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 = 62.36 L torr mol⁻¹ K⁻¹; 1 atm = 760 torr; K = °C + 273.15. Molar masses: H 1.008, He 4.003, C 12.01, O 16.00, S 32.06, Cl 35.45, Ar 39.95 g mol⁻¹. Assume every gas behaves ideally. The experiment in Question 6 is invented, and its data are calculated for practice.
Question 1 (multiple choice · foundation)
For a fixed amount of an ideal gas, which graph is a straight line that passes through the origin?
- (A) P against V, at constant temperature
- (B) V against temperature in °C, at constant pressure
- (C) P against temperature in K, at constant volume
- (D) the product PV against P, at constant temperature
Answer and explanation
Answer: (C). With n and V fixed, P = (nR/V) × T, so P is directly proportional to the kelvin temperature. The graph is a straight line through the origin.
- (A) P and V are inversely proportional, so the graph is a curve that falls and flattens.
- (B) This graph is a straight line, but it meets zero volume at −273.15 °C, not at 0 °C, so it does not pass through the origin.
- (D) At constant n and T, PV = nRT is a constant, so the graph is a horizontal line, not one through the origin.
Question 2 (multiple choice · core)
A rigid, sealed steel container holds a gas at 1.20 atm and 27 °C. The container is heated to 327 °C. What is the new pressure?
- (A) 0.600 atm
- (B) 1.20 atm
- (C) 2.40 atm
- (D) 14.5 atm
Answer and explanation
Answer: (C). n and V are fixed, so P is proportional to T in kelvin. T₁ = 300.15 K and T₂ = 600.15 K, so P₂ = 1.20 atm × 600.15 ÷ 300.15 = 2.40 atm.
- (A) inverts the ratio, as if pressure fell on heating.
- (B) assumes that a rigid container keeps the pressure constant. It keeps the volume constant.
- (D) uses Celsius temperatures: 1.20 × 327 ÷ 27 = 14.5 atm.
Question 3 (multiple choice · core)
A container holds 4.0 g of helium and 32.0 g of oxygen, O₂. The total pressure is 3.00 atm. What is the partial pressure of helium?
- (A) 0.33 atm
- (B) 1.50 atm
- (C) 2.67 atm
- (D) 3.00 atm
Answer and explanation
Answer: (B). Convert to moles: He 4.0 ÷ 4.003 = 0.999 mol; O₂ 32.0 ÷ 32.00 = 1.000 mol. X(He) = 0.999 ÷ 1.999 = 0.500, so P(He) = 0.500 × 3.00 atm = 1.50 atm.
- (A) uses the mass fraction of helium (4.0 ÷ 36.0) instead of the mole fraction.
- (C) uses the mass fraction of oxygen (32.0 ÷ 36.0), and gives it to the wrong gas.
- (D) assumes each gas exerts the total pressure. Each gas exerts only its own share.
Question 4 (multiple choice · core)
A pure gas has a density of 2.86 g L⁻¹ at 0 °C and 1.00 atm. Which gas could it be?
- (A) Ar
- (B) CO₂
- (C) SO₂
- (D) Cl₂
Answer and explanation
Answer: (C). M = dRT / P = 2.86 g L⁻¹ × 0.08206 L atm mol⁻¹ K⁻¹ × 273.15 K ÷ 1.00 atm = 64.1 g mol⁻¹. SO₂ has M = 32.06 + 2(16.00) = 64.06 g mol⁻¹.
- (A) Ar (39.95 g mol⁻¹) and (B) CO₂ (44.01 g mol⁻¹) are too light; at these conditions their densities would be only 1.78 and 1.96 g L⁻¹.
- (D) Cl₂ (70.90 g mol⁻¹) is the answer you get if you use 298.15 K instead of 273.15 K (M = 70.0 g mol⁻¹). Read the temperature carefully.
Question 5 (calculation · core)
A student collects hydrogen gas over water at 25.0 °C. The gas occupies 245.0 mL, and the total pressure inside the collection tube is 752.0 torr. The vapour pressure of water at 25.0 °C is 23.8 torr.
(a) Calculate the partial pressure of the hydrogen. (b) Calculate the amount of hydrogen collected, in moles. (c) Calculate the mass of hydrogen collected. (d) Explain why the student must subtract the vapour pressure of water.
Worked solution and suggested Marlbridge rubric
(a) The tube contains hydrogen and water vapour, so P(H₂) = P_total − P(H₂O) = 752.0 − 23.8 = 728.2 torr.
(b) Use R in torr: n = PV / RT = (728.2 torr × 0.2450 L) ÷ (62.36 L torr mol⁻¹ K⁻¹ × 298.15 K) = 9.60 × 10⁻³ mol.
(c) m = 9.596 × 10⁻³ mol × 2.016 g mol⁻¹ = 0.0193 g.
(d) The gas in the tube is a mixture. Water evaporates into it, and the water vapour exerts its own partial pressure. Only the hydrogen’s share of the total pressure counts towards n(H₂). Using 752.0 torr would give 9.91 × 10⁻³ mol, about 3 per cent too high.
| Point | What earns it |
|---|---|
| 1 | Subtracts the vapour pressure: 728.2 torr |
| 1 | Volume in litres and temperature in kelvin, with R matching the pressure unit |
| 1 | Correct amount, 9.60 × 10⁻³ mol, and mass, 0.0193 g |
| 1 | Explains that water vapour contributes a partial pressure to the total |
Converting 728.2 torr to 0.9582 atm and using R = 0.08206 L atm mol⁻¹ K⁻¹ gives the same answer and is equally valid.
Question 6 (constructed response · core)
In a fictional experiment, a sample of gas is kept at a constant pressure of 1.00 atm. Its volume is measured at five temperatures.
| Temperature (°C) | 0 | 25 | 50 | 75 | 100 |
|---|---|---|---|---|---|
| Volume (L) | 0.897 | 0.979 | 1.061 | 1.143 | 1.225 |
(a) A graph of volume against temperature in °C is a straight line. Explain why it does not pass through the origin. (b) Use the data to estimate the temperature, in °C, at which the volume would extrapolate to zero. (c) Use one data point and PV = nRT to calculate the amount of gas in the sample. (d) Describe the graph you would get if you plotted the same volumes against temperature in kelvin.
Worked solution and suggested Marlbridge rubric
(a) Volume is proportional to the kelvin temperature, not the Celsius temperature. At 0 °C the gas is at 273.15 K, so it still has a volume (0.897 L). The line meets zero volume at about −273 °C, not at 0 °C.
(b) Slope = (1.225 − 0.897) L ÷ (100 − 0) °C = 0.00328 L °C⁻¹. Starting at 0.897 L at 0 °C, the volume reaches zero after 0.897 ÷ 0.00328 = 273.5 °C of cooling, so at about −273 °C.
(c) At 50 °C: n = PV / RT = (1.00 atm × 1.061 L) ÷ (0.08206 L atm mol⁻¹ K⁻¹ × 323.15 K) = 0.0400 mol. Any point gives the same value to 3 significant figures.
(d) A straight line through the origin, because V = (nR/P) × T. Its slope is nR/P = 0.00328 L K⁻¹.
| Point | What earns it |
|---|---|
| 1 | Links the non-zero intercept to proportionality with kelvin, not Celsius, temperature |
| 1 | Finds the slope from the data |
| 1 | Extrapolates to about −273 °C (accept −270 to −276 °C) |
| 1 | n = 0.0400 mol with T in kelvin |
| 1 | Straight line through the origin for V against T in K |
Question 7 (constructed response · stretch)
Flask A (2.00 L) contains N₂ at 1.50 atm. Flask B (3.00 L) contains O₂ at 0.500 atm. The flasks are joined by a tube of negligible volume with a closed valve. Both flasks are at 25 °C. The valve is opened, and the gases mix completely at constant temperature.
(a) Calculate the partial pressure of each gas after mixing. (b) Calculate the total pressure and the mole fraction of N₂ after mixing. (c) A student says: “The partial pressure of O₂ dropped because the N₂ molecules pushed the O₂ molecules out of the way.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Each gas now spreads through the total volume of 5.00 L, with its amount and the temperature unchanged, so P₁V₁ = P₂V₂ for each gas. P(N₂) = 1.50 atm × 2.00 L ÷ 5.00 L = 0.600 atm. P(O₂) = 0.500 atm × 3.00 L ÷ 5.00 L = 0.300 atm.
(b) P_total = 0.600 + 0.300 = 0.900 atm. X(N₂) = 0.600 ÷ 0.900 = 0.667. (Check: n is proportional to PV at fixed T, so n(N₂) : n(O₂) = 3.00 : 1.50 = 2 : 1, which also gives X(N₂) = 2/3.)
(c) The claim is incorrect. In an ideal mixture each gas behaves as if the others were absent. The O₂ partial pressure fell from 0.500 to 0.300 atm only because the same amount of O₂ now occupies a larger volume (3.00 L → 5.00 L). It would fall by exactly the same amount if Flask A were empty.
| Point | What earns it |
|---|---|
| 1 | P(N₂) = 0.600 atm using the total volume |
| 1 | P(O₂) = 0.300 atm using the total volume |
| 1 | Total 0.900 atm |
| 1 | X(N₂) = 0.667 from partial pressures or mole ratio |
| 1 | Rejects the claim and explains it by the larger volume, with partial pressures independent of other gases |
Do not award the point in (c) for “the claim is wrong” without the reason.
How did you do?
- Q1 or Q6 wrong: revisit Figure 1 and “Graphs of gas behaviour” in the study guide.
- Q2 or Q4 wrong: convert to kelvin and check which variables are constant; follow Worked examples 1 and 2.
- Q3, Q5 or Q7 wrong: practise mole fractions and partial pressures with Worked example 3.
Then tick off the topic checklist.