B.3 connects the behaviour of a gas to its molecules. You need pressure P = F/A, the amount of substance n = N/N_A, the empirical gas laws and the ideal gas equation PV = nRT = Nk_BT, including changes shown on pressure–volume diagrams.
The kinetic model explains pressure as the rate of change of momentum of molecules hitting a wall, giving P = ⅓ρv². The internal energy of a monatomic ideal gas is U = (3/2)nRT. You also need the conditions of temperature, pressure and density under which a real gas behaves like an ideal one.
20 questions
133 marks
Paper 1A: 9
Paper 1B: 4
Paper 2: 7
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20 practice questions on B.3 Gas laws
1B-1A-16
Pressure·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
A force of 250 N acts perpendicular to a surface of area 0.020 m².
What is the pressure on the surface?
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Notes
Step 1P = F/A = 250 / 0.020 = 1.25 × 10⁴ Pa.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis is A/F, the reciprocal of the pressure.
BThe area has been treated as 200 cm² without converting to square metres.
CThis multiplies the force by the area. Pressure is force per unit area.
DCorrect: P = F/A = 250 / 0.020 = 1.25 × 10⁴ Pa.
Syllabus understandingB.3 — pressure as given by P = F/A where F is the force exerted perpendicular to the surface Command term: Calculate
2B-1A-17
Ideal & real gases·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksIdentify
The behaviour of a real gas can be modelled by an ideal gas.
Which row gives the conditions under which an ideal gas is the best approximation to a real gas?
DensityTemperature
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Notes
Step 1The ideal gas model assumes that the molecules have negligible volume and exert no forces on each other except in collisions. Their volume is negligible when the density is low, and the forces matter least when the molecules move fast, at high temperature.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AAt high density the molecules are close together, so their own volume is no longer negligible and intermolecular forces act.
BThese are the conditions in which a real gas departs most from ideal behaviour and is closest to liquefying.
CAt low temperature the molecules move slowly and the attractive forces between them have a significant effect, especially near condensation.
DCorrect: low density makes the molecular volume and intermolecular forces negligible, and high temperature makes the forces unimportant compared with the kinetic energy.
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; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas Command term: Identify
3B-1A-18
The empirical gas laws·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A fixed mass of an ideal gas is taken from state X to state Y. The graph shows the pressure P and the volume V of the gas. The temperature of the gas at X is 27 °C.
What is the temperature of the gas at Y?
Graph drawn to scaleShow mark scheme
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Notes
Step 1From the graph: at X, P = 1.0 × 105 Pa and V = 3.0 × 10−3 m³; at Y, P = 4.0 × 105 Pa and V = 1.0 × 10−3 m³. TX = 27 + 273 = 300 K.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2PV/T is constant: TY = 300 × (4.0 × 1.0) / (1.0 × 3.0) = 400 K.
—
Step 3In degrees Celsius: 400 − 273 = 127 °C.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThe Celsius temperature has been used in PV/T = constant: 27 × 4/3 = 36. The gas laws need the absolute temperature.
BCorrect: PV rises by a factor 4/3, so T rises from 300 K to 400 K, which is 127 °C.
C400 is the temperature in kelvin; it has not been converted back to degrees Celsius.
DThe fall in volume has been ignored: 300 × 4 = 1200 K = 927 °C. The volume falls to a third, which lowers the temperature reached.
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; changes of state of an ideal gas can be represented on pressure–volume diagrams Command term: Determine
4B-1A-19
Internal energy of a gas·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A sample of a monatomic ideal gas contains N molecules. Its pressure is P and its volume is V.
Which row gives the internal energy of the gas and the average kinetic energy of one molecule?
Internal energyAverage kinetic energy of one molecule
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Notes
Step 1U = ³⁄₂NkBT and PV = NkBT, so U = ³⁄₂PV.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2In a monatomic ideal gas all of the internal energy is random kinetic energy, so the average per molecule is U/N = 3PV / (2N).
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe factor ³⁄₂ has been dropped. PV = NkBT is not the kinetic energy of the molecules.
BThe factor has been inverted: the kinetic energy is ³⁄₂kBT per molecule, not ⅔kBT.
CCorrect: U = ³⁄₂NkBT = ³⁄₂PV, and dividing by N gives the average per molecule.
DThe total and the per-molecule energies have been interchanged. The internal energy is N times the average for one molecule.
Syllabus understandingB.3 — the relationship between the internal energy U of an ideal monatomic gas and the number of molecules or amount of substance as given by U = ³⁄₂NkBT; the equations governing the behaviour of ideal gases as given by PV = NkBTCommand term: Determine
5B-1B-05
The empirical gas laws·B.3 Gas laws
Paper 1BMedium9 marks
Data-based question9 steps to full marksDetermine
Air is trapped in a syringe connected by a short rubber tube to a pressure sensor. The plunger is moved slowly and, after a pause, the volume V is read from the syringe scale (uncertainty ±0.5 cm3) and the pressure P from the sensor. The temperature of the room is 294 K.
The student plots V against 1/P. The graph shows six of the seven points.
V / cm3
P / kPa
(1/P) / 10−3 kPa−1
50.0
101
9.90
42.0
118
8.47
35.0
141
7.09
29.0
167
5.99
24.0
196
20.0
230
4.35
16.0
277
3.61
Graph drawn to scale
(a)
Calculate the percentage uncertainty in the smallest volume reading.
(1)
(b)
Calculate the missing value of 1/P and plot the point on the graph.
(1)
(c)
Draw the line of best fit and extend it to meet the V axis.
(1)
(d)
Boyle’s law predicts a line through the origin. State the intercept on the V axis and explain what it shows about the apparatus.
(2)
(e)
Determine the amount of air, in mol, in the apparatus.
(3)
(f)
Explain why the plunger was moved slowly and a pause was made before each reading.
(1)
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Notes
Part (a)
0.5 / 16.0 × 100 = 3.1 %
✓ 1
Accept 3 %.
Part (b)
5.10 «× 10−3 kPa−1» AND plotted correctly
✓ 1
Accept 5.102.
Part (c)
single straight line of best fit extended back to 1/P = 0
✓ 1
Part (d)
intercept ≈ −3.4 cm3
✓ 1
Accept −2.5 to −4.5 cm3.
the air in the tube and sensor «≈ 3.4 cm3» is not included in the scale reading, so the total volume is V + 3.4 cm3; a systematic error
✓ 1
OWTTE.
Part (e)
gradient ≈ 5.39 cm3 per 10−3 kPa−1 = 5390 kPa cm3
✓ 1
Accept 5.23–5.55.
gradient = nRT = 5.39 J «Pa m3»
✓ 1
Conversion of kPa and cm3 to SI needed.
n = 5.39 / (8.31 × 294) = 2.2 × 10−3 mol
✓ 1
Allow ECF.
Part (f)
so that the temperature of the air stays constant «work done on the gas when compressed raises its temperature; the pause lets it return to room temperature»
✓ 1
OWTTE.
Answers: (a) 3.1 % · (b) 5.10 × 10−3 kPa−1 · (d) −3.4 cm3 · (e) 2.2 × 10−3 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 temperature as given by PV/T = constant; the equations governing the behaviour of ideal gases as given by PV = nRT Command term: Determine
6B-1B-06
The empirical gas laws·B.3 Gas laws
Paper 1BEasy8 marks
Data-based question8 steps to full marksEstimate
A glass capillary tube of uniform bore is sealed at its lower end. A short thread of oil traps a column of air, and the upper end of the tube is open to the atmosphere. The tube is fixed to a ruler and placed in a water bath. The length ℓ of the air column is measured at several temperatures θ. Each ℓ has an uncertainty of ±1 mm.
θ / °C
12.0
22.5
35.0
46.5
58.0
70.0
81.5
ℓ / mm
68
71
74
77
79
82
85
Graph drawn to scale
(a)
Explain why ℓ can be used as a measure of the volume of the trapped air.
(1)
(b)
State why the pressure of the trapped air remains constant during the experiment.
(1)
(c)
Draw the line of best fit and extend it to meet the θ axis.
(1)
(d)
Using your graph, estimate absolute zero in °C.
(1)
(e)
Calculate the percentage uncertainty in the smallest value of ℓ.
(1)
(f)
Explain why a small uncertainty in ℓ leads to a large uncertainty in the estimate in (d), and suggest one change to the experiment that would reduce it.
(2)
(g)
The student read ℓ immediately after changing the bath temperature. Suggest how this could affect the readings.
(1)
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Notes
Part (a)
the volume is Aℓ where the cross-sectional area A of the bore is constant, so V ∝ ℓ
✓ 1
Part (b)
the trapped air is always at atmospheric pressure «plus the small, constant pressure of the oil thread», because the tube is open at the top
✓ 1
OWTTE.
Part (c)
single straight line through the error bars, extended to ℓ = 0
✓ 1
Part (d)
≈ −274 °C
✓ 1
Accept −255 to −290 °C. Allow read-off from the candidate’s line.
Part (e)
1 / 68 × 100 = 1.5 %
✓ 1
Accept 1 % or 1.47 %.
Part (f)
the data cover only ≈ 70 K and the line is extrapolated ≈ 280 K beyond them, so a small change in gradient moves the intercept a long way
✓ 1
measure over a wider range of temperature «e.g. using ice–salt mixtures below 0 °C» OR use a longer air column / finer scale to reduce the uncertainty in ℓ
✓ 1
Accept any change that increases the range or reduces the percentage uncertainty in ℓ.
Part (g)
the air had not reached the bath temperature, so when heating, ℓ is too small for the recorded θ «a systematic error»
✓ 1
Accept “stir the bath and wait for ℓ to stop changing” as part of the answer.
Answers: (d) ≈ −274 °C · (e) 1.5 % (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; that Kelvin and Celsius scales are used to express temperature Command term: Estimate
7B-2-07
The empirical gas laws·B.3 Gas laws
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDraw
A fixed mass of argon, which may be treated as an ideal monatomic gas, is held in a cylinder by a piston. The gas is taken through three changes:
A → B: the gas is heated at constant pressure from 300 K to 450 K;
B → C: the piston is locked and the gas is cooled at constant volume back to 300 K;
C → A: the piston is released and the gas is compressed at constant temperature back to state A.
In state A the pressure is 1.20 × 105 Pa and the volume is 5.00 × 10−4 m3. The first graph shows state A on a pressure–volume diagram.
Graph drawn to scaleAxes for (c)
(a)
(i)
Calculate the amount of argon, in mol.
(1)
(ii)
Show that the volume in state B is 7.5 × 10−4 m3.
(1)
(iii)
Calculate the pressure in state C.
(1)
(b)
Draw, on the pressure–volume graph, the changes A → B, B → C and C → A. Label B and C.
(3)
(c)
Sketch, on the volume–temperature axes, the changes A → B, B → C and C → A. Label A, B and C.
(2)
(d)
(i)
Calculate the change in the internal energy of the argon during A → B.
(2)
(ii)
State the change in the internal energy of the argon over the complete cycle A → B → C → A, giving a reason.
(1)
(iii)
Calculate the average kinetic energy of an argon atom in state B.
Answers: (a)(i) 2.41 × 10−2 mol · (a)(iii) 8.0 × 104 Pa · (d)(i) 45 J · (d)(ii) 0 · (d)(iii) 9.3 × 10−21 J (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 equations governing the behaviour of ideal gases as given by PV = nRT; the relationship between the internal energy U of an ideal monatomic gas and the … amount of substance as given by U = 3/2 RnT (changes of state represented on pressure–volume diagrams) (with B.1 — Ek = 3/2 kBT) Command term: Draw
8B-2-08
Kinetic model & molecular speeds·B.3 Gas laws
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDiscuss
The atmosphere of Mars is mainly carbon dioxide, of molar mass 0.044 kg mol−1. At a landing site the atmospheric pressure is 610 Pa and the temperature is 210 K. Treat the atmosphere as an ideal gas.
(a)
Describe how collisions of the gas molecules with the ground give rise to a pressure on it.
(3)
(b)
(i)
Show that the density of the atmosphere at the landing site is about 0.015 kg m−3.
(2)
(ii)
Calculate the root mean square speed of the molecules at the landing site.
(2)
(iii)
By midday the temperature has risen to 270 K. Determine the percentage increase in the root mean square speed of the molecules.
(2)
(c)
Determine whether the average kinetic energy of a CO2 molecule at the landing site is greater or smaller than that of an N2 molecule in the Earth's atmosphere at 288 K.
(2)
(d)
Discuss whether the ideal gas model is a better approximation for the atmosphere at the landing site than for the Earth's atmosphere at sea level (1.0 × 105 Pa, 288 K).
(3)
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Notes
Part (a)
a molecule that strikes the ground and rebounds undergoes a change of momentum
✓ 1
the ground exerts a force on the molecule equal to its rate of change of momentum, and the molecule exerts an equal and opposite force on the ground (Newton's third law)
✓ 1
very many collisions per second produce a steady average force; pressure is this force divided by the area
✓ 1
Part (b)(i)
ρ = m/V = nM/V and n/V = P/RT, so ρ = PM/RT
✓ 1
ρ = 610 × 0.044 / (8.31 × 210) = 0.0154 kg m−3
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
P = ⅓ρv2 so vrms = √(3P/ρ)
✓ 1
Accept √(3RT/M).
vrms = √(3 × 610 / 0.0154) = 345 m s−1
✓ 1
Accept 340–350 m s−1 (0.015 kg m−3 gives 349 m s−1).
Part (b)(iii)
vrms ∝ √T
✓ 1
Accept v2 = 3RT/M quoted or implied.
√(270/210) = 1.134, an increase of 13 %
✓ 1
Accept 13–14 %. Award [1] max for 29 % (ratio of temperatures used without the square root).
Part (c)
the average kinetic energy depends only on the temperature, Ek = 3/2 kBT (not on the mass of the molecule)
✓ 1
210 K < 288 K, so the CO2 molecule has the smaller average kinetic energy (4.3 × 10−21 J compared with 6.0 × 10−21 J)
✓ 1
MP2 only scores if MP1 scores. Do not award MP2 for an argument based on molar mass.
Part (d)
at the landing site the pressure is far lower; the number of molecules per unit volume, P/kBT, is about 120 times smaller than at sea level on Earth, so the molecules are much further apart: intermolecular forces and the volume of the molecules matter less (better)
✓ 1
but the temperature is lower, so the molecules move more slowly and intermolecular forces are relatively more important (worse)
✓ 1
the effect of the much lower density dominates (the kelvin temperature is only about 27 % lower), so the model is a better approximation on Mars
✓ 1
MP3 needs a conclusion supported by a comparison of the two effects.
Answers: (b)(i) 0.0154 kg m−3 · (b)(ii) 345 m s−1 · (b)(iii) 13 % · (c) smaller (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; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas (with B.1 — Ek = 3/2 kBT) Command term: Discuss
9B-2-16
The ideal gas equation·B.3 Gas laws
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksCalculate
The two panes of glass of a double-glazed window are 16 mm apart. The space between them is 1.20 m high and 0.80 m wide and is filled with argon, a monatomic gas, and sealed at a pressure of 1.01 × 105 Pa and a temperature of 15 °C. Assume that the volume of the space does not change and that the argon behaves as an ideal gas. Molar mass of argon = 0.040 kg mol−1.
Axes for (d)
(a)
State two assumptions of the kinetic model of an ideal gas.
(2)
(b)
Show that the space contains about 0.65 mol of argon.
(1)
(c)
On a sunny day the temperature of the argon rises to 45 °C.
(i)
Calculate the pressure of the argon now.
(1)
(ii)
Calculate the increase in the internal energy of the argon.
(2)
(d)
Sketch, on the axes, a graph to show how the pressure of the argon would vary with its temperature θ from −273 °C to 50 °C if it remained an ideal gas.
(2)
(e)
Argon is used because it conducts thermal energy less well than air. Deduce the ratio (root mean square speed of argon atoms)/(root mean square speed of nitrogen molecules) when both gases are at the same temperature. Molar mass of nitrogen = 0.028 kg mol−1.
(2)
(f)
Suggest why the argon in the window can be treated as an ideal gas.
(1)
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Notes
Part (a)
any two from: the molecules/atoms have negligible volume compared with the volume of the gas; there are no intermolecular forces except during collisions; collisions are elastic; the molecules are in random motion; the duration of a collision is negligible compared with the time between collisions
Must see full substitution OR answer to at least 3 s.f. Volume = 1.20 × 0.80 × 0.016 m3 must be seen.
Part (c)(i)
P ∝ T at constant volume: P = 1.01 × 105 × 318/288 = 1.12 × 105 Pa
✓ 1
Award [0] if Celsius temperatures are used.
Part (c)(ii)
ΔU = ³⁄₂ nRΔT = 1.5 × 0.648 × 8.31 × 30
✓ 1
ΔT = 30 K must be used.
ΔU = 242 J
✓ 1
Accept 240–245 J. Allow ECF from (b). Award [2] for CNA.
Part (d)
a straight line with positive gradient
✓ 1
Do not accept a curve.
meeting the θ axis at −273 °C «P = 0» and passing through about (15 °C, 1.01 × 105 Pa)
✓ 1
Allow ECF from (c)(i) for a line through (45 °C, 1.12 × 105 Pa).
Part (e)
at the same temperature the mean kinetic energy, ³⁄₂kBT, of the particles is the same, so ½mv2 is the same and vrms ∝ 1/√m «∝ 1/√M»
✓ 1
Accept vrms = √(3RT/M).
ratio = √(0.028/0.040) = 0.84
✓ 1
Award [1 max] for 0.70 «square root omitted» or 1.20 «inverted».
Part (f)
the pressure is low «about atmospheric» and the temperature is far above the temperature at which argon liquefies, so the atoms are far apart and intermolecular forces «and the volume of the atoms» are negligible
✓ 1
OWTTE. Both a condition and a molecular consequence needed.
Answers: (b) 0.648 mol · (c)(i) 1.12 × 105 Pa · (c)(ii) 242 J · (e) 0.84 (the remaining parts are explanations — see the table above)
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 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 = nRT; the relationship between the internal energy U of an ideal monatomic gas and the amount of substance as given by U = ³⁄₂RnT; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas (linked: B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = ³⁄₂kBT) Command term: Calculate
10B-1A-36
Moles, molecules & the Avogadro constant·B.3 Gas laws
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDetermine
Oxygen has a molar mass of 0.032 kg mol−1 and the Avogadro constant is 6.02 × 1023 mol−1.
How many molecules are there in 32 g of oxygen?
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Notes
Step 1Convert first: 32 g = 0.032 kg, which is one molar mass, so n = 1.0 mol. N = nNA = 6.02 × 10²³ molecules.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
A32 is the mass in grams, not a number of molecules.
BThis multiplies the Avogadro constant by 0.032, treating the mass in kilograms as the amount in moles.
CCorrect: 32 g is 0.032 kg, which is exactly one mole, and one mole contains NA molecules.
DThis treats the sample as 1000 moles, which would come from dividing 32 g by 0.032 kg without converting the units.
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-37
Kinetic model & molecular speeds·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A fixed mass of an ideal gas is compressed to half of its original volume. At the same time its pressure increases to three times its original value.
Using P = ⅓ρv², what is the ratio vfinal / vinitial?
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Notes
Step 1Halving the volume of a fixed mass doubles the density: ρfinal = 2ρinitial.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2v² = 3P/ρ, so v² changes by a factor 3/2 and v by √1.5 = 1.2.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the pressure trebles and the density doubles, so v² rises by 3/2 and v by √1.5 = 1.2.
BThe square root has not been taken: 1.5 is the ratio of v², not of v.
CThe change in density has been ignored (√3 = 1.7). Compressing the gas to half its volume doubles its density.
DThe density ratio has been multiplied instead of divided: √(3 × 2) = 2.4. For a given pressure a denser gas has slower molecules.
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: Determine
12B-1A-38
Pressure·B.3 Gas laws
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Gas is trapped in a vertical cylinder by a frictionless piston of negligible mass and area 2.0 × 10−3 m². The volume of the gas is 60 cm³ and atmospheric pressure is 1.0 × 105 Pa.
A load of weight 100 N is placed on the piston. The gas returns to its original temperature. Which row gives the new pressure and the new volume of the gas?
New pressureNew volume
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Notes
Step 1Before the load the gas pressure equals atmospheric pressure. With the load, the piston is in equilibrium: P = 1.0 × 105 + 100 / 2.0 × 10−3 = 1.5 × 105 Pa.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2At constant temperature PV is constant: V = 60 × 1.0 × 105 / 1.5 × 105 = 40 cm³.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AOnly the load's contribution, F/A = 5.0 × 10⁴ Pa, has been used. The atmosphere still pushes down on the piston, so the gas pressure is 1.0 × 10⁵ + 5.0 × 10⁴ Pa.
BTwo errors: the atmospheric pressure has been left out of the new pressure, and the pressure ratio has been inverted.
CThe pressure is right but the ratio has been inverted. The pressure rises, so at constant temperature the volume must fall.
DCorrect: the gas pressure becomes 1.5 × 10⁵ Pa, and PV = constant gives 60 × 1.0/1.5 = 40 cm³.
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 temperature as given by PV/T = constant Command term: Determine
13B-1A-39
The empirical gas laws·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksIdentify
A fixed mass of an ideal gas is in a sealed container of fixed volume. The gas is heated.
Which statement explains why the pressure of the gas increases?
Show mark scheme
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Notes
Step 1A higher temperature means a larger mean kinetic energy, so the molecules move faster. Each collision with a wall gives a larger change in momentum, and collisions happen more often, so the force per unit area — the pressure — increases.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AMolecules do not expand when heated. In the ideal gas model their own volume is negligible.
BCorrect: faster molecules strike the walls more often and each collision gives a larger change in momentum, so the rate of change of momentum per unit area increases.
COnly half of the explanation. Faster molecules also have more momentum, so each collision with a wall gives a greater change in momentum.
DThe ideal gas model assumes no intermolecular forces except during collisions. The pressure comes from collisions with the walls.
Syllabus understandingB.3 — that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases; a qualitative explanation of the macroscopic properties of an ideal gas in terms of molecular behaviour is required Command term: Identify
14B-1A-40
The ideal gas equation·B.3 Gas laws
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A container of volume 8.0 × 10−3 m³ holds gas at 2.5 × 105 Pa and 320 K. The Boltzmann constant is 1.38 × 10−23 J K−1.
How many molecules does it contain?
Show mark scheme
Marking point
Mark
Notes
Step 1PV = NkBT, so N = PV/(kBT).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 3N = 2000 / 4.42 × 10⁻²¹ = 4.5 × 10²³ molecules, a little under one mole.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
A320 K has been converted to 47 °C and used in the equation: 2000 / (1.38 × 10−23 × 47) = 3.1 × 1024. PV = NkBT needs the absolute temperature, which 320 K already is.
CThis is the amount in moles, found using R. To get a number of molecules use kB, or multiply this by the Avogadro constant.
DPV has been multiplied by kBT instead of divided by it: 2000 × 4.42 × 10−21 = 8.8 × 10−18.
Syllabus understandingB.3 — the equations governing the behaviour of ideal gases as given by PV = NkBTCommand term: Determine
15B-1B-15
The ideal gas equation·B.3 Gas laws
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A student uses a molecular-dynamics simulation of methane gas at a constant temperature of 300 K. The simulation includes the attractive forces between the molecules and the volume of the molecules. It gives the density ρ of the gas at eight values of the pressure P.
For an ideal gas of molar mass M, ρ = PM/RT. The student plots ρ/P against P. The graph shows seven of the eight points.
P / MPa
ρ / kg m−3
(ρ/P) / kg m−3 MPa−1
1.0
6.56
6.56
2.0
13.3
6.65
3.0
20.4
6.80
4.0
27.5
6.88
5.0
35.1
6.0
42.6
7.10
7.0
50.7
7.24
8.0
58.6
7.33
Graph drawn to scale
(a)
Calculate the missing value of ρ/P and plot the point on the graph.
(1)
(b)
Draw the line of best fit and extend it to P = 0.
(1)
(c)
Explain why the value of ρ/P at P = 0 is the value that an ideal gas would have at all pressures.
(2)
(d)
Using the intercept of your line, determine the molar mass of methane.
(3)
(e)
The accepted molar mass of methane is 16.0 g mol−1. Comment on your answer to (d).
(1)
(f)
Explain, with reference to the molecular model, why ρ/P increases as P increases.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
35.1 / 5.0 = 7.02 AND plotted correctly
✓ 1
Part (b)
single straight line of best fit extended to the ρ/P axis
✓ 1
Part (c)
for an ideal gas ρ/P = M/RT, which is constant at constant T
✓ 1
as P → 0 the molecules are far apart, so intermolecular forces and the volume of the molecules are negligible and the real gas behaves as an ideal gas
✓ 1
OWTTE.
Part (d)
intercept ≈ 6.44 kg m−3 MPa−1 = 6.44 × 10−6 kg m−3 Pa−1
✓ 1
Accept 6.39–6.49. The conversion MPa → Pa is needed for MP3.
M = intercept × RT
✓ 1
M = 6.44 × 10−6 × 8.31 × 300 = 1.61 × 10−2 kg mol−1 «16.1 g mol−1»
✓ 1
Allow ECF from their intercept.
Part (e)
percentage difference ≈ 0.4 %, so very close agreement: the real gas approaches ideal behaviour as P → 0
✓ 1
Allow ECF. The comparison must be quantitative; accept any difference below 2 % described as good agreement.
Part (f)
as P increases the molecules are closer together and attractive intermolecular forces become significant
✓ 1
molecules hitting the wall are pulled back, so the pressure is less than an ideal gas of the same density would exert; ρ is larger than ideal for a given P
✓ 1
OWTTE.
Answers: (a) 7.02 kg m−3 MPa−1 · (d) 16.1 g 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 and 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 Command term: Determine
16B-1B-16
Moles, molecules & the Avogadro constant·B.3 Gas laws
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A student finds the molar mass M of butane. A gas lighter is weighed, held under water and opened so that butane bubbles into an inverted measuring cylinder full of water. When 200 ± 2 cm3 of gas has been collected, with the water levels inside and outside the cylinder equal, the lighter is closed, dried and weighed again.
The pressure of the room is 1.013 × 105 Pa and the temperature is 293 ± 1 K. The results of five trials are shown.
Trial
1
2
3
4
5
mass of butane released / g
0.461
0.478
0.470
0.482
0.466
(a)
Calculate the mean mass of butane released and its absolute uncertainty.
(2)
(b)
Show that the amount of butane collected is about 8.3 × 10−3 mol.
(1)
(c)
Determine M and its absolute uncertainty. Ignore the uncertainty in the pressure.
(3)
(d)
The accepted molar mass of butane is 58.1 g mol−1. Comment on the student’s result.
(1)
(e)
The gas in the cylinder contains some water vapour. State and explain the effect of this on the value of M.
(2)
(f)
Suggest one other source of systematic error in this procedure, and how it could be reduced.
MP3 is for matching the precision of value and uncertainty.
Part (d)
58.1 lies within the range 54.6–58.7 g mol−1, so the result is consistent with the accepted value
✓ 1
Allow ECF from (c).
Part (e)
the partial pressure of the butane is less than the room pressure «the rest is due to water vapour», so PV/RT overestimates the amount of butane
✓ 1
M = mass/n is too small «underestimated»
✓ 1
Direction must follow from the reasoning.
Part (f)
water left on the lighter makes the final mass too large, so the mass released is too small «M too small»; dry it thoroughly / leave it to dry before weighing OR some butane dissolves in the water/escapes as bubbles; use a gas syringe instead
✓ 1
Source and matching remedy needed.
Answers: (a) (0.471 ± 0.010) g · (b) 8.32 × 10−3 mol · (c) (57 ± 2) g mol−1(the remaining parts are explanations — see the table above)
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/N_A Command term: Determine
17B-2-27
Pressure·B.3 Gas laws
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
A car of mass 1240 kg rests on four identical tyres, each of which supports one quarter of the car's weight. A tyre gauge measures the pressure of the air in a tyre above atmospheric pressure (the gauge pressure).
On a cold morning, when the air in a tyre is at 10 °C, the gauge pressure is 2.20 × 105 Pa. Atmospheric pressure is 1.01 × 105 Pa. Treat the air in the tyre as an ideal gas of constant volume. In a simple model, the force between a tyre and the road equals the gauge pressure multiplied by the area of contact.
(a)
(i)
Calculate the force exerted on the road by one tyre.
(1)
(ii)
Determine the area of contact between one tyre and the road.
(2)
(iii)
Outline why the gauge pressure, rather than the total pressure of the air in the tyre, is used in this model.
(1)
(b)
After a long motorway journey the temperature of the air in the tyre is 45 °C.
(i)
Determine the gauge pressure of the tyre now.
(3)
(ii)
The driver lets air out of the hot tyre, at 45 °C, until the gauge pressure is 2.20 × 105 Pa again. Calculate the percentage of the air molecules that were let out.
(2)
(iii)
Determine the gauge pressure of this tyre when it has cooled back to 10 °C.
(2)
(c)
In practice the volume of a tyre increases slightly as it warms. State and explain the effect of this on your answer to (b)(i).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
F = 1240 × 9.81/4 = 3.04 × 103 N
✓ 1
Part (a)(ii)
A = F/P = 3.04 × 103/2.20 × 105
✓ 1
Allow ECF from (a)(i).
A = 1.38 × 10−2 m² «138 cm²»
✓ 1
Award [1 max] for 9.5 × 10−3 m² «total pressure used».
Part (a)(iii)
the atmosphere also pushes on the outside of the tyre «and on the rest of the car», so only the excess of the inside pressure over atmospheric pressure gives a resultant force
✓ 1
OWTTE.
Part (b)(i)
total pressure = 2.20 × 105 + 1.01 × 105 = 3.21 × 105 Pa AND temperatures of 283 K and 318 K used
P = NkT/V: the same number of molecules at the same temperature in a larger volume collide with each unit area of wall less often
✓ 1
OWTTE.
Answers: (a)(i) 3.04 × 103 N · (a)(ii) 1.38 × 10−2 m² · (b)(i) 2.60 × 105 Pa · (b)(ii) 11 % · (b)(iii) 1.85 × 105 Pa (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; 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 = nRTCommand term: Determine
18B-2-28
The ideal gas equation·B.3 Gas laws
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A diving cylinder has an internal volume of 12.0 × 10−3 m³. At the start of a dive it contains air at a pressure of 2.00 × 107 Pa. Throughout the dive the air in the cylinder is at the temperature of the sea water, 12 °C. Treat air as an ideal gas.
Molar mass of air = 0.029 kg mol−1.
Axes for (d)
(a)
(i)
Show that the cylinder contains about 100 mol of air.
(1)
(ii)
Calculate the mass of air in the cylinder.
(1)
(b)
The diver swims at a depth where the water pressure is 3.0 × 105 Pa. Each minute the diver breathes 15.0 × 10−3 m³ of air, measured at this pressure and at 12 °C. The diver must start to return to the surface when the pressure in the cylinder has fallen to 5.0 × 106 Pa.
(i)
Show that the diver uses about 1.9 mol of air each minute.
(1)
(ii)
Determine whether the diver can stay at this depth for 45 minutes before starting to return.
(3)
(c)
On another dive the diver stays at a depth where the water pressure is 1.5 × 105 Pa and breathes the same volume of air each minute. Explain why the air in the cylinder lasts longer on this dive.
(2)
(d)
On the axes, sketch graphs to show how the pressure of the air in the cylinder varies with time on each dive, until the pressure reaches 5.0 × 106 Pa. Label the graph for the deeper dive D and the graph for the shallower dive S.
(3)
(e)
Suggest why the ideal gas model is less accurate for the air in the full cylinder than for the air that the diver breathes.
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
amount left at 5.0 × 106 Pa = 5.0 × 106 × 12.0 × 10−3/(8.31 × 285) = 25 mol
✓ 1
usable amount = 101 − 25 = 76 mol, so time = 76/1.90 = 40 min
✓ 1
Allow ECF from (a)(i) and (b)(i). Accept 39–40 min.
40 min is less than 45 min, so the diver cannot stay for 45 minutes
✓ 1
ALTERNATIVE: at constant V and T, n ∝ P, so usable fraction = 1.50 × 107/2.00 × 107 = 0.75 ✓; volume available at 3.0 × 105 Pa = 12.0 × 10−3 × 1.50 × 107/3.0 × 105 = 0.60 m³, lasting 0.60/0.015 = 40 min ✓; less than 45 min ✓. MP3 requires a comparison of calculated values.
Part (c)
each breath has the same volume and temperature but half the pressure
✓ 1
n = PV/RT, so each breath contains half as many molecules/moles and the air is removed from the cylinder at half the rate «lasting about 80 min»
✓ 1
OWTTE.
Part (d)
two straight lines, both starting at 2.0 × 107 Pa at t = 0
✓ 1
Do not accept curves.
D reaches 5.0 × 106 Pa at about 40 min
✓ 1
Allow ECF from (b)(ii).
S has half the gradient of D, reaching 5.0 × 106 Pa at about 80 min
✓ 1
Labels D and S needed for MP2 and MP3.
Part (e)
in the cylinder the pressure/density is much higher, so the molecules are closer together and their volume and the intermolecular forces are not negligible
✓ 1
OWTTE.
Answers: (a)(i) 101 mol · (a)(ii) 2.9 kg · (b)(i) 1.90 mol · (b)(ii) 40 min: no (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; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas Command term: Determine
19B-2-29
The empirical gas laws·B.3 Gas laws
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksDetermine
The height of an office chair is supported by a gas spring: a sealed cylinder of nitrogen closed by a piston of area 3.0 × 10−4 m². When nobody sits on the chair the piston rests against a stop, the nitrogen has a volume of 1.20 × 10−4 m³ and its pressure is 1.60 × 106 Pa at 20 °C.
In this model the force exerted by the nitrogen on the piston equals the weight that the piston supports. Ignore atmospheric pressure, friction and the weight of the seat. Treat nitrogen as an ideal gas.
Graph drawn to scale
(a)
Calculate the force exerted by the nitrogen on the piston when nobody sits on the chair.
(1)
(b)
A person of mass 60 kg sits slowly on the chair, so that the nitrogen is compressed at a constant temperature of 20 °C. The graph shows this change.
(i)
Show that the pressure of the nitrogen is about 2.0 × 106 Pa when the person is supported.
(1)
(ii)
Determine the distance that the piston moves into the cylinder.
(3)
(c)
On a hot day the nitrogen is at 35 °C before anyone sits on the chair. The piston rests against the stop.
(i)
Calculate the pressure of the nitrogen.
(1)
(ii)
Determine the distance that the piston moves when the same person sits slowly on the chair, at a constant temperature of 35 °C.
(2)
(d)
Draw, on the graph, the change in (c)(ii).
(2)
(e)
Calculate the number of nitrogen molecules in the gas spring.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = PA = 1.60 × 106 × 3.0 × 10−4 = 480 N
✓ 1
Part (b)(i)
P = mg/A = 60 × 9.81/3.0 × 10−4 = 1.96 × 106 Pa
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Answers: (a) 480 N · (b)(i) 1.96 × 106 Pa · (b)(ii) 7.4 cm · (c)(i) 1.68 × 106 Pa · (c)(ii) 5.7 cm · (e) 4.7 × 1022(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; 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; changes of state of an ideal gas can be represented on pressure–volume diagrams; the equations governing the behaviour of ideal gases as given by PV = NkBT (linked: A.2 — weight Fg = mg) Command term: Determine
20B-2-36
Kinetic model & molecular speeds·B.3 Gas laws
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDetermine
A cubical box of side 0.10 m contains 1.0 × 1023 molecules of oxygen, each of mass 5.3 × 10−26 kg. In a simplified model, every molecule moves with the same speed of 500 m s−1, one third of the molecules move parallel to each edge of the box, the molecules do not collide with each other, and all collisions with the walls are elastic.
(a)
(i)
Show that the change in momentum of a molecule when it collides with a wall is about 5 × 10−23 kg m s−1.
(1)
(ii)
Calculate the time between successive collisions of one molecule with the same wall.
(1)
(iii)
Determine the average force that one molecule exerts on that wall.
(2)
(b)
(i)
Determine the pressure exerted on one wall of the box.
(3)
(ii)
Show that your answer to (b)(i) is consistent with P = ⅓ρv̄².
(2)
(c)
(i)
Determine the temperature of the gas from the average kinetic energy of its molecules.
(2)
(ii)
Show that the equation PV = NkBT gives the same temperature.
(1)
(d)
The gas is heated at constant volume until its kelvin temperature has doubled. Deduce the new speed of the molecules and the new pressure in this model.
P = ⅓ × 5.3 × 500² = 4.4 × 105 Pa, the same as (b)(i)
✓ 1
Part (c)(i)
Ek = ½mv² = ½ × 5.3 × 10−26 × 500² = 6.6 × 10−21 J
✓ 1
T = 2Ek/3kB = 320 K
✓ 1
Award [1 max] for 480 K «Ek = kBT used».
Part (c)(ii)
T = PV/NkB = 4.42 × 105 × 1.0 × 10−3/(1.0 × 1023 × 1.38 × 10−23) = 320 K
✓ 1
Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b)(i).
Part (d)
v ∝ √T, so v = 500√2 ≈ 710 m s−1
✓ 1
Do not accept 1000 m s−1.
P ∝ T «or ∝ v²», so P = 8.8 × 105 Pa
✓ 1
Allow ECF from (b)(i).
Answers: (a)(ii) 4.0 × 10−4 s · (a)(iii) 1.3 × 10−19 N · (b)(i) 4.4 × 105 Pa · (c)(i) 320 K · (d) 710 m s−1; 8.8 × 105 Pa (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 = ⅓ρv̄²; the equations governing the behaviour of ideal gases as given by PV = NkBT and PV = nRT (linked: B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = 3/2 kBT; A.2 — Newton's three laws of motion) Command term: Determine
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