The questions are on the ideal gas equation test. Take it under real conditions before reading the mark scheme.
Quick answer: An ideal gas obeys pV = nRT, where pressure, volume and kelvin temperature are linked through the amount of gas. This mark scheme gives the model answers and the marks for the 9 questions of the Physics 11-12 test on ideal gas equation, from ideal gas assumptions to no potential energy, with the equivalent answers to accept.
Marks are given for correct unit conversions, a clear formula, a correct substitution and a final answer with its unit. Accept answers within rounding error. Method marks can be earned even if an earlier numerical error carries through.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB B.1-B.3
Question 1 Ideal gas assumptions (2 marks)
Any two: molecules have negligible volume compared with the container; there are no intermolecular forces except during collisions; collisions are perfectly elastic; molecular motion is random. Two assumptions.
Marking: 1 mark for each valid assumption.
Question 2 Helium cylinder (3 marks)
- T = 290 K, V = 2.5 × 10⁻² m³. n = pV / RT = (4.0 × 10⁵ × 2.5 × 10⁻²) / (8.31 × 290) = 4.15 mol.
- N = nNA = 4.150 × 6.02 × 10²³ = 2.50 × 10²⁴ atoms.
Marking: (a) 1 mark for converting to kelvin and m³, 1 mark for the answer. (b) 1 mark.
Question 3 Heating fourfold (3 marks)
- The temperature rises by a factor of 4 and crms ∝ √T, so the speed increases by a factor √4 = 2.
- E = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 1200 = 2.48 × 10⁻²⁰ J.
Marking: (a) 1 mark. (b) 1 mark for the formula and substitution, 1 mark for the answer.
Question 4 Sealed can in a fire (2 marks)
On heating, the molecules gain kinetic energy and move faster. They hit the walls more often and with a larger change of momentum per collision. The force per unit area on the walls therefore increases, because the volume of the rigid can is constant. Pressure rises.
Marking: 1 mark for faster molecules, 1 mark for more frequent and harder collisions giving a larger force per area.
Question 5 Heating wax: melting (2 marks)
- The temperature is constant at 60 °C.
- The energy supplied is used to break the bonds between particles, increasing their potential energy rather than their kinetic energy, so the temperature does not rise.
Marking: 1 mark each.
Question 6 Heating wax: solid (2 marks)
Energy supplied in the first 100 s = Pt = 400 × 100 = 40 000 J. The temperature rises by 60 − 20 = 40 K. c = Q / (mΔT) = 40 000 / (0.50 × 40) = 2000 J/(kg K).
Marking: 1 mark for the energy, 1 mark for the substitution and answer.
Question 7 Heating wax: latent heat (2 marks)
The melting lasts from 100 s to 400 s, which is 300 s. Q = Pt = 400 × 300 = 1.2 × 10⁵ J. L = Q / m = 1.2 × 10⁵ / 0.50 = 2.4 × 10⁵ J/kg.
Marking: 1 mark for the energy, 1 mark for L.
Question 8 Kettle boiling dry (3 marks)
- E = Pt = 2200 × 90 = 1.98 × 10⁵ J.
- m = E / L = 1.98 × 10⁵ / 2.26 × 10⁶ = 0.088 kg (about 88 g).
Marking: (a) 1 mark. (b) 1 mark for the formula, 1 mark for the answer.
Question 9 No potential energy (1 mark)
The particles exert no forces on each other between collisions, so there is no potential energy stored. Internal energy is only random kinetic energy.
Marking: 1 mark.
