The gas laws connect pressure, volume, temperature and amount of gas. You need the ideal gas equation in both forms, pV = nRT and pV = NkT, the Avogadro constant and molar quantities, and the experimental laws that lead to them, including data from gas-syringe experiments.
The kinetic model explains pressure as the result of molecular collisions and links the mean kinetic energy of a molecule to absolute temperature. Questions also ask when a real gas departs from ideal behaviour and how the internal energy of an ideal gas depends on temperature alone.
42 questions
195 marks
Paper 1A: 24
Paper 1B: 7
Paper 2: 11
Full mark schemes
Showing 42 of 42 questions · 195 marks
Tick questions to build a test
18 practice questions on B.3 Gas laws
1B-1A-03
The ideal gas equation·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two samples X and Y of different ideal gases are each kept in a sealed container. The graph shows how the product pV varies with the absolute temperature T for each sample.
The mass of sample X is twice the mass of sample Y. The molar mass of gas Y is M. What is the molar mass of gas X?
pV against T for samples X and Y (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1For an ideal gas pV = nRT, so the gradient of the pV–T line is nR, proportional to the amount of substance.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2From the graph: gradient of X = 72/300 = 0.24 J K−1, gradient of Y = 24/300 = 0.080 J K−1, so nX = 3nY.
—
Step 3Molar mass = mass/amount: MX/MY = (mX/mY) × (nY/nX) = 2 × ⅓ = ⅔, so MX = 2M/3.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses only the ratio of the gradients and ignores the fact that X has twice the mass: M/3 would be correct for equal masses.
BCorrect: twice the mass shared among three times as many moles gives a molar mass ⅔ that of Y.
CThis inverts the final ratio, taking the molar mass to be proportional to amount/mass instead of mass/amount.
DThis inverts the amount ratio: the steeper line X has three times as many moles as Y, not a third as many. With nX = nY/3, (2mY)/(nY/3) gives 6M.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = nRT; the amount of substance n as given by n = N/NACommand term: Deduce
2B-1A-19
The ideal gas equation·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A small bubble of air is released at the bottom of a lake, where the pressure of the water exceeds atmospheric pressure by 2.0 × 105 Pa and the temperature is 7 °C. The bubble rises to the surface, where the pressure is atmospheric, 1.0 × 105 Pa, and the temperature is 27 °C. No air enters or leaves the bubble.
What is (volume of the bubble at the surface)/(volume of the bubble at the bottom)?
Show mark scheme
Marking point
Mark
Notes
Step 1Pressure in the bubble at the bottom = 1.0 × 105 + 2.0 × 105 = 3.0 × 105 Pa; temperatures 280 K and 300 K.
—
All 3 steps must be completed — there is no mark for a part-answer.
AThis uses the excess pressure 2.0 × 105 Pa as the pressure at the bottom: 2.0 × 300/280 = 2.1.
BThis inverts the temperature ratio: 3.0 × 280/300 = 2.8.
CThis ignores the temperature change (Boyle's law only): 3.0/1.0 = 3.0.
DCorrect: absolute pressure 3.0 × 105 Pa and absolute temperatures: 3.0 × 300/280 = 3.2.
Syllabus understandingB.3 — that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant Command term: Determine
3B-1A-24
Molecular speeds·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A container holds a mixture of helium (molar mass 4 g mol−1) and nitrogen (molar mass 28 g mol−1) at the same temperature.
What is the ratio (rms speed of the helium atoms)/(rms speed of the nitrogen molecules)?
Show mark scheme
Marking point
Mark
Notes
Step 1At the same temperature the average translational kinetic energy per molecule, ½mvrms² = (3/2)kBT, is the same for both gases.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2So vrms ∝ 1/√m.
—
Step 3Ratio = √(28/4) = √7 = 2.6.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AEqual temperatures mean equal average kinetic energies, not equal speeds; the lighter particles move faster.
BCorrect: vrms ∝ m−1/2, so the ratio is √7 = 2.6.
CThis is the ratio of the masses, 28/4 = 7.0, not of the speeds (the square root is missing).
DThis is the square of the mass ratio, 7² = 49.
Syllabus understandingB.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = (3/2)kBT; B.3 — that ideal gases are described in terms of the kinetic theory Command term: Determine
4B-1A-41
Pressure and molecular speed·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A container of volume V holds N molecules of an ideal gas, each of mass m. The root mean square speed of the molecules is v. One wall of the container is flat and has area A.
What is the magnitude of the force exerted by the gas on this wall?
Show mark scheme
Marking point
Mark
Notes
Step 1Density of the gas: ρ = Nm/V.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Kinetic model: p = ⅓ρv2 = Nmv2/(3V).
—
Step 3Force on the wall: F = pA = Nmv2A/(3V).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: F = pA with p = Nmv2/(3V).
BThis is the pressure, not the force; it has not been multiplied by the area.
CThis omits the factor ⅓; on average only one third of v2 is associated with motion perpendicular to the wall.
DThis divides the pressure by the area instead of multiplying (p = F/A rearranged wrongly).
Syllabus understandingB.3 — pressure as given by p = F/A; that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases, related to the average translational speed of molecules as given by p = ⅓ρv2Command term: Deduce
5B-1A-42
The ideal gas equation·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Container X, of volume V, holds an ideal gas at pressure p. Container Y, of volume 3V, holds the same gas at pressure 5p. Both are at the same temperature. The containers are joined by a narrow tube of negligible volume and the temperature returns to its original value.
What is the final pressure of the gas?
Show mark scheme
Marking point
Mark
Notes
Step 1At constant temperature the amount of gas is proportional to pV, and the total amount is conserved.
—
All 3 steps must be completed — there is no mark for a part-answer.
AThis adds the pressures and divides by the volume ratio, (p + 5p)/4, ignoring that each pressure acts in a different volume.
BThis is the simple mean of the two pressures, which ignores the different volumes.
CThis spreads only the gas from Y into the total volume (5p × 3V/4V) and ignores the gas already in X.
DCorrect: the total pV, 16pV, now occupies 4V.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = nRT; the amount of substance Command term: Deduce
6B-1A-43
Pressure and molecular speed·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An ideal gas is held in a rigid sealed container. Half of the molecules are allowed to escape through a valve, which is then closed. The remaining gas is heated until the root mean square speed of its molecules is twice its original value.
What is the ratio (final pressure)/(original pressure)?
Show mark scheme
Marking point
Mark
Notes
Step 1p = ⅓ρv², where v is the root mean square speed.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The volume is fixed, so halving the number of molecules halves the density: ρ → ρ/2.
—
Step 3Doubling the rms speed multiplies v² by 4, so p → ⅓(ρ/2)(4v²) = 2p.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis includes the halving of the density but ignores the change in molecular speed.
BThis takes the pressure to be proportional to v rather than v²: ½ × 2 = 1.
CCorrect: p ∝ ρv², so the ratio is ½ × 2² = 2.
DThis includes the effect of the speed but forgets that the density has halved.
Syllabus understandingB.3 — that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by p = ⅓ρv² Command term: Deduce
7B-1A-44
Internal energy of an ideal gas·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two identical containers hold helium (molar mass 4 g mol−1) and argon (molar mass 40 g mol−1) respectively, both monatomic ideal gases. The two gases have the same pressure, the same volume and the same temperature.
Which quantities are the same for the two gases?
I. The number of atoms II. The internal energy III. The root mean square speed of the atoms
Show mark scheme
Marking point
Mark
Notes
Step 1I: N = pV/kBT is the same for both, whatever the mass of the atoms.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: U = ³⁄₂NkBT = ³⁄₂pV, the same for both.
—
Step 3III: ½mv2 = ³⁄₂kBT is the same, so the lighter helium atoms are faster, by √10 ≈ 3.2. III is different.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AOmits II: the internal energy of a monatomic ideal gas depends only on N and T, i.e. on pV, not on the atomic mass.
BOmits I: equal p, V and T mean equal numbers of atoms (Avogadro).
CCorrect: same N and same U = ³⁄₂pV, but different rms speeds.
DIncludes III: equal temperatures mean equal mean kinetic energies, not equal speeds.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT; the internal energy of an ideal monatomic gas U = ³⁄₂NkBT; B.1 — Ek = ³⁄₂kBTCommand term: Deduce
8B-1A-45
The kinetic model of an ideal gas·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksState
Which statement is an assumption of the kinetic model of an ideal gas?
Show mark scheme
Marking point
Mark
Notes
Step 1The kinetic model treats the molecules as point-like particles in random motion, with negligible total volume.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Collisions (with each other and with the walls) are elastic, and there are no intermolecular forces except during collisions.
—
Step 3So the internal energy of an ideal gas is entirely the random kinetic energy of the molecules; there is no intermolecular potential energy.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: with no forces between collisions the molecules move in straight lines at constant speed between collisions.
BThe model assumes elastic collisions; a gas left in an insulated container does not cool.
CThis is the opposite of the assumption: the volume of the molecules is negligible. It is a real-gas effect at high density.
DWith no intermolecular forces there is no intermolecular potential energy; this describes a real gas.
Syllabus understandingB.3 — that ideal gases are described in terms of the kinetic theory and constitute a modelled system used to approximate the behaviour of real gases Command term: State
9B-1A-46
Changes of state on a p–V diagram·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A fixed mass of ideal gas is taken from state P to state Q along the straight line shown on the p–V diagram.
What happens to the temperature of the gas during the change from P to Q?
A fixed mass of ideal gas is taken from P to Q along the straight line.Show mark scheme
Marking point
Mark
Notes
Step 1For a fixed mass of ideal gas T ∝ pV.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3At the midpoint (2.0 × 10−3 m³, 2.0 × 105 Pa): pV = 400 J, so the temperature is higher there. The line lies above the isotherm through P and Q: T rises to a maximum and then falls back.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AP and Q lie on the same isotherm, but an isotherm is a curve (p ∝ 1/V), not a straight line; between P and Q the gas is hotter.
BThis looks only at the increase in volume and ignores the fall in pressure.
CThis looks only at the fall in pressure and ignores the increase in volume.
DCorrect: pV goes 300 J → 400 J → 300 J along the line.
Syllabus understandingB.3 — changes of state of an ideal gas represented on pressure–volume diagrams; the ideal gas law pV/T = constant Command term: Deduce
10B-1A-66
The amount of substance·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A sealed flask of volume V contains a mass m of an ideal gas of molar mass M. NA is the Avogadro constant.
What is the number of gas molecules per unit volume in the flask?
Show mark scheme
Marking point
Mark
Notes
Step 1Amount of substance n = m/M.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Number of molecules N = nNA = mNA/M.
—
Step 3Per unit volume: N/V = mNA/(MV).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: (m/M) moles × NA molecules per mole, shared over the volume V.
BThis is the total number of molecules in the flask; it has not been divided by the volume.
CThis divides by the Avogadro constant instead of multiplying: n = N/NA rearranges to N = nNA.
DThis inverts the amount of substance, taking n = M/m instead of m/M.
Syllabus understandingB.3 — the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant Command term: Determine
11B-1A-67
The ideal gas equation·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two glass bulbs of equal volume are joined by a narrow tube of negligible volume. They contain an ideal gas at pressure p and temperature 300 K.
One bulb is kept at 300 K while the other is heated to 600 K. What is the pressure of the gas when equilibrium is reached?
Show mark scheme
Marking point
Mark
Notes
Step 1The bulbs are connected, so the final pressure p′ is the same in both; the total amount of gas is unchanged but some gas moves from the hot bulb to the cold bulb.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3p′(1/600 + 1/300) = 2p/300, so p′ × 3/600 = 2p/300 and p′ = 4p/3.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis inverts the final ratio: 2/300 ÷ 3/600 = 4/3, not 3/4. Heating one bulb cannot lower the pressure.
BCorrect: conservation of the amount of gas with the two bulbs at different temperatures gives p′ = 4p/3.
CThis assumes the whole gas behaves as if at the mean temperature of 450 K; the amounts in the two bulbs are not equal, so the temperatures cannot simply be averaged.
DThis treats the hot bulb as sealed (p ∝ T for a fixed amount); gas flows out of it until the pressures are equal.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = nRT; the empirical gas law for constant volume; the amount of substance nCommand term: Deduce
12B-1A-99
Translational kinetic energy of a gas·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A room of volume 50 m³ contains air at a pressure of 1.00 × 105 Pa and a temperature of 20 °C. Treat the air as an ideal gas.
What is the total translational kinetic energy of the air molecules in the room?
Show mark scheme
Marking point
Mark
Notes
Step 1The mean translational kinetic energy of a molecule is 3⁄2kBT, so the total is E = 3⁄2NkBT.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2For an ideal gas NkBT = pV, so E = 3⁄2pV: the temperature is not needed.
—
Step 3E = 1.5 × 1.00 × 105 × 50 = 7.5 × 106 J = 7.5 MJ.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis uses the Celsius temperature in 3⁄2NkBT after finding N correctly with 293 K: 1.5 × 1.24 × 1027 × 1.38 × 10−23 × 20 = 5.1 × 105 J.
BThis is pV = NkBT, which omits the factor 3⁄2 relating the mean kinetic energy to kBT.
CCorrect: the translational kinetic energy is 3⁄2pV for any ideal gas, whatever its temperature or molar mass.
DThis comes from pV = ⅓Nmv2 with the kinetic energy taken as mv2 instead of ½mv2, giving 3pV.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by p = ⅓ρv2; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = 3⁄2kBTCommand term: Determine
13B-1A-100
Internal energy of an ideal gas·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A rigid container holds a monatomic ideal gas at a pressure of 2.0 × 105 Pa and a temperature of 300 K. The container develops a small leak. At the same time the gas is warmed, and when its temperature has reached 360 K its pressure is 1.5 × 105 Pa.
What is (final internal energy of the gas in the container)/(initial internal energy of the gas in the container)?
Show mark scheme
Marking point
Mark
Notes
Step 1For a monatomic ideal gas U = 3⁄2nRT and pV = nRT, so U = 3⁄2pV.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The container is rigid, so V is constant and U ∝ p: the temperature change and the loss of gas are both already contained in the pressure.
—
Step 3Ratio = 1.5/2.0 = 0.75. (The amount of gas falls to 0.625 of its initial value while T rises by a factor 1.2: 0.625 × 1.2 = 0.75.)
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the ratio of the amounts of gas, (p/T)final ÷ (p/T)initial = 0.625; it ignores the rise in the mean kinetic energy of each atom.
BCorrect: U = 3⁄2pV with V fixed, so the ratio equals the pressure ratio, 0.75.
CThis takes U ∝ nT with n ∝ p, i.e. U ∝ pT, counting the temperature twice: 0.75 × 1.2 = 0.90.
DThis assumes that the internal energy depends only on the temperature, 360/300 = 1.2, and ignores the gas that has leaked out.
Syllabus understandingB.3 — the internal energy U of an ideal monatomic gas as given by U = 3⁄2NkBT = 3⁄2nRT; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRTCommand term: Deduce
14B-1A-101
Graphs of the empirical gas laws·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A fixed mass of an ideal gas is used in three experiments.
In which experiments does a plot of the first quantity against the second give a straight line through the origin?
I. Pressure against kelvin temperature, at constant volume II. Volume against Celsius temperature, at constant pressure III. Pressure against 1/volume, at constant temperature
Show mark scheme
Marking point
Mark
Notes
Step 1I: at constant V, p/T = constant, so p ∝ T (kelvin): a straight line through the origin.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: at constant p, V ∝ T in kelvin; on a Celsius axis the line meets V = 0 at −273 °C, not at the origin.
—
Step 3III: at constant T, pV = constant, so p ∝ 1/V: a straight line through the origin. Answer: I and III only.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis rejects III, perhaps by picturing the curved p–V graph (a hyperbola) rather than p against 1/V.
BThis rejects I, as if p ∝ T needed the Celsius scale; on the kelvin scale the pressure law gives a line through the origin.
CThis accepts II, treating the Celsius scale as absolute: the zero of the V–θ line is at −273 °C.
DCorrect: proportionality to the kelvin temperature (I) and to 1/V (III) give lines through the origin; a Celsius axis (II) does not.
Syllabus understandingB.3 — that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant; B.1 — the Kelvin and Celsius scales Command term: Deduce
15B-1A-102
Kinetic model and molecular mass·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An ideal gas has density ρ when its pressure is p and its temperature is T.
What is the mass of one molecule of the gas?
Show mark scheme
Marking point
Mark
Notes
Step 1pV = NkBT with ρ = Nm/V (total mass Nm), so p = (ρ/m)kBT.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Rearranging: m = ρkBT/p.
—
Step 3Check with the kinetic model: ⅓ρvrms2 = p and ½mvrms2 = 3⁄2kBT give the same result.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the number of molecules per unit volume is p/(kBT), so each molecule has mass ρ ÷ p/(kBT).
BThis inverts the expression: p/(kBT) is the number of molecules per unit volume, not the mass of one molecule.
CThis uses R instead of kB: ρRT/p is the molar mass M, the mass of one mole, not of one molecule.
DThis drops the ⅓ in p = ⅓ρv2 (writing v2 = p/ρ) before combining with ½mv2 = 3⁄2kBT.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by p = ⅓ρv2; B.1 — density ρ = m/VCommand term: Deduce
16B-1A-103
Real gases and the ideal gas model·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A sample of nitrogen is at a temperature of 300 K and a pressure of 1.0 × 105 Pa. It is cooled at constant pressure to 80 K, just above the temperature at which it condenses.
Which row describes the change in the number of molecules per unit volume, and how well the ideal gas model describes the nitrogen at 80 K compared with at 300 K?
number of molecules per unit volumeideal gas model at 80 K
Show mark scheme
Marking point
Mark
Notes
Step 1N/V = p/(kBT): at constant pressure, cooling from 300 K to 80 K increases the number of molecules per unit volume by a factor 300/80 = 3.75.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2At 80 K the molecules are closer together and move more slowly, so the intermolecular attractive forces are no longer negligible compared with their kinetic energy.
—
Step 3Near condensation the ideal gas model (no intermolecular forces, negligible molecular volume) is a worse description: increases / worse.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis takes cooling to reduce the density and assumes slow molecules behave more ideally; at constant pressure N/V ∝ 1/T increases.
BThis gets the second column right but inverts the first: N/V = p/(kBT), so the number density increases as T falls.
CThe density does increase, but a denser, colder gas close to condensation is a poorer approximation to an ideal gas, not a better one.
DCorrect: the number density rises and the gas moves towards the conditions (high density, low temperature) where the ideal gas model fails.
Syllabus understandingB.3 — the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRTCommand term: Deduce
17B-1A-104
Molecular explanation of the gas laws·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
An ideal gas is enclosed in a cylinder by a piston that moves freely, so the pressure of the gas stays constant. The gas is heated slowly.
Which row describes the changes in the average momentum change of a molecule in one collision with the piston, and in the number of collisions with each unit area of the piston per second?
average momentum change per collisioncollisions per unit area per second
Show mark scheme
Marking point
Mark
Notes
Step 1The mean kinetic energy of a molecule is proportional to T, so the molecules move faster and each collision changes the momentum of a molecule by more (2mvx increases).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Pressure = (number of collisions per unit area per second) × (average momentum change per collision).
—
Step 3The pressure is constant while the momentum change per collision increases, so the collision rate per unit area must decrease (the gas expands and its number density falls).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: faster molecules give bigger impulses, but the gas expands so its number density falls, and fewer collisions per unit area per second keep the product, the pressure, constant.
BThis applies the constant-volume reasoning (faster molecules hit the walls more often) and forgets that the pressure is constant, which requires the collision rate to fall.
CThis assumes that a constant pressure means an unchanged collision rate; the pressure is the product of the rate and the momentum change per collision, and the momentum change has increased.
DThis assumes that a constant pressure means an unchanged force per collision; the higher temperature means faster molecules and a larger momentum change in each collision.
Syllabus understandingB.3 — that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by p = ⅓ρv2; a qualitative explanation of the gas laws in terms of molecular behaviour; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles Command term: Deduce
18B-1A-105
Number of molecules per unit volume·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An ultra-high vacuum chamber contains gas at a pressure of 1.0 × 10−8 Pa and a temperature of 27 °C.
What is the number of gas molecules in each cm³ of the chamber?
Show mark scheme
Marking point
Mark
Notes
Step 1N/V = p/(kBT) with T = 27 + 273 = 300 K.
—
All 3 steps must be completed — there is no mark for a part-answer.
CThis converts m−3 to cm−3 by dividing by 102 (the factor for a length) instead of 106 (the factor for a volume).
DThis is the number per m³; it has not been converted to the number per cm³.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant Command term: Determine
19B-1A-106
Root mean square speed and molar mass·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The graph shows how the root mean square speed vrms of the atoms of a monatomic ideal gas varies with √T, where T is the kelvin temperature.
What is the molar mass of the gas?
Root mean square speed vrms against √T for a monatomic gas (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1½mvrms2 = 3⁄2kBT, and multiplying by NA gives vrms2 = 3RT/M, so vrms = √(3R/M) × √T.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2From the graph, at √(T/K) = 20 (T = 400 K) vrms ≈ 700 m s−1.
—
Step 3M = 3RT/vrms2 = 3 × 8.31 × 400/7002 = 0.0204 kg mol−1 ≈ 20 g mol−1 (neon).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis omits the factor 3: RT/vrms2 = 8.31 × 400/7002 = 6.8 g mol−1.
BThis writes the kinetic energy as mvrms2 instead of ½mvrms2, so M = 3⁄2RT/vrms2 = 10 g mol−1.
CCorrect: M = 3RT/vrms2 with a point read from the line.
DThis multiplies by 2 instead of dividing when removing the ½ (M = 6RT/vrms2), giving 41 g mol−1.
Syllabus understandingB.3 — that ideal gases are described in terms of the kinetic theory and constitute a modelled system used to approximate the behaviour of real gases; that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by p = ⅓ρv2; B.1 — Ek = 3⁄2kBTCommand term: Determine
20B-1A-107
Gas trapped by a piston·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A cylinder, closed at one end, contains an ideal gas trapped by a piston of mass m and cross-sectional area A that slides without friction. Atmospheric pressure is p0. When the closed end is at the bottom, the length of the gas column is L.
The cylinder is slowly turned upside down, as shown, and the temperature of the gas does not change. The piston stays inside the cylinder. What is the new length of the gas column?
The cylinder before and after it is turned upside down (not to scale)Show mark scheme
Marking point
Mark
Notes
Step 1Closed end at the bottom: the gas supports the piston and the atmosphere above it, p1A = p0A + mg.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Upside down: the piston hangs below the gas; the atmosphere pushes up on it, so p0A = p2A + mg and p2A = p0A − mg.
—
Step 3Boyle's law with V ∝ length: p1L = p2L′, so L′ = L(p0A + mg)/(p0A − mg). (For example, with p0 = 1.01 × 105 Pa, a 2.0 kg piston of area 10 cm² turns a 0.200 m column into a 0.296 m column.)
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis inverts the pressure ratio: in Boyle's law the length is inversely proportional to the pressure, and the pressure falls when the cylinder is inverted, so the column must get longer.
BThis treats the gas pressure in the inverted cylinder as atmospheric, ignoring the weight of the hanging piston.
CCorrect: the weight of the piston adds to the gas pressure in the first position and subtracts from it in the second.
DThis ignores the weight of the piston in the first position (taking p1 = p0) and includes it only after inverting.
Syllabus understandingB.3 — pressure as given by p = F/A, where F is the force exerted perpendicular to the surface; that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant; A.2 — weight and translational equilibrium Command term: Deduce
21B-1A-108
Gas escaping through a relief valve·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A rigid container of an ideal gas is fitted with a relief valve, which lets gas escape whenever the pressure inside would otherwise exceed a set value. The gas is heated from 250 K to 500 K. The graph shows how the pressure p of the gas varies with its temperature T.
What fraction of the gas originally in the container has escaped when the temperature reaches 500 K?
Pressure against temperature for the gas in the container (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1Up to 400 K the graph is a line through the origin (p ∝ T at constant V and n): no gas escapes until the valve opens at 400 K.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2From 400 K to 500 K, p and V are constant, so n ∝ 1/T: n500/n400 = 400/500 = 0.80.
—
Step 3The fraction that has escaped is 1 − 0.80 = 0.20.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes that, because the pressure is constant, no gas escapes; at constant p and V a rising temperature requires n to fall.
BCorrect: the amount of gas left is proportional to 1/T once the valve is open, so 1 − 400/500 = 0.20.
CThis divides by the wrong temperature: (500 − 400)/400 = 0.25 is the fractional increase in T, not the fraction of the gas lost.
DThis uses the starting temperature instead of the temperature at which the valve opens: 1 − 250/500 = 0.50.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant Command term: Deduce
22B-1A-109
States on a p–V diagram·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The pressure–volume diagram shows two states X and Y of a fixed mass of an ideal gas. The temperature of the gas in state X is 300 K.
What is the temperature of the gas in state Y?
States X and Y on a pressure–volume diagram (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1From the graph: X (1.5 × 10−3 m³, 4.0 × 105 Pa), Y (5.0 × 10−3 m³, 1.5 × 105 Pa).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2For a fixed mass pV/T = constant, so TY = TX × (pYVY)/(pXVX).
AThis uses only the pressure ratio (T ∝ p, as if the volume were constant): 300 × 1.5/4.0 = 113 K.
BThis inverts the ratio of the pV products: 300 × 6.0/7.5 = 240 K.
CCorrect: T ∝ pV, and pV rises from 600 J to 750 J.
DThis uses only the volume ratio (T ∝ V, as if the pressure were constant): 300 × 5.0/1.5 = 1000 K.
Syllabus understandingB.3 — changes of state of an ideal gas represented on pressure–volume diagrams (Guidance); that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant Command term: Determine
23B-1A-110
Equal volumes of different gases·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two identical flasks are at the same temperature and the same pressure. One contains hydrogen (molar mass 2.0 g mol−1) and the other contains oxygen (molar mass 32 g mol−1). Both gases behave as ideal gases.
What is (mass of oxygen)/(mass of hydrogen)?
Show mark scheme
Marking point
Mark
Notes
Step 1N = pV/(kBT): equal p, V and T give equal numbers of molecules (and equal amounts n).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Mass = nM, so the mass ratio equals the ratio of the molar masses.
—
Step 3Ratio = 32/2.0 = 16.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis inverts the ratio: oxygen molecules are the heavier ones, so the oxygen sample has the larger mass.
BThis confuses an equal number of molecules with an equal mass.
CThis takes the square root of the molar-mass ratio, as in a ratio of root mean square speeds; the masses are directly proportional to M.
DCorrect: the same number of molecules, each 16 times heavier.
Syllabus understandingB.3 — the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRTCommand term: Deduce
24B-1A-111
Density of a gas and temperature·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A fixed mass of an ideal gas is heated at constant pressure.
Which graph shows how the density ρ of the gas varies with its kelvin temperature T?
Four sketch graphs of density against kelvin temperatureShow mark scheme
Marking point
Mark
Notes
Step 1At constant pressure V ∝ T (V/T = constant for a fixed mass).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The mass is fixed, so ρ = m/V ∝ 1/T.
—
Step 3A graph of ρ against T for ρ ∝ 1/T is a curve that falls ever more gently and never reaches the T axis: graph B.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis treats the decrease as linear, which would make the density zero at a finite temperature; ρ ∝ 1/T never reaches zero.
BCorrect: ρ = pM/(RT), so ρ ∝ 1/T at constant pressure.
CThis confuses the density with the pressure at constant volume (p ∝ T); at constant pressure the gas expands, so its density falls.
DThis assumes that a fixed mass has a fixed density; the volume increases on heating, so the density decreases.
Syllabus understandingB.3 — that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant; B.1 — density ρ = m/VCommand term: Deduce
25B-1B-06
The empirical gas laws·B.3 Gas laws
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
A rigid metal flask of volume 500 cm³ contains a fixed mass of dry air and is connected by a short tube to a digital pressure sensor. The flask is placed in a water bath whose temperature θ is raised in steps. Before the experiment the student opens the sensor to the laboratory air: the sensor displays 97.3 kPa while a calibrated barometer reads 101.3 kPa. The student does not correct the readings, which are shown in the table and the graph with the line of best fit.
θ / °C
10
30
50
70
90
p (sensor reading) / kPa
100.9
108.0
115.6
123.1
130.3
Sensor reading p against θ with the line of best fit (drawn to scale)
(a)
(i)
Determine the gradient of the line, in kPa K−1.
(2)
(b)
(i)
Determine the temperature, in °C, at which the line of best fit predicts a sensor reading of zero.
(2)
(c)
(i)
Using the sensor check described above, explain why the answer to (b) is not absolute zero, and determine a corrected value for absolute zero.
(2)
(d)
(i)
Determine, using the gradient, the amount of air in the flask.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (134.1 − 97.1)/100
✓ 1
= 0.37 kPa K−1
✓ 1
Accept 0.36–0.38. A change of 1 °C equals a change of 1 K.
Part (b)(i)
Intercept on the p axis at θ = 0 is 97.1 kPa
✓ 1
Accept 97.0–97.3 kPa.
θ = −97.1/0.37 = −262 °C
✓ 1
Allow ECF from (a). Accept −257 to −268 °C.
Part (c)(i)
Every reading is 4.0 kPa too low (zero/systematic offset): the line is shifted down without changing its gradient, so it meets p = 0 at a temperature above absolute zero
✓ 1
The offset is read from the sensor check: 101.3 − 97.3 = 4.0 kPa.
Corrected: θ = −(97.1 + 4.0)/0.37 = −273 °C
✓ 1
Allow ECF from (a) and (b). Accept −270 to −278 °C.
Part (d)(i)
p = nRT/V, so gradient = nR/V and n = 0.37 × 10³ × 5.00 × 10−4/8.31 = 2.2 × 10−2 mol
✓ 1
Allow ECF from (a). Unit conversions (kPa → Pa, cm³ → m³) are required.
Answers: (a)(i) 0.37 kPa K−1 · (b)(i) −262 °C · (c)(i) −273 °C · (d)(i) 2.2 × 10−2 mol (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant; pV = nRT; B.1 — the Kelvin and Celsius scales; Tools — zero/systematic error, extrapolation Command term: Determine
26B-1B-10
Boyle's law with a gas syringe·B.3 Gas laws
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
A gas syringe is clamped vertically with its nozzle sealed, trapping air at a constant temperature of 20 °C. A platform fixed to the plunger carries slotted masses. The plunger has a cross-sectional area of 3.14 × 10−4 m² and moves without friction. M is the total mass of the plunger, platform and slotted masses, and V is the volume of trapped air. The pressure of the trapped air is p = p0 + Mg/A, where p0 is atmospheric pressure, which the student does not know. The graph shows 1/V against M with the line of best fit, extended to the M axis.
M / kg
0.50
1.50
2.50
3.50
4.50
V / cm³
43.5
34.0
28.3
24.1
20.8
(1/V) / cm−3
0.0230
0.0294
0.0353
0.0415
0.0481
1/V against M with the line of best fit; the dashed part is the extrapolation (drawn to scale)
(a)
(i)
Determine the gradient of the line.
(2)
(b)
(i)
Using the graph, determine p0.
(2)
(c)
(i)
Test the hypothesis that the trapped air obeys Boyle's law, using three rows of the table.
(2)
(d)
(i)
Calculate the amount of trapped air.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (0.0479 − 0.0230)/(4.50 − 0.50)
✓ 1
= 6.2 × 10−3 cm−3 kg−1
✓ 1
Accept 6.0–6.4 × 10−3. Unit not required.
Part (b)(i)
The line meets 1/V = 0 (where p = 0) at M = −p0A/g; from the graph, or intercept/gradient = 0.0199/6.2 × 10−3, this is M = −3.2 kg
✓ 1
Allow ECF from (a). Accept 3.0–3.4 kg.
p0 = 3.2 × 9.81/3.14 × 10−4 = 1.0 × 105 Pa
✓ 1
Accept 0.95–1.05 × 105 Pa.
Part (c)(i)
Pressure for each row = p0 + Mg/A, e.g. for M = 0.50, 2.50, 4.50 kg: 1.16 × 105, 1.78 × 105, 2.41 × 105 Pa
✓ 1
Allow ECF from (b).
pV = 5.03, 5.04, 5.00 J, constant to within 1 %, so Boyle's law is supported
✓ 1
At least three products and a conclusion are needed. A test using two rows only scores [1 max].
Part (d)(i)
n = pV/RT = 5.02/(8.31 × 293) = 2.1 × 10−3 mol
✓ 1
Allow ECF from (c).
Answers: (a)(i) 6.2 × 10−3 cm−3 kg−1 · (b)(i) 1.0 × 105 Pa · (c)(i) pV ≈ 5.02 J · (d)(i) 2.1 × 10−3 mol (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — pressure as given by p = F/A; the empirical gas law for constant temperature; pV = nRT; A.2 — weight mg; Tools — linearising, extrapolation to an intercept, testing a hypothesis Command term: Determine
27B-1B-16
Speed of sound and molecular motion·B.3 Gas laws
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
Two microphones 1.500 m apart are placed inside a long insulated tube of air. A short pulse of sound travels along the tube, and an oscilloscope measures the time Δt for the pulse to travel from the first microphone to the second, with an uncertainty of ±0.02 ms. The air in the tube is heated to different temperatures θ. For an ideal gas the speed of sound is
v = √(γRT / M)
where γ = 1.40 for air and M is the molar mass. The graph shows v² against the kelvin temperature T with error bars and the line of best fit.
θ / °C
10
30
50
70
90
Δt / ms
4.46
4.29
4.17
4.05
3.92
v² / 105 m² s−2
1.131
1.223
1.372
1.464
v² against T with error bars and the line of best fit (drawn to scale)
(a)
(i)
Calculate the missing value of v² in the table.
(1)
(ii)
Determine the absolute uncertainty in your answer to (a)(i).
(1)
(b)
(i)
Determine the gradient of the line. Give a unit for your answer.
(2)
(c)
(i)
Determine the molar mass of air.
(2)
(ii)
The steepest and shallowest lines through the error bars have gradients 430 and 390 in the same unit. The accepted molar mass of air is 0.0290 kg mol−1. Deduce whether your answer to (c)(i) is consistent with this value.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = 1.500/4.17 × 10−3 = 359.7 m s−1, v² = 1.294 × 105 m² s−2
✓ 1
Accept 1.29 × 105.
Part (a)(ii)
Fractional uncertainty in v² = 2 × 0.02/4.17 = 0.96 %, so Δ(v²) = ±1 × 10³ m² s−2
✓ 1
Allow ECF from (a)(i). Absolute uncertainty to 1 s.f.; the doubling for the square is required.
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (1.488 − 1.081) × 105/100 = 408
✓ 1
Accept 395–420.
Unit m² s−2 K−1
✓ 1
Part (c)(i)
v² = (γR/M)T, so gradient = γR/M and M = 1.40 × 8.31/408
✓ 1
M = 0.0285 kg mol−1
✓ 1
Allow ECF from (b). Accept 0.0277–0.0295 kg mol−1; g mol−1 only if the unit is stated.
Part (c)(ii)
M lies between 1.40 × 8.31/430 = 0.0271 and 1.40 × 8.31/390 = 0.0298 kg mol−1; 0.0290 is inside this range, so the result is consistent
✓ 1
Allow ECF from (c)(i). A conclusion based only on the closeness of the two numbers, without a range, scores [0].
Answers: (a)(i) 1.294 × 105 m² s−2 · (a)(ii) ±1 × 10³ m² s−2 · (b)(i) 408 m² s−2 K−1 · (c)(i) 0.0285 kg mol−1(the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that ideal gases are described in terms of the kinetic theory; that pressure is related to the average translational speed of molecules; C.2 — the speed of a travelling wave (time of flight); Tools — linearising (v² against T), gradient with unit, error bars and steepest/shallowest lines, power-law uncertainty Command term: Determine
28B-1B-17
Real and ideal gases·B.3 Gas laws
Paper 1BEasy7 marks
Data-based question7 steps to full marksDetermine
Methane is stored as compressed gas in vehicle fuel cylinders. The compression factor Z of a gas is defined as Z = pV/(nRT); for an ideal gas Z = 1. The table gives values of Z for methane at 300 K, measured at different pressures p. A fuel cylinder of internal volume 5.00 × 10−2 m³ is filled with methane to 8.0 MPa at 300 K. The molar mass of methane is 16.0 g mol−1.
p / MPa
0.5
1.0
2.0
4.0
6.0
8.0
10.0
Z
0.993
0.982
0.968
0.933
0.907
0.882
0.857
(a)
(i)
Calculate the mass of methane in the cylinder predicted by the ideal gas equation.
(2)
(b)
(i)
Determine, using the table, the actual mass of methane in the cylinder.
(2)
(c)
(i)
Calculate the percentage by which the ideal gas equation underestimates the mass of methane.
(1)
(d)
(i)
Explain, with reference to the molecules, why Z is less than 1 and decreases as the pressure increases.
Z = 0.882 at 8.0 MPa, so n = pV/(ZRT) = 160/0.882 = 182 mol
✓ 1
Allow ECF from (a).
Mass = 182 × 0.0160 = 2.91 kg
✓ 1
Accept 2.90–2.92 kg.
Part (c)(i)
(2.91 − 2.57)/2.91 × 100 = 12 %
✓ 1
Allow ECF from (a) and (b). Accept 12–13 % (13 % if calculated relative to the ideal value).
Part (d)(i)
At higher pressure the gas is denser, so the molecules are closer together on average and the intermolecular attractive forces are no longer negligible
✓ 1
A molecule moving towards the wall is pulled back by its neighbours, so the momentum change per collision (the force on the wall) is reduced; the pressure is less than nRT/V, more so the higher the density
✓ 1
Accept "softer/fewer collisions with the walls".
Answers: (a)(i) 2.57 kg · (b)(i) 2.91 kg · (c)(i) 12 % (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that ideal gases … constitute a modelled system used to approximate the behaviour of real gases; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas; pV = nRT; Tools — interpreting unfamiliar data Command term: Determine
29B-1B-32
Avogadro constant from electrolysis·B.3 Gas laws
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
A student uses the electrolysis of water to determine the Avogadro constant. A constant current of 0.500 A (±1 %) passes between two electrodes in dilute acid. The hydrogen produced at one electrode is dried and collected in a gas syringe, where its pressure is 1.00 × 105 Pa and its temperature is 293 K. Each hydrogen molecule produced requires two electrons to pass through the circuit.
The table and the graph show the volume V of hydrogen collected against the time t since the current was switched on, with the line of best fit.
t / s
120
240
360
480
600
720
V / cm³
6.3
13.6
21.4
28.9
36.7
43.9
Volume of hydrogen collected against time, with the line of best fit (drawn to scale)
(a)
(i)
Determine the gradient of the line. Give a unit for your answer.
(2)
(b)
(i)
Determine a value for the Avogadro constant from these results.
(2)
(c)
(i)
The uncertainty in the gradient is ±2 %. The uncertainties in the pressure and the temperature are negligible. Determine the absolute uncertainty in your answer to (b) and comment on whether your value agrees with the accepted value of the Avogadro constant.
(2)
(d)
(i)
The line of best fit does not pass through the origin. Suggest a reason for this and state why it does not affect your answer to (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (49.1 − (−1.3))/(800 − 0) = 0.0630
✓ 1
Accept 0.061–0.065.
Unit cm³ s−1
✓ 1
Accept equivalent units, e.g. 6.3 × 10−8 m³ s−1.
Part (b)(i)
Molecules of hydrogen produced per second = I/(2e) = 0.500/(2 × 1.60 × 10−19) = 1.56 × 1018 s−1
✓ 1
Amount produced per second = p × gradient/(RT) = 1.00 × 105 × 6.30 × 10−8/(8.31 × 293) = 2.59 × 10−6 mol s−1, so NA = 1.56 × 1018/2.59 × 10−6 = 6.0 × 1023 mol−1
✓ 1
Allow ECF from (a). Accept 5.8–6.3 × 1023 mol−1. The volume must be converted to m³.
Allow ECF from (b). Percentage uncertainties are added; the 1 % uncertainty in the current must be included.
The range (6.0 ± 0.2) × 1023, i.e. 5.8 × 1023 to 6.2 × 1023 mol−1 includes 6.02 × 1023 mol−1, so the value agrees with the accepted value
✓ 1
Allow ECF from (b) and the uncertainty. The conclusion must be based on the range, not on closeness alone. Accept a range of 5.9–6.2 × 1023 mol−1 from unrounded values.
Part (d)(i)
The line meets the V axis at about −1.3 cm³: the first 1.3 cm³ or so of the hydrogen produced did not reach the syringe (e.g. it dissolved in the acid or stayed as bubbles on the electrode); this is a systematic error that shifts every volume by the same amount, and NA was found from the gradient only
✓ 1
Both the reason and the use of the gradient are needed. Accept an intercept of −1.0 to −1.7 cm³. "Random error" or "human error" scores [0].
Answers: (a)(i) 0.0630 cm³ s−1 · (b)(i) 6.0 × 1023 mol−1 · (c)(i) ±2 × 1022 mol−1; agrees (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; B.5 — direct current as a flow of charge carriers as given by I = Δq/Δt; D.2 — elementary charge e; Tools — gradient with unit, propagation of uncertainties, systematic error (intercept) Command term: Determine
30B-1B-33
Testing Boyle's law with a log–log graph·B.3 Gas laws
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
A student traps a fixed mass of air in a vertical glass tube above a column of oil. A pump forces the oil up the tube, which reduces the volume V of the air; a gauge measures the pressure p of the air with an uncertainty of ±5 kPa. After each change the student waits 30 s before taking the readings. The room temperature is 20 °C.
The student suggests that p = CVk, where C and k are constants, and plots lg p against lg V. The graph shows the data with error bars, the line of best fit (solid) and the steepest and shallowest lines consistent with the error bars (dashed).
V / cm³
p / kPa
lg (V / cm³)
lg (p / kPa)
20.0
262
1.301
2.418
25.0
205
1.398
2.312
30.0
174
1.477
2.241
35.0
152
1.544
40.0
128
1.602
2.107
45.0
117
1.653
2.068
50.0
103
1.699
2.013
lg p against lg V with error bars, the line of best fit and the steepest and shallowest lines (drawn to scale)
(a)
(i)
Calculate the missing value in the table.
(1)
(b)
(i)
Determine k and its uncertainty.
(2)
(c)
(i)
Deduce whether the data support Boyle's law.
(1)
(d)
(i)
The experiment is repeated with the same mass of air at 60 °C. Predict how the new line would differ from the line on the graph.
(2)
(e)
(i)
Suggest why the student waits 30 s after each change before taking the readings.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
lg 152 = 2.182
✓ 1
Accept 2.18.
Part (b)(i)
k = gradient of the line of best fit from two well-separated points on the line, e.g. (2.016 − 2.417)/(1.70 − 1.30) = −1.00
✓ 1
Accept −0.97 to −1.03. The negative sign is required.
Uncertainty = half the difference between the gradients of the steepest (−1.06) and shallowest (−0.951) lines = ±0.05
✓ 1
Accept ±0.04 to ±0.07.
Part (c)(i)
Boyle's law (pV = constant) requires k = −1; −1 lies within −1.00 ± 0.05, so the data support Boyle's law
✓ 1
Allow ECF from (b). The value −1 and a comparison with the uncertainty range are both required.
Part (d)(i)
The gradient is unchanged (−1): at the new constant temperature pV is still constant
✓ 1
Allow ECF from (c).
pV ∝ T, so the line is shifted upwards by lg(333/293) = 0.056 (the intercept increases by 0.056)
✓ 1
Kelvin temperatures are required; lg(60/20) = 0.48 scores [0] for this mark.
Part (e)(i)
Compressing the air does work on it and raises its temperature; waiting allows the air to return to room temperature, so that the temperature is the same for every reading (as Boyle's law requires)
✓ 1
"To let the reading settle" alone scores [0].
Answers: (a)(i) 2.182 · (b)(i) k = −1.00 ± 0.05 · (d)(i) same gradient; shifted up by 0.056 (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant; the empirical gas law for constant temperature; Tools — linearising a power law with logarithms, gradient, steepest and shallowest lines, testing a hypothesis Command term: Determine
31B-1B-34
Real gases and the ideal-gas limit·B.3 Gas laws
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
A rigid steel vessel of internal volume 1.00 × 10−3 m³ is kept in a liquid bath whose temperature T is to be measured. Known amounts n of carbon dioxide are admitted to the vessel and the pressure p is measured each time. Carbon dioxide does not behave as an ideal gas at these pressures.
The table shows the results and the quantity pV/n. The graph shows pV/n against p with the line of best fit, extended to p = 0.
n / mol
p / MPa
(pV/n) / J mol−1
0.20
0.573
2865
0.40
1.123
2808
0.60
1.661
2768
0.80
2.177
1.00
2.683
2683
1.20
3.166
2638
1.40
3.640
2600
pV/n against p for carbon dioxide in the vessel, with the line of best fit (drawn to scale)
(a)
(i)
Calculate the missing value in the table.
(1)
(b)
(i)
Determine the temperature of the bath.
(2)
(c)
(i)
Determine the gradient of the line and show that its unit can be expressed as m³ mol−1.
(2)
(d)
(i)
A student instead calculates the temperature from the last row of the table alone, using pV = nRT. Determine the percentage by which this underestimates the temperature.
(1)
(e)
(i)
Explain why extrapolating the line to p = 0 gives the correct temperature even though carbon dioxide is not an ideal gas at the pressures used.
J mol−1 Pa−1 = (N m) mol−1/(N m−2) = m³ mol−1, so the gradient = −8.5 × 10−5 m³ mol−1
✓ 1
Allow ECF from the gradient. The conversion of MPa to Pa is required for the numerical value.
Part (d)(i)
T = 3.640 × 106 × 1.00 × 10−3/(1.40 × 8.31) = 313 K, which is 11 % below 350 K
✓ 1
Allow ECF from (b). Accept 10–12 %.
Part (e)(i)
As p → 0 (at constant T) the density → 0: the molecules are very far apart, so the intermolecular forces and the volume of the molecules become negligible and the gas obeys pV = nRT exactly
✓ 1
Both the low density and its consequence for the forces (or molecular volume) are required.
Answers: (a)(i) 2721 J mol−1 · (b)(i) 350 K · (c)(i) −8.5 × 10−5 m³ mol−1 · (d)(i) 11 % (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas; that ideal gases are described in terms of the kinetic theory and constitute a modelled system used to approximate the behaviour of real gases; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; Tools — extrapolation to an intercept, gradient with unit in fundamental SI units, comparing results Command term: Determine
32B-2-03
The ideal gas equation·B.3 Gas laws
Paper 2Medium9 marks
Short answer & extended response9 steps to full marksDetermine
A car of mass 1250 kg rests on four identical tyres on level ground. Each tyre contains air at an absolute pressure of 3.20 × 105 Pa and a temperature of 12 °C. The volume of air in each tyre is 2.50 × 10−2 m³ and can be treated as constant. Atmospheric pressure is 1.01 × 105 Pa. Treat the air as an ideal gas.
(a)
(i)
Calculate the amount of air, in mol, in one tyre.
(1)
(b)
(i)
Where a tyre touches the road it is flattened over an area called the contact patch. The upward force of the road on each tyre is balanced by the net force of the air pressures on the flattened part of the tyre. Determine the area of the contact patch of one tyre.
(3)
(c)
(i)
After a long drive the air in the tyre is at 40 °C. Determine the new pressure of the air.
(1)
(d)
(i)
Deduce the ratio (area of the contact patch after the drive)/(area of the contact patch before the drive).
(2)
(e)
(i)
Explain, in terms of the motion of the molecules, why the pressure of the air increases when its temperature increases at constant volume.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
T = 285 K; n = pV/RT = 3.20 × 105 × 2.50 × 10−2/(8.31 × 285) = 3.38 mol
✓ 1
Kelvin temperature required: 80 mol (12 K used) scores 0.
Part (b)(i)
Force on one tyre = 1250 × 9.81/4 = 3.07 × 103 N
✓ 1
The weight is shared by four tyres.
Net pressure across the tyre wall = 3.20 × 105 − 1.01 × 105 = 2.19 × 105 Pa
✓ 1
The atmosphere pushes on the outside of the patch.
A = 3.07 × 103/2.19 × 105 = 1.40 × 10−2 m²
✓ 1
Accept 1.4 × 10−2 m². Award [2 max] for 9.6 × 10−3 m² (absolute pressure) or 5.6 × 10−2 m² (weight not shared).
Part (c)(i)
At constant volume p ∝ T: p = 3.20 × 105 × 313/285; p = 3.51 × 105 Pa
✓ 1
Allow ECF from (a) for the kelvin conversion; award [0] for 3.20 × 105 × 40/12.
Part (d)(i)
The force on each tyre is unchanged, so the area is inversely proportional to the pressure difference (p − patm)
✓ 1
This idea must be explicit.
Ratio = 2.19/(3.51 − 1.01) = 0.87
✓ 1
Allow ECF from (b) and (c). Award [1 max] for 3.20/3.51 = 0.91 (absolute pressures).
Part (e)(i)
At a higher temperature the mean kinetic energy, and so the mean speed, of the molecules is greater, so each collision with the wall produces a greater change of momentum
✓ 1
The molecules also hit the walls more often, so the rate of change of momentum, the force on unit area of the wall and hence the pressure are greater
✓ 1
"The molecules move faster so they hit harder" alone scores [1].
Answers: (a)(i) 3.38 mol · (b)(i) 1.40 × 10−2 m² · (c)(i) 3.51 × 105 Pa · (d)(i) 0.87 (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — pressure as given by p = F/A where F is the force exerted perpendicular to the surface; pV = nRT and the empirical gas laws; a qualitative explanation of the macroscopic properties of an ideal gas in terms of molecular behaviour; A.2 — the normal force and free-body diagrams in translational equilibrium Command term: Determine
33B-2-12
Kinetic model of an atmosphere·B.3 Gas laws
Paper 2Medium9 marks
Short answer & extended response9 steps to full marksDetermine
The Moon has no significant atmosphere. The graph shows how the surface temperature at a point on the Moon's equator varies during one lunar day, which lasts 29.5 Earth days.
Mass of the Moon = 7.35 × 1022 kg; radius of the Moon = 1.74 × 106 m; mass of a helium-4 atom = 6.64 × 10−27 kg.
Surface temperature T at a point on the lunar equator against time t (drawn to scale)
(a)
(i)
For an ideal gas of N molecules, each of mass m, in a volume V, p = ⅓ρv² where v is the root mean square speed. Show that the mean translational kinetic energy of a molecule is (3/2)kBT.
(2)
(b)
(i)
Helium-4 atoms are released from the lunar rocks. Determine the root mean square speed of these atoms at the highest temperature shown on the graph.
(2)
(c)
(i)
Calculate the escape speed from the surface of the Moon.
(1)
(d)
(i)
Explain why helium released on the day side escapes from the Moon, although its root mean square speed is less than the escape speed.
(2)
(e)
(i)
Deduce, without calculating speeds, the ratio (rms speed of argon-40 atoms)/(rms speed of helium-4 atoms) at the same temperature, and comment on whether argon is retained by the Moon more easily than helium.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ρ = Nm/V, so pV = ⅓Nmv²
✓ 1
Equating with pV = NkBT gives ⅓mv² = kBT, so ½mv² = (3/2)kBT
✓ 1
Part (b)(i)
Highest temperature read from the graph = 390 K
✓ 1
Accept 380–395 K.
v = √(3 × 1.38 × 10−23 × 390/6.64 × 10−27) = 1.56 × 103 m s−1
The rms speed is an average: the atoms have a wide spread of speeds, and the escape speed is only about 1.5 times the rms speed, so a significant fraction of the atoms move faster than the escape speed
✓ 1
Allow ECF from (b) and (c).
With no atmosphere, these fast atoms leave without collisions; the remaining atoms keep being heated (every lunar day) and new fast atoms are produced, so over a long time almost all the helium escapes
✓ 1
Part (e)(i)
Same temperature, so the same mean kinetic energy: v ∝ 1/√m, ratio = √(4/40) = 0.32
✓ 1
The argon rms speed is about 490 m s−1, about one-fifth of the escape speed, so a far smaller fraction of argon atoms can escape: argon is retained much more easily
✓ 1
Allow ECF from (b) and (c).
Answers: (b)(i) 1.56 × 103 m s−1 · (c)(i) 2.37 × 103 m s−1 · (e)(i) 0.32 (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that pressure is related to the average translational speed of molecules as given by p = ⅓ρv²; pV = NkBT; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles, Ek = (3/2)kBT; D.1 — the escape speed vesc = √(2GM/r) Command term: Determine
34B-2-24
The ideal gas equation·B.3 Gas laws
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine
A diver breathes air from a tank of volume 12.0 × 10−3 m³ while swimming at a constant depth of 20 m in sea water of density 1025 kg m−3. The air she breathes is delivered at the pressure of the surrounding water. Atmospheric pressure is 1.01 × 105 Pa and the temperature of the air, in the tank and in her lungs, is 288 K. The graph shows how the pressure p in the tank varies with time t.
Pressure p of the air in the tank against time t (drawn to scale)
(a)
(i)
Show that the pressure at a depth h in water of density ρ is patm + ρgh.
(2)
(b)
(i)
Calculate the pressure of the water at a depth of 20 m.
(1)
(c)
(i)
Use the graph to determine the amount of air, in mol, that leaves the tank each minute.
(2)
(d)
(i)
Determine the volume of air that she breathes in each minute, measured at the pressure of the surrounding water.
(1)
(e)
(i)
Deduce, without using the gas equation again, how long the same tank would last (from 20.0 MPa to 5.0 MPa) at a depth of 40 m, for the same volume of air breathed each minute.
(2)
(f)
(i)
At 20 m her lungs contain 5.0 × 10−3 m³ of air. Explain why she must breathe out while rising to the surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A column of water of cross-sectional area A and height h has weight ρAhg
✓ 1
Pressure = weight/area + pressure of the atmosphere on top = patm + ρgh
✓ 1
Part (b)(i)
p = 1.01 × 105 + 1025 × 9.81 × 20 = 3.02 × 105 Pa
✓ 1
Allow ECF from (a).
Part (c)(i)
Rate of fall of pressure = 15.0 MPa/29.8 min = 0.50 MPa min−1
Each breath of the same volume at a higher pressure contains more air: the amount used per minute is proportional to the surrounding pressure, p at 40 m = 5.03 × 105 Pa
✓ 1
Allow ECF from (a) and (b).
Time = 29.8 × 3.02/5.03 = 18 min
✓ 1
Accept 17–19 min.
Part (f)(i)
If she held her breath, the volume of the air would increase as the pressure falls (Boyle's law): 5.0 × 3.02/1.01 = 15 × 10−3 m³
✓ 1
Allow ECF from (b).
This is about three times the original volume, far more than the lungs can hold, so they would be damaged
✓ 1
Answers: (b)(i) 3.02 × 105 Pa · (c)(i) 2.5 mol min−1 · (d)(i) 0.020 m³ · (e)(i) 18 min (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — pressure p = F/A; the ideal gas law pV = nRT and the empirical gas laws (constant temperature); the amount of substance; A.2 — weight Fg = mg and equilibrium of forces Command term: Determine
35B-2-25
Kinetic model of an ideal gas·B.3 Gas laws
Paper 2Hard9 marks
Short answer & extended response9 steps to full marksExplain
In a vacuum chamber, a beam of argon atoms, each of mass 6.64 × 10−26 kg, strikes a flat plate of area 1.0 × 10−4 m² at right angles. Every second 2.0 × 1020 atoms hit the plate, each arriving with a speed of 550 m s−1 and rebounding elastically.
(a)
(i)
Determine the force exerted on the plate by the beam.
(2)
(b)
(i)
Calculate the pressure on the plate.
(1)
(c)
(i)
The beam is switched off and the plate is surrounded instead by argon gas at 300 K. Show that the root mean square speed of the atoms of the gas is about 430 m s−1.
(1)
(d)
(i)
In the kinetic-theory result p = ⅓ρv², explain the origin of the factor ⅓.
(2)
(e)
(i)
The gas pressure is adjusted to the value found in (b). Determine the number of atoms per unit volume in the beam and in the gas, and explain why they are different.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Change of momentum of one atom = 2mv = 2 × 6.64 × 10−26 × 550 = 7.30 × 10−23 N s
✓ 1
Award [0] for mv (rebound ignored) unless the factor 2 appears later.
F = rate of change of momentum = 2.0 × 1020 × 7.30 × 10−23 = 1.46 × 10−2 N
✓ 1
Part (b)(i)
p = 1.46 × 10−2/1.0 × 10−4 = 146 Pa
✓ 1
Allow ECF from (a).
Part (c)(i)
½mv² = (3/2)kBT, so v = √(3 × 1.38 × 10−23 × 300/6.64 × 10−26) = 432 m s−1
✓ 1
An answer to at least 3 s.f. is required.
Part (d)(i)
The atoms move in random directions, with v² = vx² + vy² + vz²; only the component perpendicular to the wall (vx) changes the momentum normal to the wall
✓ 1
With no preferred direction the three mean squares are equal, so the mean of vx² is ⅓ of the mean of v²
✓ 1
Part (e)(i)
Beam: n = (atoms per second)/(Av) = 2.0 × 1020/(1.0 × 10−4 × 550) = 3.6 × 1021 m−3
In the beam every atom moves straight at the plate at 550 m s−1; in the gas the atoms move in random directions (only the perpendicular component counts, and only half move towards the plate) and more slowly, so about ten times as many atoms per unit volume are needed for the same pressure
✓ 1
Two of the three ideas (direction, fraction moving towards the plate, speed) for the mark.
Answers: (a)(i) 1.46 × 10−2 N · (b)(i) 146 Pa · (c)(i) 432 m s−1 · (e)(i) 3.6 × 1021 m−3; 3.5 × 1022 m−3(the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases, and from that analysis pressure is related to the average translational speed of molecules as p = ⅓ρv²; pV = NkBT; B.1 — Ek = (3/2)kBT; A.2 — F = Δp/Δt and impulse Command term: Explain
36B-2-37
Pressure and the kinetic model·B.3 Gas laws
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A vacuum desiccator is a glass vessel of internal volume 5.0 × 10−3 m³. It is closed by a flat glass lid of mass 1.5 kg that rests on the rim of a circular opening of diameter 0.20 m. A pump removes air from the desiccator, which stays at 293 K. The graph shows how the pressure p of the air inside varies with time t after the pump is switched on.
Atmospheric pressure = 1.01 × 105 Pa. Treat the air as an ideal gas.
Pressure of the air in the desiccator against time (drawn to scale)
(a)
(i)
Determine the number of molecules removed by the pump between t = 0 and t = 200 s.
(3)
(ii)
By drawing a tangent to the graph, determine the rate at which molecules are being removed at t = 40 s.
(2)
(b)
(i)
Explain, in terms of the molecules, why the pressure of the air falls as molecules are removed at constant temperature.
(2)
(c)
(i)
At t = 200 s the desiccator is sealed. Determine the minimum vertical force that must be applied to lift the lid. Treat the pressure difference as acting over the area of the opening.
(2)
(ii)
The lid is instead slid sideways off the rim. The coefficient of static friction between the lid and the rim is 0.20. Determine the horizontal force needed to start the lid sliding.
(2)
(d)
(i)
Explain why the air in the sealed desiccator is better described by the ideal gas model than the air outside it.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p at 200 s read from the graph ≈ 2.7 kPa, so Δp = 101 − 2.7 = 98.3 kPa
Allow ECF from the tangent gradient. Accept 1.0–1.2 × 1021 s−1.
Part (b)(i)
The temperature is constant, so the mean kinetic energy (and mean speed) of the molecules is unchanged: each collision with the wall transfers the same momentum on average
✓ 1
There are fewer molecules per unit volume, so fewer collisions with each unit area of wall per second and a smaller rate of change of momentum, i.e. a smaller force per unit area
✓ 1
"Fewer collisions" without "per unit time" (or "per second") scores [1 max].
Part (c)(i)
Force from the pressure difference = (1.01 × 105 − 2.7 × 103) × π × 0.102 = 3.09 × 103 N
✓ 1
Allow ECF from (a)(i).
Minimum force = 3.09 × 103 + 1.5 × 9.81 = 3.1 × 103 N
✓ 1
The weight of the lid must be added (or shown to be negligible, 15 N).
Part (c)(ii)
The normal force on the lid is its weight plus the force from the pressure difference = 3.10 × 103 N
✓ 1
Allow ECF from (c)(i). This hidden step is required.
F = 0.20 × 3.10 × 103 = 620 N
✓ 1
Award [1 max] for 2.9 N (normal force taken as the weight only).
Part (d)(i)
At the same temperature the density is about 37 times smaller (density ∝ p at constant T), so the molecules are much further apart and the intermolecular forces and the volume of the molecules are negligible
✓ 1
Both the lower density and its consequence for the intermolecular forces (or molecular volume) are required. Allow ECF from (a)(i).
Answers: (a)(i) 1.2 × 1023 · (a)(ii) 1.1 × 1021 s−1 · (c)(i) 3.1 × 103 N · (c)(ii) 620 N (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — pressure as given by p = F/A; the amount of substance n = N/NA; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas; A.2 — weight, the normal force and static friction Ff ≤ μsFNCommand term: Determine
37B-2-50
Helium canister and party balloons·B.3 Gas laws
Paper 2Easy9 marks
Short answer & extended response9 steps to full marksDetermine
A disposable canister of internal volume 1.40 × 10−2 m³ contains helium at a pressure of 1.71 × 106 Pa and a temperature of 288 K. The canister is used to fill party balloons. Each filled balloon contains helium at a pressure of 1.05 × 105 Pa, a temperature of 288 K and a volume of 8.2 × 10−3 m³. Helium is a monatomic gas of molar mass 4.0 g mol−1. Treat helium as an ideal gas.
(a)
(i)
Show that the amount of helium in the canister is about 10 mol.
(1)
(b)
(i)
Determine the number of balloons that can be filled completely from the canister, at 288 K.
(2)
(c)
(i)
Calculate the internal energy of the helium in one filled balloon.
(1)
(d)
(i)
The rubber of each balloon has a mass of 3.0 g. The density of the surrounding air is 1.23 kg m−3. Neglecting the volume of the rubber, determine the resultant upward force on a filled balloon when it is released.
(2)
(e)
(i)
Calculate the number of helium atoms in one balloon.
(1)
(f)
A filled balloon is left in a car in the sun. The pressure of the helium stays at 1.05 × 105 Pa while its temperature rises from 288 K to 318 K.
(i)
Calculate the increase in the internal energy of the helium, and explain why the energy transferred to the helium is greater than this increase.
Buoyancy force = ρairVg = 1.23 × 8.2 × 10−3 × 9.81 = 0.0989 N
✓ 1
Weight = (0.360 × 4.0 × 10−3 + 3.0 × 10−3) × 9.81 = 0.0435 N, so the resultant force = 0.055 N upwards
✓ 1
Allow ECF from (b). The weight of the helium must be included: award [1 max] for 0.070 N.
Part (e)(i)
N = nNA = 0.360 × 6.02 × 1023 = 2.2 × 1023
✓ 1
Allow ECF from (b).
Part (f)(i)
ΔU = 3⁄2nRΔT = 1.5 × 0.360 × 8.31 × 30 = 135 J
✓ 1
Allow ECF from (b). ALT: (c) × 30/288 = 135 J.
The helium expands (V ∝ T at constant pressure) and so does work on its surroundings; the energy supplied must provide this work as well as ΔU
✓ 1
Reference to work done by the expanding gas is required.
Answers: (a)(i) 10.0 mol · (b)(i) 26 · (c)(i) 1290 J · (d)(i) 0.055 N · (e)(i) 2.2 × 1023 · (f)(i) 135 J (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant; the internal energy U of an ideal monatomic gas as given by U = 3⁄2NkBT = 3⁄2nRT; A.2 — buoyancy Fb = ρVg and weight; B.4 — the first law of thermodynamics (work done by an expanding gas) Command term: Determine
38B-2-51
Argon-filled double glazing·B.3 Gas laws
Paper 2Easy12 marks
Short answer & extended response12 steps to full marksDetermine
A double-glazed window consists of two glass panes, each 1.20 m × 0.80 m, separated by a sealed gap 16 mm wide. The gap is filled with argon, a monatomic gas of molar mass 40 g mol−1, at a pressure of 1.01 × 105 Pa and a temperature of 20 °C, and then sealed. Treat argon as an ideal gas and assume that the volume of the gap does not change.
(a)
(i)
Calculate the amount of argon in the gap.
(1)
(ii)
Calculate the number of argon atoms in the gap.
(1)
(b)
On a sunny day the argon reaches a temperature of 45 °C. The outside surfaces of the panes remain at atmospheric pressure, 1.01 × 105 Pa.
(i)
Determine the resultant force exerted on one pane by the gases on its two sides.
(3)
(c)
(i)
Determine the increase in the internal energy of the argon when its temperature rises from 20 °C to 45 °C.
(2)
(ii)
State why the thermal energy absorbed by the argon is equal to your answer to (c)(i).
(1)
(d)
Assume that energy is transferred across the gap only by conduction. The inner pane is at a temperature 20 K higher than the outer pane. The thermal conductivity of argon is 0.018 W m−1 K−1 and that of air is 0.026 W m−1 K−1.
(i)
Determine the percentage reduction in the rate of thermal energy transfer across the gap when air is replaced by argon.
(2)
(ii)
At the same temperature, the molecules of air (mean molar mass 29 g mol−1) and the atoms of argon have the same mean translational kinetic energy. Suggest, with reference to the kinetic model, one reason why argon conducts thermal energy less well than air.
Allow ECF from the pressure found in the first marking point.
F = Δp × A = 8.6 × 103 × 0.96 = 8.3 × 103 N (outwards)
✓ 1
Award [1 max] for 1.1 × 105 N (the absolute pressure used).
Part (c)(i)
ΔU = 3⁄2nRΔT with ΔT = 25 K
✓ 1
Allow ECF from (a)(i). A temperature difference of 25 K must be used (318 − 293).
ΔU = 1.5 × 0.637 × 8.31 × 25 = 199 J
✓ 1
Accept 198–199 J.
Part (c)(ii)
The volume is constant, so the argon does no work (W = 0); by the first law Q = ΔU
✓ 1
Allow ECF from (c)(i).
Part (d)(i)
Rate with argon = kAΔT/Δx = 0.018 × 0.96 × 20/0.016 = 21.6 W
✓ 1
Rate with air = 31.2 W, so the reduction = 31 %
✓ 1
Accept 31 % from the ratio of the conductivities, (0.026 − 0.018)/0.026.
Part (d)(ii)
Same mean kinetic energy, so vrms ∝ 1/√M: the heavier argon atoms move more slowly, vrmsAr/vrmsair = √(29/40) = 0.85 of the speed of the air molecules
✓ 1
Energy is carried across the gap by moving particles and passed on in collisions, so slower particles transfer energy across the gap more slowly
✓ 1
"Argon is a better insulator" alone scores [0].
Answers: (a)(i) 0.637 mol · (a)(ii) 3.8 × 1023 · (b)(i) 8.3 × 103 N · (c)(i) 199 J · (d)(i) 31 % (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant; the internal energy U of an ideal monatomic gas as given by U = 3⁄2NkBT = 3⁄2nRT; B.1 — thermal conduction ΔQ/Δt = kAΔT/Δx and Ek = 3⁄2kBT; B.4 — the first law of thermodynamics Command term: Determine
39B-2-52
A computer model of a gas·B.3 Gas laws
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine
A computer model of a gas places 500 neon atoms, each of mass 3.35 × 10−26 kg, in a cubic box of side L = 10.0 nm. The atoms move randomly and collide elastically with each other and with the walls. The model is run at a temperature of 300 K. One wall of the box is labelled W.
(a)
(i)
An atom moves with a velocity component vx perpendicular to W. Between collisions with W it travels to the opposite wall and back without colliding with other atoms. Show that the mean force exerted on W by this atom is mvx2/L.
(2)
(b)
For atoms moving randomly, the mean value of vx2 is one third of the mean value of v2.
(i)
Using (a), determine the pressure exerted on W by all the atoms.
(3)
(ii)
Show that your answer to (b)(i) is consistent with the ideal gas law.
(1)
(c)
(i)
The model is changed so that every collision with W is inelastic: an atom leaves W with 0.90 of the perpendicular speed it had when it arrived. All other collisions remain elastic. Explain what happens to the pressure on W over time.
(2)
(d)
(i)
The model is run again with the same 500 atoms at 300 K, all collisions elastic, but with a box of side 20.0 nm. Deduce the factor by which the force on W and the pressure on W change.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Each collision with W reverses vx, so the change in momentum of the atom is 2mvx
✓ 1
The time between successive collisions with W is 2L/vx, so the mean force = rate of change of momentum = 2mvx ÷ (2L/vx) = mvx2/L
✓ 1
Newton's second law in the form F = Δp/Δt (or the impulse) must be used.
Total force on W = 500 × m × (mean vx2)/L = 500 × 3.35 × 10−26 × (6092/3)/1.00 × 10−8 = 2.07 × 10−10 N
✓ 1
Allow ECF from (a) and the rms speed.
p = F/L2 = 2.07 × 10−10/(1.00 × 10−8)2 = 2.07 × 106 Pa
✓ 1
Allow ECF from the force.
Part (b)(ii)
NkBT/V = 500 × 1.38 × 10−23 × 300/(1.00 × 10−8)3 = 2.07 × 106 Pa, the same as (b)(i)
✓ 1
Allow ECF from (b)(i). The volume (1.00 × 10−24 m³) must be seen.
Part (c)(i)
Immediately: each collision changes the momentum by 1.9mvx instead of 2mvx, so the pressure on W falls at once (by about 5 %)
✓ 1
Allow ECF from (a).
Kinetic energy is lost at every collision with W, so the mean kinetic energy and hence the temperature of the gas fall continuously; the atoms move ever more slowly and the pressure keeps decreasing (towards zero)
✓ 1
Reference to the loss of kinetic energy (temperature) is required.
Part (d)(i)
From (a) the force from each atom ∝ 1/L (mean vx2 unchanged at the same temperature), so the force on W is halved
✓ 1
Allow ECF from (a).
The area of W is 4 times larger, so the pressure is ½ ÷ 4 = ⅛ of its original value (consistent with p ∝ 1/V ∝ 1/L3)
✓ 1
Award [1 max] for "pressure ⅛" with no reference to the force.
Answers: (b)(i) 2.07 × 106 Pa · (b)(ii) 2.07 × 106 Pa · (d)(i) force × ½; pressure × ⅛ (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by p = ⅓ρv2; that ideal gases are described in terms of the kinetic theory and constitute a modelled system used to approximate the behaviour of real gases; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; A.2 — Newton's second law in the form F = Δp/Δt and impulse; B.1 — Ek = 3⁄2kBTCommand term: Determine
40B-2-53
The atmosphere of Mars·B.3 Gas laws
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksDetermine
The atmosphere of Mars is almost entirely carbon dioxide, of molar mass M = 0.0440 kg mol−1, which may be treated as an ideal gas. The gravitational field strength near Mars is 3.72 N kg−1 and may be taken as constant up to 40 km. The graph shows how the pressure p of the atmosphere varies with height h above the surface.
Pressure of the Martian atmosphere against height above the surface (drawn to scale)
(a)
(i)
Show that the density of an ideal gas of molar mass M is ρ = pM/(RT).
(1)
(ii)
The temperature at the surface is 215 K. Calculate the density of the atmosphere at the surface.
(1)
(b)
(i)
Consider a thin horizontal layer of the atmosphere of thickness Δh and area A, in equilibrium. Show that the pressure changes across the layer by Δp = −ρgΔh.
(2)
(c)
(i)
By drawing a tangent to the graph at h = 10 km, determine the temperature of the atmosphere at this height.
(3)
(d)
(i)
Calculate the number of molecules per unit volume at a height of 10 km.
(1)
(e)
A scientific balloon is filled with helium, of molar mass 0.0040 kg mol−1. The envelope is not stretched, so the helium is always at the same pressure and temperature as the surrounding atmosphere. The helium does not leak out. The volume of the envelope material is negligible.
(i)
Show that the buoyancy force on the balloon does not change as it rises.
(2)
(ii)
The mass of the envelope and the payload is 2.5 kg. Determine the least volume of the balloon at the surface for it to lift off.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A mass m of gas contains n = m/M moles, so pV = (m/M)RT and ρ = m/V = pM/(RT)
✓ 1
Part (a)(ii)
ρ = 640 × 0.0440/(8.31 × 215) = 0.0158 kg m−3
✓ 1
Allow ECF from (a)(i). Accept 630–650 Pa read from the graph.
Part (b)(i)
The weight of the layer is ρAΔh × g
✓ 1
Equilibrium: (pressure below − pressure above) × A = weight, so the pressure decreases upwards: Δp = −ρgΔh
✓ 1
The direction (pressure lower at the top) must be justified by the equilibrium of forces.
Part (c)(i)
Gradient of the tangent at 10 km ≈ −24.5 Pa km−1 = −0.0245 Pa m−1
✓ 1
Accept −22 to −27 Pa km−1.
ρ = −(Δp/Δh)/g = 0.0245/3.72 = 6.6 × 10−3 kg m−3
✓ 1
Allow ECF from (b) and the gradient.
T = pM/(ρR) = 247 × 0.0440/(6.6 × 10−3 × 8.31) = 199 K
✓ 1
Allow ECF from (a)(i). Accept 180–220 K. p at 10 km read from the graph: accept 240–255 Pa.
The volume of the balloon is V = nHeRT/p, where p and T are those of the surrounding CO2
✓ 1
Buoyancy force = ρVg = (pM/RT) × (nHeRT/p) × g = nHeMg, which is independent of p and T, so it is constant
✓ 1
Allow ECF from (a)(i).
Part (e)(ii)
Lift-off when nHeMg ≥ (2.5 + nHe × 0.0040)g, so nHe = 2.5/(0.0440 − 0.0040) = 62.5 mol
✓ 1
Allow ECF from (e)(i). The weight of the helium must be included: award [1 max] for 56.8 mol, giving 159 m³.
V = nRT/p = 62.5 × 8.31 × 215/640 = 174 m³
✓ 1
Allow ECF from (a)(ii). Accept 170–178 m³.
Answers: (a)(ii) 0.0158 kg m−3 · (c)(i) 199 K · (d)(i) 9.0 × 1022 m−3 · (e)(ii) 174 m³ (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; pressure as given by p = F/A, where F is the force exerted perpendicular to the surface; the amount of substance n as given by n = N/NA, where N is the number of molecules and NA is the Avogadro constant; A.2 — weight, buoyancy Fb = ρVg and translational equilibrium; Tools — the tangent to a curve Command term: Determine
41B-2-54
A bicycle pump on a p–V diagram·B.3 Gas laws
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine
A bicycle pump has a barrel that holds 150 cm³ of air when the handle is pulled out fully. At the start of each stroke the barrel is full of air at atmospheric pressure, 1.00 × 105 Pa, and 293 K. As the handle is pushed in, the air is compressed until its pressure equals that of the air in the tyre; a valve then opens and the rest of the stroke pushes the air into the tyre. At the end of the stroke the barrel volume is zero.
During one stroke the pressure of the air in the tyre is 3.00 × 105 Pa. Assume that the stroke is slow enough for the air to stay at 293 K.
Axes for the answer to (b)
(a)
(i)
Calculate the volume of air in the barrel at the moment the valve opens.
(1)
(b)
(i)
On the axes, draw a graph to show how the pressure p of the air in the barrel varies with its volume V during one stroke.
(2)
(c)
(i)
The volume of the tyre is 2.10 × 10−3 m³ and does not change. Determine the number of strokes needed to raise the pressure in the tyre from 2.00 × 105 Pa to 3.00 × 105 Pa, with the air in the tyre at 293 K.
(3)
(d)
In practice the handle is pushed in quickly and the air in the barrel becomes warmer during the compression.
(i)
Explain, in terms of the molecules, why the temperature of the air increases.
(2)
(ii)
State and explain the effect of the quick compression on the volume of air in the barrel when the valve opens.
(1)
(iii)
State and explain whether the quick compression changes the amount of air delivered to the tyre in each stroke.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Boyle's law: V = 150 × 1.00/3.00 = 50.0 cm³
✓ 1
Part (b)(i)
Curve from (150 cm³, 1.00 × 105 Pa) to (50 cm³, 3.00 × 105 Pa) with a magnitude of gradient that increases as V decreases, passing close to (100, 1.50) and (75, 2.00)
✓ 1
Allow ECF from (a). A straight line between the end points scores [0] for this mark.
Horizontal line at 3.00 × 105 Pa from V = 50 cm³ to V = 0
✓ 1
Allow ECF from (a).
Part (c)(i)
Amount of air delivered in one stroke = pV/(RT) = 1.00 × 105 × 1.50 × 10−4/(8.31 × 293) = 6.16 × 10−3 mol
✓ 1
The volume must be converted to m³.
Increase in pressure per stroke = nRT/Vtyre = 6.16 × 10−3 × 8.31 × 293/2.10 × 10−3 = 7.14 × 103 Pa
✓ 1
Allow ECF from the amount per stroke. ALT: each stroke adds 150 cm³ of air at 1.00 × 105 Pa, so Δp = 1.00 × 105 × 150/2100.
Number of strokes = 1.00 × 105/7.14 × 103 = 14.0, so 14 strokes
✓ 1
Allow ECF from the pressure increase.
Part (d)(i)
Molecules that collide with the piston moving inwards rebound with a greater speed (the piston does work on the gas)
✓ 1
So the mean kinetic energy of the molecules increases, and the temperature is a measure of the mean kinetic energy
✓ 1
"The molecules collide more often" alone scores [0].
Part (d)(ii)
The valve opens at a larger volume than 50 cm³: at each volume the warmer air has a greater pressure (p ∝ T/V), so the pressure reaches 3.00 × 105 Pa sooner
✓ 1
Allow ECF from (a).
Part (d)(iii)
No change: the barrel fills with the same amount of air at 1.00 × 105 Pa and 293 K at the start of each stroke, and all of it is pushed into the tyre by the end of the stroke
✓ 1
Allow ECF from (c). The reason (same intake, barrel emptied) is required.
Answers: (a)(i) 50.0 cm³ · (c)(i) 14 (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — changes of state of an ideal gas represented on pressure–volume diagrams (Guidance); that the ideal gas law equation can be derived from the empirical gas laws for constant pressure, constant volume and constant temperature as given by pV/T = constant; the equations governing the behaviour of ideal gases as given by pV = NkBT and pV = nRT; B.4 — work done on a gas (qualitative); B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles Command term: Determine
42B-2-65
Hot-air balloon·B.3 Gas laws
Paper 2Hard20 marks
Short answer & extended response20 steps to full marksDetermine
A hot-air balloon has an envelope of fixed volume 2.80 × 103 m³ that is open at the bottom, so the air inside it is always at the pressure of the surrounding atmosphere, 1.01 × 105 Pa. The surrounding air is at 15 °C. Treat air as an ideal gas of molar mass 0.029 kg mol−1.
The total mass of the envelope, basket, burner, fuel and passengers is 620 kg. Neglect the volume of everything except the envelope.
Temperature T of the air in the envelope against time t after the burner is turned off (drawn to scale)
(a)
(i)
Show that the amount of air in the envelope before it is heated is about 1.2 × 105 mol.
(1)
(ii)
Calculate the mass of this air.
(1)
(iii)
The air in the envelope is now heated. Explain why the mass of air in the envelope decreases.
(2)
(b)
(i)
The density of the surrounding air is ρ0 and its temperature is T0. The air in the envelope has temperature T. Show that the resultant upward force on the balloon is F = ρ0Vg(1 − T0/T) − msg, where ms = 620 kg.
(2)
(ii)
Determine the least temperature, in °C, of the air in the envelope for the balloon to float.
(3)
(c)
(i)
Estimate the energy needed to heat the air from 15 °C to the temperature in (b)(ii), by assuming that the air heated is the mass of air in the envelope at the end of the heating. The specific heat capacity of air at constant pressure is 1.0 × 103 J kg−1 K−1. State why this estimate is too small.
(2)
(ii)
The burner burns propane of energy density 4.6 × 107 J kg−1, and 75 % of the energy released is transferred to the air. Determine the mass of propane needed for the heating in (c)(i).
(1)
(d)
With the air at the temperature found in (b)(ii), the burner is turned off. The graph shows how the temperature T of the air in the envelope then varies with time t. Assume that the mass of air in the envelope stays at the value used in (c)(i).
(i)
Determine the rate at which the air loses thermal energy immediately after the burner is turned off.
(3)
(ii)
The envelope has a surface area of 960 m² and an emissivity of 0.75, and is at the temperature of the air inside it. It absorbs radiation from its surroundings at a rate equal to 0.75σAT0⁴. Determine the net rate at which the envelope loses energy by radiation.
(2)
(iii)
Suggest, with reference to your answers to (d)(i) and (d)(ii), how the remaining energy is lost.
(1)
(e)
(i)
The air in the envelope is kept at a constant temperature as the balloon rises. The temperature of the atmosphere is approximately constant over the height of the flight, but its pressure and density decrease with height. Explain why the balloon stops rising at a certain height.
Allow ECF from (b)(ii). Using 352 K gives 6.3 × 105 W emitted.
Net loss = 3.4 × 105 W
✓ 1
Accept 3.3–3.5 × 105 W. Award [1 max] for 6.2–6.3 × 105 W (absorption ignored).
Part (d)(iii)
Radiation accounts for about 75 % of the loss; the rest is conducted through the fabric and carried away by convection in the surrounding air (and by hot air leaking out)
✓ 1
Allow ECF from (d)(i) and (d)(ii).
Part (e)(i)
Using (b)(i), the resultant force is ρ0Vg(1 − T0/T) − msg with T0/T unchanged, so the lift is proportional to the density ρ0 of the outside air
✓ 1
Allow ECF from (b)(i). Both the buoyancy force and the weight of the hot air fall in the same proportion.
As ρ0 decreases with height the lift falls until it equals the weight msg; there the resultant force is zero and the balloon stops rising
✓ 1
Answers: (a)(i) 1.18 × 105 mol · (a)(ii) 3430 kg · (b)(ii) 79 °C · (c)(i) 1.8 × 108 J · (c)(ii) 5.2 kg · (d)(i) 4.6 × 105 W · (d)(ii) 3.4 × 105 W (the remaining parts are explanations — see the table above)
Syllabus understandingB.3 — the ideal gas equation pV = nRT and the amount of substance n; B.1 — Q = mcΔT; the Stefan–Boltzmann law for a body with emissivity; conduction, convection and radiation; B.2 — emissivity; A.2 — buoyancy Fb = ρVg and translational equilibrium; A.3 — energy density of fuel sources and efficiency; Tools — the tangent to a curve Command term: Determine
No questions match that combination yet
Try widening the difficulty, or choose “All subtopics”. The bank is still growing.
