These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: R = 0.08206 L atm mol⁻¹ K⁻¹. Gases W, Y and Z are invented, and all their measured values are invented for practice.
Question 1 (multiple choice · foundation)
Under which conditions does a sample of nitrogen gas behave most like an ideal gas?
- (A) High temperature and low pressure
- (B) Low temperature and high pressure
- (C) Low temperature and low pressure
- (D) High temperature and high pressure
Answer and explanation
Answer: (A). At low pressure the molecules are far apart, so both their attractions and their own volume are negligible. At high temperature their kinetic energy is large compared with the attractions between them.
- (B) is the condition closest to condensation, where deviations are largest.
- (C) Low pressure helps, but at low temperature slow molecules are more affected by attractions.
- (D) High temperature helps, but at high pressure the molecules are crowded together, so particle volume and attractions both matter.
Question 2 (multiple choice · core)
Four gases are at 25 °C and 3 atm. Which gas shows the largest deviation from ideal behaviour?
- (A) Hydrogen, H₂
- (B) Methane, CH₄
- (C) Argon, Ar
- (D) Sulfur dioxide, SO₂
Answer and explanation
Answer: (D). At a moderate pressure the main cause of deviation is attraction between particles. SO₂ is a bent, polar molecule, so it has dipole–dipole forces, and with 32 electrons it also has the strongest London dispersion forces of the four. Its molecules attract each other most, so its measured pressure falls furthest below the ideal value.
- (A) H₂ has only 2 electrons and very weak dispersion forces: it is close to ideal.
- (B) CH₄ is nonpolar with 10 electrons: dispersion forces only, and weak ones.
- (C) Ar is nonpolar with 18 electrons. Its dispersion forces are stronger than those of H₂ and CH₄, but it has no dipole and fewer electrons than SO₂.
Question 3 (multiple choice · core)
A rigid container holds a known amount of a gas at 220 K. The measured pressure is lower than the value calculated with P = nRT/V. Which statement best explains this?
- (A) Attractions between the gas particles reduce the frequency and force of their collisions with the walls.
- (B) The gas particles take up part of the container’s volume.
- (C) Kinetic energy is lost to heat in every collision between gas particles.
- (D) At low temperature the particles move faster than the ideal gas law predicts.
Answer and explanation
Answer: (A). A measured pressure below the ideal value means attractions are the dominant effect. Particles are pulled back by their neighbours as they approach the walls, so they hit the walls less often and less forcefully. The low temperature makes this worse, because slow particles are more affected by attractions.
- (B) Particle volume makes the measured pressure higher, not lower, than ideal.
- (C) Lost kinetic energy is not the cause. If the particles kept losing energy, the temperature would keep falling, but the gas stays at 220 K. Real-gas deviations come from attractions and particle volume.
- (D) The average speed depends only on temperature and mass; lower temperature means slower, not faster, particles.
Question 4 (multiple choice · core)
1.00 mol of a gas is compressed to 1000 atm at 273 K. The ideal gas law predicts a volume of 22.4 mL. How does the actual volume compare, and why?
- (A) Greater than 22.4 mL, because the particles themselves occupy a large share of the space.
- (B) Less than 22.4 mL, because attractions pull the particles closer together.
- (C) Equal to 22.4 mL, because the kinetic molecular theory applies at all pressures.
- (D) Greater than 22.4 mL, because attractions push the particles further apart.
Answer and explanation
Answer: (A). V_ideal = nRT/P = (1.00)(0.08206)(273) / 1000 = 0.0224 L = 22.4 mL. At such an extreme pressure the particles fill much of that space, so the gas cannot be squeezed as far as the ideal law predicts. Particle volume dominates and the real volume is larger.
- (B) describes the effect of attractions, which dominates at moderate pressures, not at extreme ones.
- (C) The KMT assumptions of negligible particle volume and no attractions fail at very high pressure.
- (D) gives the right direction for the wrong reason: attractions pull particles together; they never push them apart.
Question 5 (constructed response · core)
A rigid 0.500 L vessel contains 0.800 mol of gas Z at 200 K. The measured pressure is 21.7 atm.
(a) Calculate the pressure predicted by the ideal gas law. (b) Calculate PV/nRT for the sample and the percentage by which the measured pressure differs from the ideal value. (c) Identify the dominant cause of the deviation. Justify your answer at the particle level. (d) The vessel is heated to 400 K. Predict whether PV/nRT moves closer to 1 or further from 1. Justify your prediction.
Worked solution and suggested Marlbridge rubric
(a) P_ideal = nRT/V = (0.800 mol)(0.08206 L atm mol⁻¹ K⁻¹)(200 K) / 0.500 L = 26.3 atm.
(b) PV/nRT = 21.7 / 26.26 = 0.826. The measured pressure is (26.26 − 21.7) / 26.26 × 100 = 17.4% lower than ideal.
(c) Attractions between the particles. The ratio is below 1. Particles near a wall are pulled back by neighbouring particles, so they collide with the walls less often and with less force than ideal particles would.
(d) Closer to 1. At 400 K the particles have twice the average kinetic energy, so the attractions are small by comparison and have less effect on their motion. The gas is further from the conditions that cause condensation.
| Point | What earns it |
|---|---|
| 1 | 26.3 atm with working |
| 1 | Ratio 0.826 and 17.4% (or a correct carried-forward value) |
| 1 | Attractions identified and linked to fewer or weaker wall collisions |
| 1 | “Closer to 1” with the reason that kinetic energy becomes large compared with the attractions |
Do not award (c) for “intermolecular forces” alone without the link to collisions and pressure.
Question 6 (constructed response · core)
The table shows PV/nRT for 1.00 mol of gas W at two temperatures.
| P (atm) | 10 | 50 | 100 | 200 |
|---|---|---|---|---|
| PV/nRT at 200 K | 0.97 | 0.86 | 0.77 | 0.93 |
| PV/nRT at 400 K | 1.00 | 1.00 | 1.00 | 1.05 |
(a) At which temperature does gas W behave more ideally? Explain why, in terms of particle motion. (b) Explain why PV/nRT at 200 K falls from 0.97 to 0.77 as the pressure rises from 10 atm to 100 atm. (c) Explain why PV/nRT at 200 K rises again between 100 atm and 200 atm.
Worked solution and suggested Marlbridge rubric
(a) At 400 K: the values stay within about 0.05 of 1. The particles have greater average kinetic energy, so the attractions between them have little effect on their motion.
(b) As the pressure increases, the particles are pushed closer together. Closer particles feel stronger attractions, which pull them back from the walls. The measured pressure (or volume) falls further below the ideal value, so the ratio drops.
(c) At higher pressure the particles fill a growing share of the container. The free space for motion is smaller than the container volume, which raises the real volume (or pressure) relative to ideal. This particle-volume effect now grows faster than the effect of attractions, so the ratio rises back towards 1. Attractions still dominate at 200 K and 200 atm (0.93 is below 1), but at even higher pressures the ratio would go above 1.
| Point | What earns it |
|---|---|
| 1 | 400 K, with the link to greater kinetic energy compared with attractions |
| 1 | (b) closer particles → more effect of attractions → measured value below ideal |
| 1 | (c) particle volume becomes significant and its effect grows faster than that of attractions |
Accept “the gas is further from condensation at 400 K” in (a) if it is linked to kinetic energy or attractions.
Question 7 (constructed response · stretch)
1.00 mol of gas Y is held at 300 K and 600 atm. Its measured volume is 62.0 mL. The molecules of gas Y themselves take up about 40 mL per mole.
(a) Calculate the volume predicted by the ideal gas law, in mL. (b) Calculate PV/nRT and state which effect dominates. (c) Explain, using the information about the molecules, why the measured volume is larger than the ideal volume. (d) A student claims: “Because gas Y’s molecules take up 40 mL per mole, gas Y will also deviate strongly from ideal behaviour at 1 atm and 300 K.” Evaluate the claim with a calculation.
Worked solution and suggested Marlbridge rubric
(a) V_ideal = nRT/P = (1.00)(0.08206)(300) / 600 = 0.04103 L = 41.0 mL.
(b) PV/nRT = V_measured / V_ideal = 62.0 / 41.03 = 1.51. Above 1, so particle volume dominates.
(c) The ideal gas law assumes all 41.0 mL is free space. But the molecules themselves take up about 40 mL, nearly the whole ideal volume. The gas cannot be compressed into a space that small and still leave room for the molecules to move, so the real volume is much larger than predicted.
(d) At 1 atm and 300 K, V_ideal = (1.00)(0.08206)(300) / 1 = 24.6 L = 24 600 mL. The molecules then take up 40 / 24 600 × 100 = 0.16% of the volume, which is negligible. The claim is incorrect: particle volume matters only when the gas is compressed into a volume not much bigger than the molecules themselves.
| Point | What earns it |
|---|---|
| 1 | 41.0 mL |
| 1 | Ratio 1.51 and particle volume identified |
| 1 | Explanation comparing the 40 mL taken up by the molecules with the 41.0 mL ideal volume |
| 1 | Claim rejected with the 0.16% (or 40 mL vs about 24.6 L) comparison |
Award (d) only if a numerical comparison is given.
How did you do?
- Q1 or Q6(a) wrong: re-read “Effect 1” and “Both effects together” in the study guide.
- Q2 wrong: revise “Predicting which gas deviates more” and the intermolecular forces from Topic 3.1.
- Q3, Q4, Q5 or Q7 wrong: practise the direction of each effect with Figure 2 and Worked examples 1 and 2.
Then tick off the topic checklist.