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: H 1.008, He 4.003, C 12.01, O 16.00, Ne 20.18, Xe 131.29 g mol⁻¹; T (K) = T (°C) + 273. Treat every gas as ideal.
Question 1 (multiple choice · foundation)
A sample of argon gas is sealed in a rigid metal can and heated. Which statement best explains why the pressure in the can increases?
- (A) The argon atoms get larger, so they fill more of the can.
- (B) Heating produces more argon atoms, so there are more collisions.
- (C) The atoms move faster on average, so they hit the walls more often and with more force.
- (D) The attractions between argon atoms become stronger, so the atoms pull on the walls.
Answer and explanation
Answer: (C). Average kinetic energy is proportional to the Kelvin temperature. Faster atoms strike each unit of wall area more often, and each impact delivers more force, so the pressure rises.
- (A) Atoms do not change size when heated; in KMT their volume is negligible anyway.
- (B) The can is sealed, so the number of atoms is fixed.
- (D) KMT assumes no attractions between particles, and pressure comes from collisions, not from pulling.
Question 2 (multiple choice · core)
A sample of nitrogen gas is cooled from 400 K to 250 K. How does its Maxwell–Boltzmann speed distribution change?
- (A) The peak moves to a lower speed and becomes taller; the area under the curve is unchanged.
- (B) The peak moves to a lower speed and becomes shorter; the area under the curve is unchanged.
- (C) The peak moves to a higher speed and becomes taller; the area under the curve increases.
- (D) The peak stays at the same speed; the area under the curve decreases.
Answer and explanation
Answer: (A). At a lower temperature the molecules are slower on average, so the peak moves left. The speeds are spread over a narrower range, and because the number of molecules (the area) is unchanged, the curve must be taller.
- (B) A narrower curve with the same area cannot also be shorter.
- (C) Cooling lowers the average speed, so the peak cannot move to a higher speed. The area only changes if molecules are added or removed.
- (D) The peak position depends on temperature, and cooling does not remove molecules.
Question 3 (multiple choice · core)
Hydrogen, H₂, and oxygen, O₂, are at the same temperature. How does the average speed of the H₂ molecules compare with that of the O₂ molecules?
- (A) About 16 times greater
- (B) About 4 times greater
- (C) The same
- (D) About one quarter as great
Answer and explanation
Answer: (B). At the same temperature the average kinetic energies are equal, so v(H₂) / v(O₂) = √(M(O₂) / M(H₂)) = √(32.00 / 2.016) = √15.87 = 3.98, about 4.
- (A) is the ratio of the masses (15.87), not the speeds. You need the square root because KE depends on v².
- (C) confuses equal average kinetic energy with equal speed.
- (D) inverts the ratio: the lighter molecule is the faster one.
Question 4 (multiple choice · core)
A gas is heated from 100 °C to 200 °C. By what factor does the average kinetic energy of its particles increase?
- (A) 0.79
- (B) 1.13
- (C) 1.27
- (D) 2.00
Answer and explanation
Answer: (C). Average KE ∝ Kelvin temperature. T₁ = 373 K and T₂ = 473 K, so the factor is 473 / 373 = 1.27.
- (A) is 373 / 473: the ratio is upside down, which would mean the gas cooled.
- (B) is √(473 / 373), the factor for the average speed, not the kinetic energy.
- (D) divides the Celsius temperatures (200 / 100). Celsius values are not proportional to kinetic energy.
Question 5 (constructed response · core)
Two containers are at 300 K. One holds helium, He, and the other holds methane, CH₄. A student plots the Maxwell–Boltzmann speed distribution of each gas on the same axes. Curve X has its peak at 558 m s⁻¹. Curve Y is lower and wider, with its peak at 1116 m s⁻¹.
(a) Identify which curve belongs to helium. Justify your answer. (b) Compare the average kinetic energy of a helium atom with that of a methane molecule at 300 K. (c) Show that the ratio of the peak speeds is consistent with the molar masses of the two gases.
Worked solution and suggested Marlbridge rubric
(a) Curve Y is helium. The two gases are at the same temperature, so their particles have the same average kinetic energy. Helium atoms (4.003 g mol⁻¹) are lighter than methane molecules (16.04 g mol⁻¹), so they must move faster to carry the same kinetic energy. The curve peaking at the higher speed belongs to helium.
(b) They are equal. Average kinetic energy depends only on the Kelvin temperature, and both gases are at 300 K.
(c) Expected speed ratio: v(He) / v(CH₄) = √(16.04 / 4.003) = √4.01 = 2.00. Observed: 1116 / 558 = 2.00. The two agree.
| Point | What earns it |
|---|---|
| 1 | Curve Y identified as helium, with the reason that lighter particles move faster at equal average KE |
| 1 | Average kinetic energies equal, because the temperatures are equal |
| 1 | Square-root ratio calculated (2.00) and compared with 1116 / 558 = 2.00 |
Do not award (a) for “helium is lighter” alone; the link to equal average kinetic energy is needed.
Question 6 (constructed response · core)
A rigid 2.00 L flask contains a gas at 250 K and 0.900 atm. The flask is heated to 350 K.
(a) Calculate the new pressure. (b) Calculate the factor by which the average speed of the gas particles increases. (c) Use the kinetic molecular theory to explain the change in pressure. Refer to two separate effects of heating on the collisions. (d) A student draws particle diagrams of the flask before and after heating. State two features the “after” diagram should keep the same as the “before” diagram, and one feature that should be different.
Worked solution and suggested Marlbridge rubric
(a) At constant V and n, P ∝ T: P₂ = 0.900 atm × (350 K / 250 K) = 1.26 atm.
(b) Average KE rises by 350 / 250 = 1.40. Since KE ∝ v², the speed rises by √1.40 = 1.18.
(c) The particles move faster on average. (1) They reach the walls more often, so there are more collisions per second with each unit of wall area. (2) Each collision is more forceful, because the particles hit at higher speed. Both effects raise the force per unit area, so the pressure rises.
(d) Same: the number of particles; the size of the flask (and the size of the particles). Different: the velocity arrows should be longer on average (still in random directions, with a range of lengths).
| Point | What earns it |
|---|---|
| 1 | 1.26 atm with working |
| 1 | √1.40 = 1.18 for the speed (not 1.40) |
| 1 | Two effects in (c): more frequent collisions and more forceful collisions |
| 1 | Correct features in (d): same particle number and container, longer arrows on average |
Award the point in (c) only if both effects are given.
Question 7 (constructed response · stretch)
A container at constant temperature holds 0.60 mol of neon, Ne, and 0.20 mol of xenon, Xe. The total pressure is 2.00 atm.
(a) Calculate the factor by which the average speed of a neon atom exceeds that of a xenon atom. (b) A student claims: “Xenon exerts the larger partial pressure, because its atoms are much heavier and hit the walls harder.” Calculate both partial pressures and use the kinetic molecular theory to evaluate the claim. (c) The Maxwell–Boltzmann distributions of the two gases are plotted as fraction of atoms against kinetic energy. How do the two curves compare? Explain.
Worked solution and suggested Marlbridge rubric
(a) v(Ne) / v(Xe) = √(131.29 / 20.18) = √6.506 = 2.55.
(b) Mole fractions: X(Ne) = 0.60 / 0.80 = 0.75; X(Xe) = 0.25. Partial pressures: P(Ne) = 0.75 × 2.00 atm = 1.50 atm; P(Xe) = 0.25 × 2.00 atm = 0.50 atm. The claim is incorrect. A xenon atom is heavier, but it moves more slowly (2.55 times), so it hits the walls less often. At the same temperature both kinds of atom have the same average kinetic energy, so on average each atom contributes equally to the pressure. Partial pressure therefore depends on the number of atoms (mole fraction), and there are three times as many neon atoms.
(c) The two curves are identical. The distribution of kinetic energies depends only on the temperature, not on the mass of the particles. (On a speed axis they would differ: xenon would peak at a lower speed with a taller, narrower curve.)
| Point | What earns it |
|---|---|
| 1 | Speed ratio 2.55 with the square-root relationship |
| 1 | Partial pressures 1.50 atm (Ne) and 0.50 atm (Xe) |
| 1 | Claim rejected with KMT reasoning: heavier atoms are slower, so they collide less often; equal average KE means equal contribution per atom |
| 1 | Identical kinetic-energy curves, because the energy distribution depends only on temperature |
Accept a valid alternative in (b) that says partial pressure is independent of particle identity in an ideal mixture and links this to equal average kinetic energy.
How did you do?
- Q1 or Q6 wrong: re-read “Pressure from collisions” and Figure 1 in the study guide.
- Q3, Q4 or Q5(c) wrong: practise the square-root speed ratio and kelvin conversion in Worked examples 1 and 2.
- Q2 or Q7(c) wrong: compare Figures 2 and 3, and remember that the area under the curve stays the same.
Then tick off the topic checklist.