OCR A Level Physics A: Ideal gases and kinetic theory (H556) – 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 cover section 5.1.4 (Ideal gases) of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, part of the full A level. Revise first with the study guide or the revision notes. The course hub is OCR A Level Physics and the printable checklist lists each statement. Use NA = 6.02 × 10²³ mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.

Questions

1. A sealed flask contains 6.4 g of oxygen gas, O₂, of molar mass 32 g mol⁻¹.

(a) Calculate the amount of oxygen in moles. [1] (b) Calculate the number of molecules. [1] (c) Calculate the number of atoms. [1]

2. State three assumptions of the kinetic theory model of an ideal gas. [3]

3. Explain how the molecules of a gas produce a steady pressure on the walls of a container. [4]

4. A cylinder of volume 0.030 m³ holds an ideal gas at 1.8 × 10⁶ Pa and 285 K.

(a) Calculate the amount of gas in moles. [2] (b) Calculate the number of molecules. [1] (c) Some gas leaks out until the pressure is 1.2 × 10⁶ Pa at the same temperature. Calculate the amount of gas lost. [2]

5. Describe an experiment to test Boyle’s law for air, stating how you keep the conditions constant and how you analyse the results. [4]

6. A sealed flask of air at constant volume was heated in a water bath. The pressure p was measured at several temperatures θ.

θ / °C	10	30	50	70	90
p / kPa	97.0	104.0	111.0	118.0	125.0

(a) Calculate the gradient of a graph of p against θ. [2] (b) Use the graph to estimate absolute zero in °C. [2] (c) Suggest why the estimate differs from the accepted value of absolute zero and how the experiment could be improved. [2]

7. (a) Hydrogen gas at 1.0 × 10⁵ Pa has density 0.090 kg m⁻³. Calculate the mean square speed and the r.m.s. speed of its molecules. [2] (b) Five molecules have speeds of 2.0, 3.0, 5.0, 7.0 and 8.0 m s⁻¹. Calculate their mean speed, mean square speed and r.m.s. speed. [3]

8. (a) Use pV = ⅓Nm<c²> and pV = NkT to show that the mean kinetic energy of a molecule is 3/2 kT. [3] (b) Show that k = R/NA has the value 1.38 × 10⁻²³ J K⁻¹. [1]

9. Two Maxwell-Boltzmann distribution curves are drawn for the same sample of gas, one at 300 K and one at 600 K.

(a) Describe the general shape of the curve at 300 K. [2] (b) State and explain how the 600 K curve differs in the position of its peak, its height and the area under it. [3]

10. Neon atoms have mass 3.35 × 10⁻²⁶ kg and helium atoms 6.64 × 10⁻²⁷ kg.

(a) Calculate the r.m.s. speed of neon atoms at 450 K. [3] (b) Compare the mean kinetic energy of a neon atom with that of a helium atom at the same temperature. [1] (c) Calculate the ratio of the r.m.s. speed of helium to that of neon at the same temperature. [1]

11. (a) Explain why the internal energy of an ideal gas is equal to the total kinetic energy of its molecules. [1] (b) 3.0 mol of ideal gas is heated from 250 K to 400 K. Calculate the internal energy at each temperature and the increase. [3] (c) State what happens to the internal energy if the pressure is doubled at constant temperature, and explain. [1]

12. A container of volume 4.0 × 10⁻³ m³ holds 0.15 mol of argon (molar mass 0.040 kg mol⁻¹) at 330 K. Treat argon as an ideal gas.

(a) Calculate the pressure. [2] (b) Calculate the number of atoms. [1] (c) Calculate the mean kinetic energy of one atom. [2] (d) Calculate the internal energy of the gas. [2] (e) Calculate the r.m.s. speed of the atoms. [2]

Answers

1. (a) n = 6.4 / 32 = 0.20 mol [1] (b) N = 0.20 × 6.02 × 10²³ = 1.2 × 10²³ molecules [1] (c) Two atoms per molecule, so 2.4 × 10²³ atoms [1] Examiner insight: Part (c) is where answers lose credit, because the question names the particle, so counting molecules again does not answer it.

2. Any three from the following, one mark each [3]: a large number of molecules in random, rapid motion; molecules occupy negligible volume compared with the gas; collisions are perfectly elastic; each collision lasts a negligible time compared with the gaps between collisions; negligible forces between molecules except during collisions. Examiner insight: Write “negligible volume compared with the volume of the gas”, since “no volume” overstates the assumption.

3. A molecule hits the wall and rebounds, so its momentum changes. [1] The rate of change of momentum is the force on the molecule, so by Newton’s third law there is an equal force on the wall. [1] Very many molecules collide with the wall at random, so the average force is steady. [1] Pressure is this force divided by the area of the wall. [1] Examiner insight: Four marks need four linked steps, so name momentum, force, the large number of collisions and force per unit area.

4. (a) n = pV / RT = (1.8 × 10⁶ × 0.030) / (8.31 × 285) [1] = 23 mol [1] (b) N = 22.8 × 6.02 × 10²³ = 1.4 × 10²⁵ molecules [1] (c) At the same V and T, n is proportional to p, so n = 22.8 × (1.2 / 1.8) = 15.2 mol [1]; lost = 22.8 − 15.2 = 7.6 mol [1] Examiner insight: Keep the unrounded value of n for later parts, because early rounding to 23 mol would give a different final digit.

5. Trap a fixed mass of air in a sealed tube connected to a pressure gauge. [1] Change the volume in steps, waiting after each so the air returns to room temperature (change it slowly, so constant temperature is maintained). [1] Record p and V for each step in a table. [1] Plot p against 1/V: a straight line through the origin shows pV is constant. [1] Examiner insight: Say how temperature is kept constant, because a list of apparatus without that condition earns little.

6. (a) gradient = (125.0 − 97.0) / (90 − 10) [1] = 0.35 kPa °C⁻¹ [1] (b) p = 0 when θ = 10 − 97.0 / 0.35 [1] = −267 °C [1] (c) The line is extrapolated a long way beyond the data, so a small error in gradient makes a large error in the intercept. [1] Use a wider temperature range, keep the flask fully immersed and use a very short connecting tube. [1] Examiner insight: Extrapolation answers should give a number with its unit, and a comparison with the accepted value of about −273 °C.

7. (a) <c²> = 3p / ρ = 3 × 1.0 × 10⁵ / 0.090 = 3.3 × 10⁶ m² s⁻² [1]; c_rms = √(3.33 × 10⁶) = 1.8 × 10³ m s⁻¹ [1] (b) mean speed = 25.0 / 5 = 5.0 m s⁻¹ [1]; <c²> = (4 + 9 + 25 + 49 + 64) / 5 = 30.2 m² s⁻² [1]; c_rms = √30.2 = 5.5 m s⁻¹ [1] Examiner insight: Square first, then average, then take the root; taking the root before averaging is the usual error.

8. (a) ⅓Nm<c²> = NkT [1], so m<c²> = 3kT [1], so ½m<c²> = 3/2 kT, the mean kinetic energy of a molecule [1] (b) k = 8.31 / 6.02 × 10²³ = 1.38 × 10⁻²³ J K⁻¹ [1] Examiner insight: On a “show that” question, every line of the algebra is needed, including the cancelling of N.

9. (a) The curve starts at the origin, [1] rises to a peak and falls with a long tail at high speed, so it is not symmetrical. [1] (b) The peak moves to a higher speed [1] and is lower, with the curve broader, [1] because the molecules are spread over a wider range of speeds. The area is the same, because it represents the same number of molecules. [1] Examiner insight: Mention that the area is unchanged, and describe the shape change in words rather than leaving the sketch to speak for itself.

10. (a) c_rms = √(3kT / m) = √(3 × 1.38 × 10⁻²³ × 450 / 3.35 × 10⁻²⁶) [1] = √(5.56 × 10⁵) [1] = 7.46 × 10² m s⁻¹ [1] (b) They are equal, because mean kinetic energy is 3/2 kT. [1] (c) ratio = √(3.35 × 10⁻²⁶ / 6.64 × 10⁻²⁷) = 2.25 [1] Examiner insight: Part (b) is a reasoning mark, so state that mean kinetic energy depends only on temperature.

11. (a) There are negligible forces between the molecules except in collisions, so there is no potential energy. [1] (b) U = 3/2 nRT: at 250 K, 3/2 × 3.0 × 8.31 × 250 = 9.3 × 10³ J [1]; at 400 K, 1.5 × 10⁴ J [1]; increase = 5.6 × 10³ J [1] (c) It is unchanged, because U depends only on temperature for a fixed amount of ideal gas. [1] Examiner insight: Calculate the increase from the unrounded values, which gives 5.6 × 10³ J and not 5.7 × 10³ J.

12. (a) p = nRT / V = (0.15 × 8.31 × 330) / (4.0 × 10⁻³) [1] = 1.0 × 10⁵ Pa [1] (b) N = 0.15 × 6.02 × 10²³ = 9.0 × 10²² atoms [1] (c) 3/2 kT = 1.5 × 1.38 × 10⁻²³ × 330 [1] = 6.83 × 10⁻²¹ J [1] (d) U = 3/2 nRT = 1.5 × 0.15 × 8.31 × 330 [1] = 6.2 × 10² J [1] (e) m = 0.040 / 6.02 × 10²³ = 6.64 × 10⁻²⁶ kg; c_rms = √(3kT / m) [1] = 4.5 × 10² m s⁻¹ [1] Examiner insight: A good check is that N times the mean kinetic energy per atom gives the same internal energy as 3/2 nRT.

Where marks are usually lost

• Using °C where kelvin is needed.
• Not converting cm³ or kPa to SI units.
• Rounding intermediate values and drifting from the final answer.
• Giving “no volume” instead of negligible volume as an assumption.
• Leaving out the momentum, force and area chain when explaining pressure.
• Averaging the speeds before squaring when finding r.m.s. speed.
• Drawing Maxwell-Boltzmann curves that are symmetrical or reach the axis at high speed.
• Treating 3/2 kT as the energy of the whole sample.
• Omitting how temperature is kept constant in a Boyle’s law experiment.

Next steps

• Re-read the revision notes.
• Return to the study guide.
• Visit the OCR A Level Physics hub and the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), first assessment 2017, published by OCR: section 5.1.4, statements (a) to (i).