IB DP Chemistry Structure 1.5: Ideal gases – Revision Notes

Use these notes once you have learned the unit; the full teaching and longer worked examples are in the ideal gases study guide.

Scope: Structure 1.5, “Ideal gases”, as set out in the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023), understandings 1.5.1, 1.5.2, 1.5.3 and 1.5.4. No AHL statements exist for this sub-topic, so SL and HL candidates revise the same material. The notes apply to the course examined from 2025 onwards.

When you finish, attempt the ideal gases question set. Track progress on the DP Chemistry checklist, find other units from the IB DP Chemistry hub, and spot gaps with a free diagnostic. A one-page view of all five sub-topics sits in the Structure 1 summary.

Structure 1.5.1 The ideal gas model in four lines

• Particles move constantly and randomly.
• The particles’ own volume is negligible compared with the container.
• No intermolecular forces act between particles.
• Collisions are elastic: total kinetic energy is conserved.

Pressure is caused by particle collisions with the container walls. Raising the Kelvin temperature raises the average kinetic energy of the particles (Structure 1.1), so collisions with the walls become harder and more frequent.

Structure 1.5.2 Real gases: when and why the model fails

Condition	What changes	Assumption that fails
High pressure	Particles pushed close together; their own volume becomes a significant part of the container	Negligible particle volume
Low temperature	Particles move slowly, so attractions between them have time to act; wall collisions are softer	No intermolecular forces

• Most ideal: high temperature, low pressure, small non-polar particles (e.g. He, Ne, H₂).
• Least ideal: low temperature, high pressure, particles with strong intermolecular forces: hydrogen bonding (NH₃, H₂O) or large dipoles, or large, polarisable particles with strong London forces.
• No mathematical treatment is required for this understanding. Explain in words.

Must-know distinction. Volume effect versus force effect. At high pressure, say the particles’ volume is no longer negligible. At low temperature, say the attractive forces are no longer negligible. Using the wrong one is the commonest slip.

Structure 1.5.3 Molar volume and gas relationships

Molar volume, Vm: the volume of one mole of gas. For an ideal gas it depends only on T and P, not on the identity of the gas.

• At STP (273.15 K and 100 kPa): Vm = 22.7 dm³ mol⁻¹ for an ideal gas.
• At any other conditions: Vm = RT/P.
• n = V/Vm, with V and Vm in the same units.

Fixed mass of gas: relationships and graphs

Held constant	Relationship	Straight line through origin	Other graph shapes
T	P ∝ 1/V	P against 1/V	P against V: downward curve, never touching the axes; PV against P: horizontal line
P	V ∝ T (kelvin)	V against T/K	V against t/°C: straight line, intercept on the temperature axis near −273 °C
V	P ∝ T (kelvin)	P against T/K	P against t/°C: straight line, same intercept

The guide does not assess the names of these laws. Learn the relationships and graph shapes.

Method: reading a gas graph

1. Read the axis labels and units (°C or K? V or 1/V?).
2. Decide which variable was held constant.
3. A straight line through the origin = direct proportion.
4. Extrapolate a V–t/°C or P–t/°C line to V = 0 or P = 0
   to estimate absolute zero.
5. Use two well-separated points on the line for a gradient,
   never two adjacent data points.

Structure 1.5.4 Equations and units

Equation	Use it when	Units
PV = nRT	One set of conditions; finding n, V, P, T or (via n = m/M) the molar mass	R = 8.31 J K⁻¹ mol⁻¹ with pressure (Pa), volume (m³), temperature (K)
P₁V₁/T₁ = P₂V₂/T₂	Same sample of gas, conditions change	Any consistent P and V units; T must be in K
M = mRT/(PV)	Molar mass of a gas or vapour from mass, volume, T and P	m in g gives M in g mol⁻¹
Vm = RT/P	Molar volume at given conditions	m³ mol⁻¹; multiply by 10³ for dm³ mol⁻¹

According to the guide, the data booklet supplies R, both gas equations and the STP molar volume; volume and pressure must be in SI units.

Conversions

kPa × 10³ = Pa        MPa × 10⁶ = Pa
dm³ × 10⁻³ = m³       cm³ × 10⁻⁶ = m³
t/°C + 273.15 = T/K

Method: PV = nRT in steps

1. List P, V, T (and n or m) with units.
2. Convert to Pa, m³, K. Write the converted values down.
3. Rearrange symbolically first: n = PV/RT, V = nRT/P,
   T = PV/nR, P = nRT/V.
4. Substitute, calculate, keep extra figures in working.
5. Round the final answer to 3 significant figures, with units.
6. Sense-check: does a lab volume look like a lab volume?

Worked reminders

Moles from a volume at STP. 3.40 dm³ of an ideal gas at STP: n = 3.40 ÷ 22.7 = 0.150 mol.

Combined gas law, solving for T. A sample at 100 kPa and 20.0 °C is compressed to half its volume and its pressure becomes 230 kPa. Then T₂ = T₁ × (P₂V₂)/(P₁V₁) = 293.15 × (230 × 0.5)/(100 × 1) = 337 K, which is 64.0 °C. Convert back to °C only at the end, and only if asked.

Proportional reasoning without numbers. For a fixed mass, doubling the Kelvin temperature at constant volume doubles the pressure. Doubling the Celsius temperature does not.

Must-know distinctions

• Ideal gas versus real gas. Ideal is a model with four assumptions; real gases approach it at high T and low P.
• Molar volume at STP versus at other conditions. 22.7 dm³ mol⁻¹ applies only at 273.15 K and 100 kPa.
• Kelvin versus Celsius. Proportionality only works with kelvin. A rise from 20 °C to 40 °C is a rise from 293 K to 313 K, about 7%, not double.
• PV = nRT versus the combined gas law. The first needs SI units and n; the second only needs consistent units and the same sample.
• Avogadro’s law versus molar volume. Avogadro’s law (Structure 1.4.6) gives volume ratios between gases; molar volume turns a volume into an amount.

Quick self-test

1. State the assumption of the ideal gas model about collisions.
2. Which assumption fails most at high pressure?
3. Convert −18.0 °C to kelvin.
4. Name a type of intermolecular force that makes ammonia deviate more from ideal behaviour than methane under the same conditions.
5. Calculate the amount of gas in 0.500 dm³ at 27.0 °C and 150 kPa.
6. Calculate the molar volume of an ideal gas at 100.0 °C and 100 kPa.
7. A fixed mass of gas has its volume halved and its Kelvin temperature doubled. By what factor does the pressure change?
8. A gas occupies 240 cm³ at 20.0 °C and 100 kPa. Find its volume at 120.0 °C and 300 kPa.
9. A 0.880 g sample of gas occupies 0.500 dm³ at 25.0 °C and 100 kPa. Calculate its molar mass.
10. Which graph gives a straight line through the origin for a fixed mass of gas at constant temperature?
11. Sketch-describe the graph of PV against P for an ideal gas at constant temperature.

Answers

1. All collisions are perfectly elastic (no loss of total kinetic energy).
2. That the volume of the particles is negligible.
3. 255 K (255.15 K).
4. Hydrogen bonding (ammonia is also polar, so dipole–dipole forces add to it).
5. n = (1.50 × 10⁵ × 5.00 × 10⁻⁴) ÷ (8.31 × 300.15) = 3.01 × 10⁻² mol.
6. Vm = (8.31 × 373.15) ÷ (1.00 × 10⁵) = 3.10 × 10⁻² m³ mol⁻¹ = 31.0 dm³ mol⁻¹.
7. P = nRT/V; T × 2 and V ÷ 2 gives × 4.
8. V₂ = 240 × (100/300) × (393.15/293.15) = 107 cm³.
9. n = (1.00 × 10⁵ × 5.00 × 10⁻⁴) ÷ (8.31 × 298.15) = 2.018 × 10⁻² mol; M = 0.880 ÷ 0.02018 = 43.6 g mol⁻¹.
10. P against 1/V.
11. A horizontal straight line: PV stays constant as P changes.

Where marks are usually lost

• Substituting temperature in °C into PV = nRT or the combined gas law.
• Mixing kPa with cm³, or Pa with dm³, in PV = nRT, giving an answer out by a factor of 10³ or more.
• Using 22.7 dm³ mol⁻¹ when the question’s conditions are not STP.
• Naming high temperature or low pressure as the conditions for deviation; these are the conditions where gases are most ideal.
• Explaining high-pressure deviation with intermolecular forces only, without mentioning particle volume.
• Stating “no forces” for an ideal gas without saying intermolecular, which leaves the meaning unclear.
• Sketching V against t/°C through the origin, or P against V as a straight line.
• Taking a gradient from two adjacent data points instead of from the best-fit line.
• Rounding n early, then carrying the rounded value into a molar-mass division.
• Dropping units, or giving m³ when the question asks for cm³.

Official syllabus

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