OxfordAQA International AS and A-level Physics: Ideal gases and kinetic theory (9630) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.11.3 (ideal gases) and 3.11.4 (kinetic theory). Section 3.11 is International A-level only.

Read the study guide and revision notes first. See also the course page, the checklist and the free diagnostics. Take 0 °C as 273 K, k as 1.38 × 10⁻²³ J K⁻¹, R as 8.31 J mol⁻¹ K⁻¹ and N_A as 6.02 × 10²³ mol⁻¹.

Questions

1. (a) Convert 25 °C and −73 °C to kelvin. [2] (b) State what is meant by absolute zero. [1]

2. A syringe traps 80 cm³ of air at 1.00 × 10⁵ Pa. The plunger is pushed in until the volume is 32 cm³. The temperature is constant. (a) Calculate the new pressure. [2] (b) Explain why the plunger should be pushed in slowly. [1] (c) State which graph of pressure gives a straight line through the origin. [1]

3. A sealed rigid flask contains air at 1.02 × 10⁵ Pa and 293 K. It is heated to 373 K. (a) Calculate the final pressure. [2] (b) Explain, in terms of molecules, why the pressure rises. [2]

4. A flask of volume 2.5 × 10⁻³ m³ contains 4.0 g of helium at 300 K. The molar mass of helium is 4.0 g mol⁻¹. Treat helium as an ideal gas. (a) Calculate the number of moles. [1] (b) Calculate the pressure. [2] (c) Calculate the mass of one helium atom. [1] (d) Calculate the number of atoms in the flask. [1]

5. A bubble of volume 0.80 cm³ forms at the bottom of a lake, at 3.0 × 10⁵ Pa and 6 °C. It rises to the surface, at 1.0 × 10⁵ Pa and 15 °C. Assume the gas is ideal and none dissolves. Calculate the volume at the surface. [4]

6. A gas in a cylinder expands against a piston at a constant pressure of 1.2 × 10⁵ Pa. Its volume increases from 3.0 × 10⁻³ m³ to 4.5 × 10⁻³ m³. (a) Calculate the work done. [2] (b) In a second expansion the pressure falls uniformly from 3.0 × 10⁵ Pa to 1.0 × 10⁵ Pa while the volume rises from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. Calculate the work done. [2] (c) State whether work is done on the gas or by the gas in part (a). [1]

7. Smoke in a lit glass cell is viewed through a microscope. Describe what is seen and explain what it shows about air. [4]

8. (a) State four assumptions made in the kinetic theory of an ideal gas. [4] (b) Give the reason an ideal gas has no potential energy, so that its internal energy is purely kinetic. [1]

9. Show that, for N molecules of mass m in a cube of side l, the pressure p and volume V satisfy pV = ⅓Nm(c_rms)². Start from one molecule moving with velocity component c_x towards a wall. [6]

10. A container of volume 0.020 m³ holds 4.0 × 10²³ molecules, each of mass 5.3 × 10⁻²⁶ kg. The mean square speed is 2.5 × 10⁵ m² s⁻². (a) State the root mean square speed. [1] (b) Calculate the pressure. [2] (c) Calculate the temperature. [2] (d) Calculate the average kinetic energy of a molecule. [1]

11. A vessel holds 0.40 mol of neon at 350 K. The molar mass of neon is 0.0200 kg mol⁻¹. (a) Calculate the average kinetic energy of a neon atom. [2] (b) Calculate the internal energy of the gas. [2] (c) Calculate c_rms for neon at this temperature. [3] (d) Helium has a molar mass of 0.0040 kg mol⁻¹. Calculate the temperature at which neon atoms have the same c_rms as helium atoms at 350 K. [2]

12. In a Charles’s law experiment, dry air is trapped by a liquid index in a uniform capillary tube sealed at one end. The tube is placed in a stirred water bath. The length l of the air column is measured at different temperatures θ.

θ / °C	10	30	50	70	90
l / mm	84.5	91.0	97.5	102.5	109.0

(a) Explain why l can be used in place of volume, and why the pressure of the air is constant. [2] (b) A line of best fit has gradient 0.3025 mm °C⁻¹ and cuts the l axis at 81.8 mm. Use these values to estimate absolute zero on the Celsius scale. [3] (c) Calculate the percentage difference between your estimate and −273 °C. [1] (d) Give two precautions that improve the reliability of the readings. [2]

Answers

1. (a) 25 + 273 = 298 K [1]. −73 + 273 = 200 K [1]. [2] (b) The lowest possible temperature, 0 K (about −273 °C), where an ideal gas would have zero volume (or pressure) when extrapolated [1]. [1] Examiner insight: Define absolute zero as 0 K, the lowest temperature. “Very cold” is not a definition.

2. (a) Boyle’s law: p₂ = p₁V₁ / V₂ [1] = 1.00 × 10⁵ × 80 / 32 = 2.5 × 10⁵ Pa [1]. [2] (b) Fast compression warms the gas, so T would change [1]. [1] (c) p against 1/V [1]. [1] Examiner insight: Units of volume cancel in a ratio, so cm³ is safe here. Show the ratio 80 / 32 so the method is visible.

3. (a) Volume is constant, so p₂ = p₁T₂ / T₁ [1] = 1.02 × 10⁵ × 373 / 293 = 1.30 × 10⁵ Pa [1]. [2] (b) Higher temperature means a higher average speed (kinetic energy) [1], so molecules hit the walls harder and more often, giving a larger force per unit area [1]. [2] Examiner insight: “Harder” and “more often” are separate ideas. Give both.

4. (a) n = 4.0 / 4.0 = 1.0 mol [1]. [1] (b) p = nRT / V [1] = (1.0 × 8.31 × 300) / 2.5 × 10⁻³ = 1.0 × 10⁶ Pa (9.97 × 10⁵ Pa) [1]. [2] (c) m = M / N_A = 0.0040 / 6.02 × 10²³ = 6.6 × 10⁻²⁷ kg [1]. [1] (d) N = nN_A = 6.0 × 10²³ [1]. [1] Examiner insight: Pair grams with g mol⁻¹, or convert both to SI.

5. The mass is fixed, so p₁V₁ / T₁ = p₂V₂ / T₂ [1]. T₁ = 279 K and T₂ = 288 K [1]. V₂ = (3.0 × 10⁵ × 0.80 × 288) / (279 × 1.0 × 10⁵) [1] = 2.5 cm³ (2.48 cm³) [1]. [4] Examiner insight: Show the kelvin values on the page.

6. (a) Work = pΔV [1] = 1.2 × 10⁵ × 1.5 × 10⁻³ = 180 J [1]. [2] (b) Average pressure = ½(3.0 + 1.0) × 10⁵ = 2.0 × 10⁵ Pa [1], so work = 2.0 × 10⁵ × 2.0 × 10⁻³ = 400 J [1]. [2] (c) The gas expands, so it does work by the gas on its surroundings [1]. [1] Examiner insight: For a straight-line graph use the trapezium area.

7. Bright specks of smoke move in small, jerky, random paths [1]. The particles are visible but air molecules are not [1]. The molecules, moving randomly, hit each particle with unequal numbers on opposite sides at any moment, so the net force changes size and direction [1]. This is evidence that air consists of molecules in continuous random motion [1]. [4] Examiner insight: The visible smoke particles move. The invisible molecules cause it.

8. (a) Four from: very many molecules in random motion [1]; negligible molecular volume compared with the container [1]; no intermolecular forces except in collisions [1]; elastic collisions [1]. Also acceptable: collision time negligible compared with the time between collisions; Newton’s laws apply. [4] (b) There are no forces between molecules except during collisions [1], so no potential energy and the internal energy is the sum of the molecular kinetic energies. [1] Examiner insight: Compare “negligible volume” with the container.

9. The rebound is elastic, so the momentum change is 2mc_x [1]. The molecule travels 2l between hits on one wall, so the time between hits is 2l / c_x [1]. Force = rate of change of momentum = 2mc_x / (2l / c_x) = mc_x² / l [1]. For N molecules, the total force is F = (Nm / l)⟨c_x²⟩, and the pressure is p = F / l² = Nm⟨c_x²⟩ / V [1]. Motion is random, so ⟨c²⟩ = ⟨c_x²⟩ + ⟨c_y²⟩ + ⟨c_z²⟩ = 3⟨c_x²⟩ [1]. Hence pV = ⅓Nm⟨c²⟩ = ⅓Nm(c_rms)² [1]. [6] Examiner insight: In a “show that” question write every step, including the reason for the factor of three, because the final line is given.

10. (a) c_rms = √(2.5 × 10⁵) = 5.0 × 10² m s⁻¹ [1]. [1] (b) p = Nm(c_rms)² / 3V [1] = (4.0 × 10²³ × 5.3 × 10⁻²⁶ × 2.5 × 10⁵) / (3 × 0.020) = 8.8 × 10⁴ Pa [1]. [2] (c) T = pV / Nk [1] = (8.83 × 10⁴ × 0.020) / (4.0 × 10²³ × 1.38 × 10⁻²³) = 3.2 × 10² K [1]. [2] (d) Kinetic energy = 3kT/2 = 1.5 × 1.38 × 10⁻²³ × 320 = 6.6 × 10⁻²¹ J [1]. [1] Examiner insight: Carry unrounded values from part (b) into part (c). Early rounding can shift the last figure.

11. (a) Kinetic energy = 3kT/2 [1] = 1.5 × 1.38 × 10⁻²³ × 350 = 7.2 × 10⁻²¹ J [1]. [2] (b) U = 3nRT/2 [1] = 1.5 × 0.40 × 8.31 × 350 = 1.7 × 10³ J [1]. [2] (c) m = 0.0200 / 6.02 × 10²³ = 3.32 × 10⁻²⁶ kg [1]. (c_rms)² = 2 × 7.245 × 10⁻²¹ / 3.32 × 10⁻²⁶ = 4.36 × 10⁵ [1], so c_rms = 6.6 × 10² m s⁻¹ [1]. [3] (d) c_rms is proportional to √(T/M), so T / 0.0200 = 350 / 0.0040 [1] and T = 1.8 × 10³ K (1750 K) [1]. [2] Examiner insight: In part (d) the ratio method is quicker.

12. (a) The tube has a uniform cross-section, so the volume is proportional to the length [1]. The index is free to move, so the pressure of the air equals atmospheric pressure [1]. [2] (b) l = 0 where 81.8 + 0.3025θ = 0 [1], so θ = −81.8 / 0.3025 [1] = −2.70 × 10² °C [1]. [3] (c) Difference = (273 − 270.4) / 273 × 100 = about 1% [1]. [1] (d) Two from: allow time for the air to reach the bath temperature before reading [1]; keep the whole air column below the water surface [1]. Also acceptable: stir the water; place the thermometer next to the tube. [2] Examiner insight: Give a precaution with its reason. “Be careful” is too vague to earn credit.

Where marks are usually lost

• Using Celsius temperatures in any gas-law equation.
• Using cm³, dm³ or grams in pV = nRT without converting.
• Using the mass of a mole where the mass of one molecule is needed.
• Quoting (c_rms)² as a speed, or calling c_rms the mean speed.
• Saying “faster molecules” to explain a change at constant temperature.
• Saying that the molecules themselves can be seen in Brownian motion.

Next steps

• Recheck the revision notes and the study guide.
• Visit the course page and the checklist.
• Take the free 10-minute diagnostics.
• Next, book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA. Sections 3.11.3 and 3.11.4.