IB DP Chemistry Structure 1.5: Ideal gases – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

The questions below test Structure 1.5, “Ideal gases”, from the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023). Each of understandings 1.5.1, 1.5.2, 1.5.3 and 1.5.4 appears at least once. Because ideal gases have no AHL statements, SL and HL candidates can attempt every question, for the course examined from 2025 onwards.

Learn the unit first from the ideal gases study guide and the condensed notes. Other DP Chemistry units are listed on the course hub, the checklist tracks what you have covered, and a free diagnostic points to weak topics.

Use these data throughout: R = 8.31 J K⁻¹ mol⁻¹. Molar volume of an ideal gas at STP (273.15 K, 100 kPa) = 22.7 dm³ mol⁻¹. T/K = t/°C + 273.15. Ar values: H 1.01, C 12.01, N 14.01, O 16.00, Na 22.99, Zn 65.38. Round final answers to three significant figures unless a question says otherwise.

Questions

1. The ideal gas model treats every collision between gas particles as elastic.

(a) State what “elastic” means in this context. [1] (b) State the two assumptions the model makes about the particles’ volume and the forces between them. [2]

2. Real gases do not always behave as the ideal gas model predicts.

(a) Explain why a real gas deviates from ideal behaviour at high pressure. [2] (b) Explain why a real gas deviates from ideal behaviour at low temperature. [2]

3. Samples of nitrogen, N₂, and water vapour, H₂O, are held at 420 K and 150 kPa. Deduce, with a reason, which gas deviates more from ideal behaviour. [3]

4. An ideal gas is at 50.0 °C and 105 kPa.

(a) Calculate the molar volume of the gas, in dm³ mol⁻¹, under these conditions. [2] (b) Calculate the volume, in dm³, occupied by 0.0800 mol of the gas under the same conditions. [1]

5. A fixed mass of an ideal gas is studied.

(a) Describe the shape of a graph of pressure against volume at constant temperature. [1] (b) Describe the shape of a graph of volume against temperature in kelvin at constant pressure. [1] (c) Describe the graph of volume against temperature in °C at constant pressure, including where the extended line meets the temperature axis. [2]

6. The volume of a fixed mass of gas was changed at constant temperature, and the pressure measured.

V / cm³	20.0	25.0	40.0	50.0
P / kPa	150	120	75.0	60.0

(a) Show that these data are consistent with pressure being inversely proportional to volume. [2] (b) Predict the pressure when the volume is 32.0 cm³. [1] (c) The temperature during the experiment was 300 K. Calculate the amount of gas in the syringe. [2]

7. A 10.0 dm³ steel cylinder contains oxygen, O₂, at 20.0 MPa and 18.0 °C. Calculate the mass of oxygen in the cylinder, in kg. [3]

8. A gas syringe contains 75.0 cm³ of air at 19.0 °C and 100 kPa. The plunger is pushed in quickly so that the volume becomes 60.0 cm³ and the pressure reads 140 kPa. Calculate the temperature of the air immediately after compression, in °C. [3]

9. A 0.200 g sample of a volatile liquid is injected into a gas syringe held at 99.0 °C. It vaporises completely, giving 104 cm³ of vapour at 101 kPa.

(a) Calculate the molar mass of the liquid. [3] (b) The liquid is one of propanone (C₃H₆O), ethanol (C₂H₅OH) or hexane (C₆H₁₄). Identify it, showing your reasoning. [1] (c) The experimental molar mass is slightly higher than the true value. Suggest a reason related to the behaviour of real gases. [2]

10. A student adds 0.392 g of zinc to excess dilute sulfuric acid and collects the hydrogen in a gas syringe.

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

(a) Calculate the amount, in mol, of zinc used. [1] (b) Calculate the volume of hydrogen, in cm³, expected at 24.0 °C and 99.5 kPa. [3] (c) Calculate the volume this hydrogen would occupy at STP. [1] (d) The measured volume was smaller than the value in (b). Suggest one experimental reason. [1] (e) State why hydrogen is expected to behave almost ideally under these conditions. [1]

11. In one design of air bag, sodium azide decomposes to produce nitrogen.

2NaN₃(s) → 2Na(s) + 3N₂(g)

The bag must fill to 60.0 dm³ at 30.0 °C and 125 kPa.

(a) Calculate the amount of nitrogen needed. [2] (b) Calculate the mass of sodium azide required. [2] (c) Calculate the volume of the gas if it cooled to 5.0 °C at constant pressure. [2] (d) Explain why the ideal gas equation gives a reliable estimate for nitrogen under these conditions. [2]

Answers

1. (a) The total kinetic energy of the particles is the same before and after a collision [1]. [1] (b) The volume of the particles is negligible compared with the volume of the container [1]; there are no intermolecular forces between the particles [1]. [2] Examiner insight: “No forces” on its own is vague; say intermolecular forces, and say negligible volume rather than “no volume”.

2. (a) At high pressure the particles are pushed close together [1], so the volume of the particles themselves is no longer negligible compared with the container volume [1]. [2] (b) At low temperature the particles move more slowly [1], so the attractive intermolecular forces between them are no longer negligible and reduce the force of collisions with the walls [1]. [2] Examiner insight: Match each condition to its failed assumption; intermolecular forces alone do not answer the high-pressure part.

3. Water vapour deviates more [1]. Water molecules are polar and form hydrogen bonds with each other [1], whereas nitrogen is non-polar with only weak London (dispersion) forces, so attractions between H₂O molecules are much stronger [1]. [3] Examiner insight: Compare both gases; describing only water’s forces leaves the comparison incomplete.

4. (a) T = 323.15 K and P = 1.05 × 10⁵ Pa; Vm = RT/P = (8.31 × 323.15) ÷ (1.05 × 10⁵) [1] = 2.558 × 10⁻² m³ mol⁻¹ = 25.6 dm³ mol⁻¹ [1]. [2] (b) V = 0.0800 × 25.58 = 2.05 dm³ [1]. [1] Examiner insight: 22.7 dm³ mol⁻¹ is wrong here because the conditions are not STP.

5. (a) A curve: pressure falls as volume rises, steeply at first then levelling off [1]. [1] (b) A straight line through the origin [1]. [1] (c) A straight line with a positive gradient that does not pass through the origin [1]; extended back, it meets the temperature axis at about −273 °C (0 K) [1]. [2] Examiner insight: P against V is never a straight line, and V against °C never passes through the origin; label axes with units when sketching.

6. (a) PV for each pair: 20.0 × 150 = 3000; 25.0 × 120 = 3000; 40.0 × 75.0 = 3000; 50.0 × 60.0 = 3000 kPa cm³ [1]. PV is constant, so P is inversely proportional to V [1]. [2] (b) P = 3000 ÷ 32.0 = 93.8 kPa [1]. [1] (c) PV = 3000 × 10³ Pa × 10⁻⁶ m³ = 3.00 J; n = PV/RT = 3.00 ÷ (8.31 × 300) [1] = 1.20 × 10⁻³ mol [1]. [2] Examiner insight: “Show that” needs every product shown, not one pair; a conclusion drawn from a single calculation is not a demonstration.

7. P = 2.00 × 10⁷ Pa, V = 1.00 × 10⁻² m³, T = 291.15 K [1]. n = PV/RT = (2.00 × 10⁷ × 1.00 × 10⁻²) ÷ (8.31 × 291.15) = 82.66 mol [1]. m = 82.66 × 32.00 = 2645 g = 2.65 kg [1]. [3] Examiner insight: MPa is 10⁶ Pa, not 10³ Pa; use M(O₂) = 32.00 g mol⁻¹, not 16.00, and give kilograms as asked.

8. T₁ = 292.15 K [1]. T₂ = T₁ × P₂V₂ / (P₁V₁) = 292.15 × (140 × 60.0) ÷ (100 × 75.0) = 327.2 K [1]. t = 327.2 − 273.15 = 54.1 °C [1]. [3] Examiner insight: Work in kelvin and convert to °C only at the end; substituting 19.0 °C directly gives a wrong 21.3 °C.

9. (a) P = 1.01 × 10⁵ Pa, V = 1.04 × 10⁻⁴ m³, T = 372.15 K [1]. n = PV/RT = (1.01 × 10⁵ × 1.04 × 10⁻⁴) ÷ (8.31 × 372.15) = 3.397 × 10⁻³ mol [1]. M = 0.200 ÷ 3.397 × 10⁻³ = 58.9 g mol⁻¹ [1]. [3] (b) M values: propanone 58.09, ethanol 46.08, hexane 86.20 g mol⁻¹; the closest is propanone [1]. [1] (c) Attractions between vapour molecules are not negligible (the vapour is not far above its boiling point) [1], so the real volume is smaller than an ideal gas would occupy, giving a smaller calculated n and a larger M [1]. [2] Examiner insight: In (c), trace the cause through n to M; “the gas is not ideal” alone does not explain the direction of the error.

10. (a) n(Zn) = 0.392 ÷ 65.38 = 5.996 × 10⁻³ mol [1]. [1] (b) n(H₂) = n(Zn) = 5.996 × 10⁻³ mol (1 : 1 ratio) [1]. V = nRT/P = (5.996 × 10⁻³ × 8.31 × 297.15) ÷ (9.95 × 10⁴) [1] = 1.488 × 10⁻⁴ m³ = 149 cm³ [1]. [3] (c) V = 5.996 × 10⁻³ × 22.7 = 0.136 dm³ = 136 cm³ [1]. [1] (d) Any one: gas escaped before the syringe was connected; the plunger stuck; the zinc was impure [1]. [1] (e) Hydrogen molecules are small and non-polar, so intermolecular forces are very weak (and the conditions are near room temperature and pressure) [1]. [1] Examiner insight: Convert the answer from m³ to cm³ as asked; 1.49 × 10⁻⁴ is correct only with the unit m³ attached.

11. (a) n = PV/RT = (1.25 × 10⁵ × 6.00 × 10⁻²) ÷ (8.31 × 303.15) [1] = 2.98 mol [1]. [2] (b) n(NaN₃) = 2.977 × 2/3 = 1.985 mol [1]; M(NaN₃) = 65.02 g mol⁻¹, so mass = 1.985 × 65.02 = 129 g [1]. [2] (c) V₂ = V₁ × T₂/T₁ = 60.0 × 278.15 ÷ 303.15 [1] = 55.1 dm³ [1]. [2] (d) Nitrogen is non-polar with only weak London forces [1]; the pressure is close to atmospheric and the temperature is far above its boiling point, so particle volume and attractions are negligible [1]. [2] Examiner insight: Use the mole ratio 2 : 3 the right way round; multiplying by 3/2 instead of 2/3 gives 290 g, a sign the ratio was inverted.

Where marks are usually lost

• Substituting °C instead of kelvin, especially in the combined gas law where the error is less obvious.
• Converting MPa as if it were kPa, or cm³ as if it were dm³.
• Applying 22.7 dm³ mol⁻¹ at conditions other than STP.
• Giving the failed assumption for the wrong condition in deviation questions.
• “Show that” answers that check only one data pair.
• Inverting mole ratios when linking gas volumes to reacting masses.
• Leaving the final answer in m³ when cm³ or dm³ was asked for.

Next steps

• Ideal gases revision notes
• Ideal gases study guide
• All DP Chemistry units on the hub
• DP Chemistry topic checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

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