2 (negligible volume) and 3 (no forces). The guide asks you to explain this in words only; no mathematical treatment is required.

High pressure. Squeezing a gas pushes its particles close together. The particles now take up a significant fraction of the container, so the space they can move through is smaller than the container volume. The model has ignored this, so it no longer predicts the volume well.

Low temperature. Cooler particles move more slowly. Attractions between neighbouring particles now have time to act, pulling particles towards each other as they approach a wall. They hit the wall less hard, so the measured pressure is lower than the model predicts. If cooling continues, the attractions win and the gas condenses.

The type of gas matters too. Under the same conditions, a gas whose particles attract each other more strongly deviates more. Larger, more polarisable particles have stronger London (dispersion) forces; polar molecules add dipole–dipole forces; molecules such as NH₃ and H₂O can form hydrogen bonds (Structure 2.2). Small, non-polar particles such as He deviate least.

Worked example 1: ranking deviation

Three gases, helium, methane (CH₄) and ammonia (NH₃), are each held at 400 K and 3000 kPa. Rank them from most ideal to least ideal and explain your order.

Most ideal:  He   - single small atoms, very few electrons,
                    only very weak London forces
Middle:      CH₄  - non-polar, but larger with more electrons,
                    so stronger London forces than He
Least ideal: NH₃  - polar and forms hydrogen bonds between
                    molecules, so the strongest attractions

The high pressure makes all three less ideal than they would be at 100 kPa. The ranking comes from the strength of the intermolecular forces.

Summary of conditions. A gas comes closest to ideal behaviour at high temperature and low pressure, and deviates most at low temperature and high pressure.

Structure 1.5.3 Molar volume and the gas relationships

Molar volume

The molar volume, Vm, is the volume occupied by one mole of gas. For an ideal gas, Vm depends only on temperature and pressure, not on which gas it is. This is Avogadro’s law (Structure 1.4.6) in another form: equal volumes at the same temperature and pressure hold equal numbers of particles.

At standard temperature and pressure (STP), 273.15 K (0 °C) and 100 kPa, any ideal gas has a molar volume of 22.7 dm³ mol⁻¹. At any other conditions, work Vm out from the ideal gas equation: Vm = RT/P.

amount of gas, n = V ÷ Vm      (V and Vm in the same units)

Worked example 2: molar volume away from STP

Calculate the molar volume of an ideal gas at 35.0 °C and 95.0 kPa, and the volume of 0.150 mol of gas under these conditions. Use R = 8.31 J K⁻¹ mol⁻¹.

T  = 35.0 + 273.15 = 308.15 K
P  = 95.0 kPa = 9.50 × 10⁴ Pa
Vm = RT / P = (8.31 × 308.15) ÷ (9.50 × 10⁴)
   = 2.696 × 10⁻² m³ mol⁻¹ = 27.0 dm³ mol⁻¹
V  = n × Vm = 0.150 × 26.96 = 4.04 dm³

The value exceeds 22.7 dm³ mol⁻¹ because the gas is warmer and at lower pressure than STP.

The three relationships for a fixed mass of gas

Hold the amount of gas fixed and change one variable while keeping a second constant. The guide does not assess the names of these laws, so learn the relationships and the graph shapes instead.

Held constant	Relationship	Graph that gives a straight line through the origin
Temperature	P ∝ 1/V (PV is constant)	P against 1/V
Pressure	V ∝ T, with T in kelvin	V against T/K
Volume	P ∝ T, with T in kelvin	P against T/K

Graph shapes you should be able to sketch and read:

• P against V at constant temperature is a curve that falls steeply and then levels off, never touching either axis.
• PV against P at constant temperature is a horizontal line for an ideal gas.
• V against temperature in °C (constant pressure) is a straight line that misses the origin. Extended backwards, it meets the temperature axis close to −273 °C, which is 0 K.

In the laboratory, a gas syringe sealed with a pressure sensor lets you change volume at constant temperature. A sealed flask or syringe in a water bath lets you change temperature.

Worked example 3: finding absolute zero from data

A fixed mass of air is trapped in a syringe at constant pressure. These volumes were recorded:

t / °C	10.0	30.0	50.0	70.0	90.0
V / cm³	51.8	55.5	59.2	62.8	66.5

Use the data to estimate the temperature at which the volume would become zero.

The points lie on a straight line (volume rises by about
3.7 cm³ for every 20.0 °C).

gradient = (66.5 − 51.8) ÷ (90.0 − 10.0) = 0.184 cm³ °C⁻¹
intercept at 0 °C: 51.8 − (0.18375 × 10.0) = 49.96 cm³
V = 0 when t = −49.96 ÷ 0.18375 = −272 °C

The estimate, −272 °C, is close to absolute zero, −273.15 °C. A real gas would condense long before then, so the line extrapolates ideal behaviour.

Structure 1.5.4 The ideal gas equation and the combined gas law

The three relationships, together with Avogadro’s law, combine into the ideal gas equation:

PV = nRT
P  pressure in Pa
V  volume in m³
n  amount of gas in mol
R  gas constant, 8.31 J K⁻¹ mol⁻¹
T  temperature in K

For a fixed amount of gas moving from one set of conditions to another, nR is constant, which gives the combined gas law:

P₁V₁ / T₁ = P₂V₂ / T₂

According to the guide, the data booklet supplies R, both equations and the molar volume at STP, and volume and pressure must be in SI units. In the combined gas law any consistent volume unit and pressure unit will cancel, but T must always be in kelvin.

Conversions you will need:

kPa → Pa      × 10³          MPa → Pa      × 10⁶
dm³ → m³      × 10⁻³         cm³ → m³      × 10⁻⁶
°C  → K       + 273.15

Worked example 4: amount and mass in a cylinder

A 12.0 dm³ cylinder holds argon at 2.50 MPa and 20.0 °C. Calculate the amount and the mass of argon. (Ar of Ar = 39.95)

P = 2.50 × 10⁶ Pa    V = 12.0 × 10⁻³ m³    T = 293.15 K
n = PV / RT = (2.50 × 10⁶ × 12.0 × 10⁻³) ÷ (8.31 × 293.15)
  = 12.31 mol  →  12.3 mol
m = n × M = 12.31 × 39.95 = 492 g

Worked example 5: molar mass from experimental data

The guide links this equation to finding the molar mass of a gas from experimental results. Gas from a lighter is released into a dry gas syringe. The lighter loses 0.236 g and the syringe collects 98.0 cm³ of gas at 22.0 °C and 101.0 kPa. Calculate the molar mass and suggest the identity of the gas.

P = 1.010 × 10⁵ Pa    V = 98.0 × 10⁻⁶ m³    T = 295.15 K
n = PV / RT = (1.010 × 10⁵ × 98.0 × 10⁻⁶) ÷ (8.31 × 295.15)
  = 4.036 × 10⁻³ mol
M = m / n = 0.236 ÷ 4.036 × 10⁻³ = 58.5 g mol⁻¹

The value is close to butane, C₄H₁₀ (M = 58.12 g mol⁻¹). The small difference comes from experimental error and from the gas not being perfectly ideal.

Worked example 6: combined gas law

A balloon holds 3.20 m³ of helium at 101 kPa and 15.0 °C. It rises to a height where the pressure is 35.0 kPa and the temperature is −40.0 °C. Calculate its new volume, assuming no gas escapes.

T₁ = 288.15 K    T₂ = 233.15 K
V₂ = P₁V₁T₂ / (T₁P₂)
   = (101 × 3.20 × 233.15) ÷ (288.15 × 35.0)
   = 7.47 m³

The pressure falls to about a third, nearly tripling the volume; the cooling shrinks it slightly. Checking directions like this catches inverted ratios.

Common errors

• Using °C in PV = nRT or the combined gas law. A temperature of 0 °C would give a zero volume or a division by zero. Always add 273.15.
• Leaving kPa and dm³ in PV = nRT. kPa with dm³ happens to cancel (10³ × 10⁻³), but kPa with cm³ or Pa with dm³ gives answers out by a factor of a thousand. Convert both to Pa and m³ every time.
• Using 22.7 dm³ mol⁻¹ at non-STP conditions. It is only the molar volume at 273.15 K and 100 kPa.
• Writing “high temperature” as a cause of deviation. Deviation is greatest when the gas is cold and compressed.
• Explaining deviation with the wrong assumption. High pressure is mainly about particle volume; low temperature is mainly about intermolecular attractions.

Where to go next

Next, condense this unit with the Structure 1.5 revision notes and then try the Structure 1.5 questions. For the wider component, see the Structure 1 summary notes.

Official syllabus

This page is based on the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Structure 1.5.