67 free, high-quality past-paper-style questions on Gas laws for IB Physics HL, each with a full mark scheme. Covers: The kinetic model of an ideal gas; The amount of substance; Pressure and molecular speed; and more. For when you have finished the past papers, or want to practise one topic at a time.B.3 at Standard Level →Read the B.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.293 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1B-1A-45The kinetic model of an ideal gasPaper 1AEasy1 mark
Which statement is an assumption of the kinetic model of an ideal gas?
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Step 1The kinetic model treats the molecules as point-like particles in random motion, with negligible total volume.
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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.
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Step 3So the internal energy of an ideal gas is entirely the random kinetic energy of the molecules; there is no intermolecular potential energy.
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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.
3IBP2H-014-00-01Pressure and molecular speedPaper 1AEasy1 mark
A sample of ideal gas is at 27 °C. To what temperature must the gas be heated so that the average translational kinetic energy of its molecules doubles?
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The average translational kinetic energy is proportional to the absolute (kelvin) temperature: E = (3/2)kT. 27 °C = 300 K, so doubling the energy requires 600 K, which is 600 − 273 = 327 °C. Doubling the Celsius value (54 °C) ignores the kelvin scale. 600 °C forgets to convert back to Celsius. 151 °C (424 K) wrongly assumes the energy is proportional to √T.
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Answer C
Answer: C
Common error: Any proportionality involving temperature in the gas laws or kinetic theory uses kelvin. Convert to kelvin first, scale, then convert back if the answer is asked for in °C.
4IBP2H-014-00-04The kinetic model of an ideal gasPaper 1AEasy1 mark
Which statement is not an assumption of the kinetic model of an ideal gas?
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The kinetic model assumes elastic collisions, no intermolecular forces except during collisions, and negligible molecular volume. It does not assume equal speeds: the molecules have a distribution of speeds, which is why a root mean square speed is used. Collisions continually change individual speeds.
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Answer D
Answer: D
Common error: Temperature fixes the mean kinetic energy, not the speed of every molecule. The speeds are spread over a wide range at any temperature.
5IBP2H-014-01-01Pressure and molecular speedPaper 1AEasy1 mark
A fixed mass of ideal gas is held in a rigid, sealed container. The gas is heated until the root-mean-square (rms) speed of its molecules has doubled. What is the ratio new pressure ÷ original pressure?
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Kinetic theory gives p = ⅓ρc2rms. In a rigid sealed container the density ρ is constant, so p ∝ c2rms. Doubling crms multiplies the pressure by 22 = 4. The answer 1 wrongly assumes a fixed volume means a fixed pressure. The answer 2 treats p as proportional to speed rather than speed squared. The answer 16 squares the factor twice.
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Answer C
Answer: C
Common error: Pressure depends on the square of the rms speed. It is also proportional to absolute temperature, so doubling crms means the kelvin temperature has quadrupled.
6IBP2H-014-01-04The ideal gas equationPaper 1AEasy1 mark
A sealed gas syringe has a freely moving, frictionless piston, so the trapped air stays at atmospheric pressure. The syringe is warmed in a water bath from 20 °C to 40 °C. What is the approximate ratio final volume ÷ initial volume of the trapped air?
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At constant pressure and fixed amount of gas, V ∝ T, where T is the absolute (kelvin) temperature. T1 = 293 K and T2 = 313 K, so V2/V1 = 313/293 ≈ 1.07. 2.0 comes from using Celsius temperatures (40/20). 0.94 inverts the ratio (293/313). 0.50 uses Celsius temperatures and also inverts the ratio (20/40).
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Answer B
Answer: B
Common error: The gas laws give proportionality with temperature only in kelvin. Doubling a Celsius temperature does not double the volume.
7IBP3H-B3-20r12-01The ideal gas equationPaper 1AEasy1 mark
A rigid sealed vessel contains a fixed mass of volcanic gas sampled at 27 °C. The vessel is heated to 327 °C. The pressure of the gas becomes
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At constant volume and fixed mass, p ∝ T with T in kelvin: 300 K to 600 K, so the pressure doubles (B). A is the ratio 327/27, using Celsius values. C assumes p ∝ T2. D inverts the ratio, but pressure rises with temperature.
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Answer B
Answer: B
Common error: Gas-law ratios need the temperature in kelvin; ratios of Celsius values are meaningless.
8IBP3H-B3-20r12-02The ideal gas equationPaper 1AEasy1 mark
A cargo tank on a ship has volume 0.020 m3 and contains an ideal gas at pressure 2.0×105 Pa and temperature 300 K. The Boltzmann constant is 1.38×10−23 J K−1. Which is the best estimate of the number of molecules in the tank?
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N = pV/(kBT) = (2.0×105 × 0.020)/(1.38×10−23 × 300) = 9.66e+23, i.e. 9.7×1023 (A). B takes the volume as 20 instead of 0.020 m3 (unit conversion). C leaves out the temperature. D uses 27 instead of 300 K, i.e. a temperature that has not been converted.
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Answer A
Answer: A
Exam tip: Convert every quantity to SI units before substituting into pV = NkBT.
9IBP3H-B3-20r12-03The kinetic model of an ideal gasPaper 1AEasy1 mark
The gas in a sealed, rigid canister in a theatre fog machine is heated. Which statement correctly explains why the pressure of the gas increases?
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Higher temperature means a larger mean kinetic energy and speed, so collisions with the walls are more frequent and each gives a larger momentum change; the force per unit area rises (B). A is wrong because the molecules are treated as points with negligible volume. C is wrong because the number of molecules in a sealed canister is fixed. D is wrong because the ideal-gas model neglects intermolecular forces.
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Answer B
Answer: B
Common error: Pressure in the kinetic model comes from molecular collisions with the walls, not from repulsion between molecules.
10B-1A-19The ideal gas equationPaper 1AMedium1 mark
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)?
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Step 1Pressure in the bubble at the bottom = 1.0 × 105 + 2.0 × 105 = 3.0 × 105 Pa; temperatures 280 K and 300 K.
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12B-1A-41Pressure and molecular speedPaper 1AMedium1 mark
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?
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Step 1Density of the gas: ρ = Nm/V.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Kinetic model: p = ⅓ρv2 = Nmv2/(3V).
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Step 3Force on the wall: F = pA = Nmv2A/(3V).
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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).
13B-1A-42The ideal gas equationPaper 1AMedium1 mark
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?
Diagram not to scale
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Step 1At constant temperature the amount of gas is proportional to pV, and the total amount is conserved.
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14B-1A-43Pressure and molecular speedPaper 1AMedium1 mark
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)?
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Step 1p = ⅓ρv², where v is the root mean square speed.
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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.
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Step 3Doubling the rms speed multiplies v² by 4, so p → ⅓(ρ/2)(4v²) = 2p.
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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.
15B-1A-44Internal energy of an ideal gasPaper 1AMedium1 mark
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
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Step 1I: N = pV/kBT is the same for both, whatever the mass of the atoms.
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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.
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Step 3III: ½mv2 = ³⁄₂kBT is the same, so the lighter helium atoms are faster, by √10 ≈ 3.2. III is different.
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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.
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.
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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.
17IBP2H-014-00-02Internal energy of an ideal gasPaper 1AMedium1 mark
A neon lamp manufacturer fills one tube with helium (molar mass 4 g mol−1) and another with argon (molar mass 40 g mol−1). Both gases are at the same temperature. What is
(root mean square speed of helium atoms) / (root mean square speed of argon atoms)?
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At the same temperature the atoms have the same average kinetic energy, ½mc2 = (3/2)kT, so crms ∝ 1/√m. The ratio is √(40/4) = √10 ≈ 3.2. Choosing 1 confuses equal kinetic energy with equal speed; 10 forgets the square root; 1/√10 inverts the relationship, as if heavier atoms moved faster.
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Answer B
Answer: B
Exam tip: Same temperature means same mean kinetic energy, not same speed. Lighter particles must move faster to carry the same energy.
18IBP2H-014-00-03Boyle's law with a gas syringePaper 1AMedium1 mark
A student investigates Boyle's law using a sealed gas syringe connected to a pressure sensor. The plunger is moved slowly so that the temperature of the trapped air remains constant. Pressure p and volume V are recorded. Which graph would be a straight line passing through the origin?
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At constant temperature and fixed amount of gas, pV = constant, so p = constant × (1/V). A graph of p against 1/V is therefore a straight line through the origin. p against V is a curve (hyperbola). pV against V is a horizontal line, which does not pass through the origin. 1/p = V/constant, so 1/p against 1/V is a curve, not a straight line.
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Answer B
Answer: B
Exam tip: To test an inverse proportionality, rearrange it into the form y = mx and check that the plotted quantities match y and x exactly.
19IBP2H-014-00-05The kinetic model of an ideal gasPaper 1AMedium1 mark
Carbon dioxide is stored for use in a fire extinguisher. Under which conditions, and for which reason, does the gas deviate most from ideal behaviour?
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Ideal behaviour requires negligible molecular volume and negligible intermolecular forces. At high pressure the molecules are close together, so their own volume and the attractive forces between them become significant; at low temperature they move slowly, so attractive forces have more effect during encounters. Frequent wall collisions alone do not cause non-ideal behaviour, and low pressure with high temperature is the condition where real gases are most nearly ideal.
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Answer B
Answer: B
Exam tip: Remember the pairing: real gases are most ideal at low pressure and high temperature, and least ideal at high pressure and low temperature, close to liquefaction.
20IBP2H-014-00-08Internal energy of an ideal gasPaper 1AMedium1 mark
A cylinder used to fill party balloons contains helium, treated as an ideal monatomic gas, at a pressure of 2.0 ×105 Pa in a volume of 0.030 m3. What is the internal energy of the helium?
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For an ideal monatomic gas the internal energy is entirely translational kinetic energy: U = (3/2)NkT = (3/2)nRT = (3/2)pV = 1.5 × 2.0 ×105 × 0.030 = 9.0 ×103 J. 6.0 ×103 J is just pV, omitting the factor 3/2; 3.0 ×103 J uses ½ instead of 3/2; 1.8 ×104 J uses a factor of 3, a confusion with the 1/3 in p = ⅓ρc2.
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Answer C
Answer: C
Exam tip: Because nRT = pV, the internal energy of an ideal monatomic gas can be found directly from pressure and volume, without knowing the temperature.
21IBP2H-014-01-02Internal energy of an ideal gasPaper 1AMedium1 mark
Sample X is 2.0 mol of helium at 300 K. Sample Y is 1.0 mol of argon at 600 K. Both behave as ideal monatomic gases. The molar mass of argon is ten times that of helium. What is internal energy of X ÷ internal energy of Y?
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For an ideal monatomic gas U = (3/2)nRT. This does not depend on the molar mass. UX/UY = (2.0 × 300)/(1.0 × 600) = 1. 0.2 assumes U is proportional to the mass of the sample (8 g compared with 40 g). 0.5 compares the temperatures only. 2 compares the amounts of gas only.
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Answer C
Answer: C
Exam tip: The internal energy of an ideal gas is the total random kinetic energy of its particles. It depends on both the amount of gas and the absolute temperature, but not on the particle mass.
22IBP2H-014-01-03Changes of state on a p–V diagramPaper 1AMedium1 mark
A fixed mass of ideal gas can exist in any of four states, shown as points on a p–V diagram: A: 100 kPa, 3.0 L B: 200 kPa, 2.0 L C: 300 kPa, 1.0 L D: 150 kPa, 1.5 L In which state is the temperature of the gas greatest?
The four states A, B, C and D
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For a fixed mass of ideal gas, T ∝ pV. The products pV are: A: 300 kPa L B: 400 kPa L C: 300 kPa L D: 225 kPa L B has the largest pV, so B has the highest temperature. C has the highest pressure and A has the largest volume, but neither has the largest product.
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Answer B
Answer: B
Exam tip: On a p–V diagram, isotherms further from the origin correspond to higher temperatures. To compare temperatures, compare the product pV, not p or V separately.
23IBP2H-014-01-05Pressure and molecular speedPaper 1AMedium1 mark
The speed of sound in a monatomic ideal gas depends on how fast disturbances are passed between molecules by their random motion. Which change would increase the speed of sound in the gas?
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Sound is passed on by molecular collisions, so its speed scales with the rms molecular speed, √(3RT/M). Raising T increases the molecular speeds and therefore the speed of sound. At constant temperature, changing the pressure or the density changes p and ρ in the same proportion. The ratio p/ρ = RT/M, and so the molecular speed, stays the same. A monatomic gas of larger molar mass at the same T has slower molecules, so sound travels more slowly in it.
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Answer B
Answer: B
Common error: Sound in a gas is not faster at higher pressure or density if the temperature is unchanged. For a given type of gas, only T and the molar mass M affect the molecular speed.
24IBP2H-014-01-07Pressure and molecular speedPaper 1AMedium1 mark
Dry air in a sports hall has a density of 1.2 kg m−3 and a pressure of 1.0 × 105 Pa. Assume the air behaves as an ideal gas. What is the rms speed of the air molecules?
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From p = ⅓ρc2rms: crms = √(3p/ρ) = √(3 × 1.0 × 105/1.2) = √(2.5 × 105) = 500 m s−1. 290 m s−1 omits the factor 3. 410 m s−1 uses a factor 2 instead of 3. 2.5 × 105 m s−1 forgets the square root.
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Answer C
Answer: C
Exam tip: Check the order of magnitude. Air molecules at room temperature move at a few hundred metres per second, which is comparable to the speed of sound.
25IBP3H-B3-20r12-04Molecular speedsPaper 1AMedium1 mark
Sulfur dioxide (molar mass 64 g mol−1) and carbon dioxide (molar mass 44 g mol−1) are emitted together from a vent and are at the same temperature. What is the value of (r.m.s. speed of SO2 molecules) ÷ (r.m.s. speed of CO2 molecules)?
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At equal temperature the mean kinetic energy is equal, so vrms ∝ 1/√M. The ratio is √(44/64) = 0.829 = 0.83 (C). A is 44/64 without the square root. B is √(64/44), which inverts the ratio. D is 64/44, inverted and without the square root.
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Answer C
Answer: C
Exam tip: Equal mean kinetic energy means the heavier molecules are slower: vrms ∝ M⁻¹/².
26IBP3H-B3-20r12-05Real and ideal gasesPaper 1AMedium1 mark
A gas cylinder on a cargo ship holds gas at 8.0 MPa and 300 K. A leak allows 25% of the molecules to escape, and the remaining gas cools to 270 K. The volume of the cylinder does not change. What is the new pressure?
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p = nRT/V with V fixed, so p2 = 8.0 × 0.75 × (270/300) = 5.4 MPa (D). A allows for the loss of gas only. B allows for the cooling only. C uses 300/270 for the temperature, which inverts the effect of cooling.
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Answer D
Answer: D
Exam tip: With fixed volume, pressure is proportional to both the amount of gas and its kelvin temperature, so apply both factors.
27IBP3H-B3-20r12-06Pressure and molecular speedPaper 1AMedium1 mark
In a theatre, a fixed mass of gas in a cylinder with a piston is compressed slowly at constant temperature until its volume is one third of its original volume. Which row correctly describes the changes in the r.m.s. speed of the molecules and in the rate of collisions with the walls per unit area of wall?
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The temperature is constant, so the mean kinetic energy and r.m.s. speed are unchanged. The number of molecules per unit volume triples, so the collision rate per unit wall area triples, which matches p ∝ 1/V (C). A assumes compression heats the gas, but the temperature is fixed. B ignores the higher number density. D has no basis: a factor 9 would come from squaring the density ratio.
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Answer C
Answer: C
Common error: Constant temperature means constant mean kinetic energy; pressure rises because the number density rises, not because the molecules speed up.
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
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Step 1For an ideal gas pV = nRT, so the gradient of the pV–T line is nR, proportional to the amount of substance.
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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.
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Step 3Molar mass = mass/amount: MX/MY = (mX/mY) × (nY/nX) = 2 × ⅓ = ⅔, so MX = 2M/3.
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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.
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?
Diagram not to scale
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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.
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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.
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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.
30IBP2H-014-00-06Molecular speedsPaper 1AHard1 mark
Two rigid containers have equal volumes and are at the same pressure. Container A holds hydrogen (molar mass 2 g mol−1) at kelvin temperature T. Container B holds oxygen (molar mass 32 g mol−1) at kelvin temperature 2T. Both gases are ideal. What is (mass of gas in A) / (mass of gas in B)?
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From pV = nRT with equal p and V, n ∝ 1/T, so nA/nB = 2T/T = 2. Mass = nM, so mA/mB = 2 × (2/32) = 1/8. The value 1/16 ignores the temperature difference; 2 is only the ratio of amounts, ignoring molar mass; 8 inverts the ratio.
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Answer B
Answer: B
Exam tip: Split the problem into two steps: use pV = nRT to find the ratio of amounts, then multiply by the molar masses to get the mass ratio.
31IBP2H-014-00-07Pressure and molecular speedPaper 1AHard1 mark
A rigid container holds an ideal gas. Half of the molecules leak out, and the remaining gas is then heated until the root mean square speed of its molecules is twice its original value. What is (final pressure) / (initial pressure)?
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p = NkT/V with V fixed. Since ½mc2 ∝ T, doubling crms multiplies T by 4. Halving N then gives pfinal/pinitial = ½ × 4 = 2. The value 1 comes from treating T as proportional to speed rather than speed squared; 4 ignores the leak; 1/2 ignores the heating.
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Answer C
Answer: C
Common error: Temperature is proportional to the square of the rms speed. Doubling the speed quadruples the kelvin temperature.
32IBP2H-014-01-06Molecular speedsPaper 1AHard1 mark
In the upper atmosphere, hydrogen molecules (H2) and nitrogen molecules (N2) are at the same temperature. Over long periods, hydrogen escapes into space much more readily than nitrogen. Which statement best explains this?
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At the same temperature, all molecules have the same mean kinetic energy, (3/2)kT. Hydrogen molecules are 14 times less massive than nitrogen molecules, so their rms speed is √14 ≈ 3.7 times greater. A much larger fraction of the speed distribution therefore lies above the escape speed. The escape speed is independent of the mass of the escaping particle, so option C is wrong. Option A contradicts the equal mean kinetic energy at equal T. Option D is false and irrelevant.
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Answer B
Answer: B
Exam tip: Equal temperature means equal mean kinetic energy, not equal speed. The speed then follows from ½mv² = (3/2)kT.
33IBP2H-014-01-08Internal energy of an ideal gasPaper 1AHard1 mark
A rigid helium cylinder used for filling party balloons slowly leaks. Half of the gas atoms escape while the temperature of the remaining gas stays constant. Treat the helium as an ideal monatomic gas. I. The mean kinetic energy of the remaining atoms is unchanged. II. The internal energy of the remaining gas is halved. III. The rms speed of the remaining atoms is halved. Which statements are correct?
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Statement I is correct. The mean kinetic energy per atom is (3/2)kT, which depends only on T. Statement II is correct. The internal energy of an ideal monatomic gas is U = (3/2)NkT, so halving N at constant T halves U. Statement III is wrong. The rms speed, \(\sqrt{3kT/m}\), depends only on T and the atomic mass, so it is unchanged. The pressure halves because fewer atoms strike the walls, not because each atom moves more slowly. The answer is I and II only.
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Answer B
Answer: B
Common error: A fall in pressure does not always mean slower particles. Per-particle quantities such as mean kinetic energy and rms speed depend only on T. Totals such as internal energy depend on both N and T.
34IBP3H-B3-20r12-07Molecular speedsPaper 1AHard1 mark
Argon (molar mass 40 g mol−1) in a vessel is at 300 K. Helium (molar mass 4.0 g mol−1) is to be cooled until its r.m.s. speed equals the r.m.s. speed of the argon. What is the temperature of the helium?
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vrms2 = 3RT/M, so equal speeds need T/M equal: T = 300 × (4.0/40) = 30 K (D). A, 300×√10, treats T ∝ M1/2 inverted. B, 300×10, inverts the ratio. C, 300×√0.1 = 95 K, takes T ∝ √M, but the mean-square speed (not the speed) is proportional to T.
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Answer D
Answer: D
Common error: T is proportional to vrms², not to vrms; square the speed relationship before comparing temperatures.
35IBP3H-B3-20r12-08The ideal gas equationPaper 1AHard1 mark
A fixed mass of gas in a ship's engine-room tank is heated from 300 K to 450 K while its volume increases by 20%. By what factor does the pressure change?
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pV/T is constant, so p2/p1 = (T2/T1)(V1/V2) = 1.50/1.20 = 1.25 (D). A allows for the volume change only (1/1.2). B allows for the temperature change only. C multiplies by 1.2 instead of dividing, which is wrong because larger volume lowers the pressure.
✓ 1
Answer D
Answer: D
Exam tip: Write p₁V₁/T₁ = p₂V₂/T₂ and rearrange for the unknown rather than combining the factors by intuition.
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
Data from the table
(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)
37B-1B-16Speed of sound and molecular motionPaper 1BMedium7 marks
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)
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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)
38B-1B-06The empirical gas lawsPaper 1BHard7 marks
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)
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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)
39B-1B-10Boyle's law with a gas syringePaper 1BHard7 marks
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)
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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)
40IBP2H-014-00-09The ideal gas equationPaper 2Easy6 marks
Aircraft tyres are filled with nitrogen. Before a flight, one tyre contains gas at an absolute pressure of 1.40 ×106 Pa and a temperature of 15 °C. The volume of the tyre is 0.12 m3 and may be assumed constant. During landing, heat from the brakes raises the temperature of the gas to 75 °C. Treat the nitrogen as an ideal gas.
(a)
Calculate the amount of nitrogen, in mol, in the tyre.
(2)
(b)
Calculate the pressure of the nitrogen after landing.
(2)
(c)
Explain, in terms of the molecules, why the pressure increases.
(2)
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Notes
Part (a)
n = pV/(RT) = (1.40 ×106 × 0.12)/(8.31 × 288) ; n = 70 mol (accept 70.2 mol)
✓ 2
Part (b)
Constant V and n so p/T is constant, using kelvin temperatures 288 K and 348 K ; p = 1.40 ×106 × 348/288 = 1.69 ×106 Pa
✓ 2
Part (c)
Higher temperature means greater mean kinetic energy / greater average speed of the molecules ; molecules hit the walls more often and with greater change of momentum per collision, so the force per unit area on the walls increases
✓ 2
Common error: Using 75/15 instead of 348/288 gives 7.0 ×10⁶ Pa, five times the starting pressure and more than four times the correct answer. Always convert to kelvin before using a temperature ratio.
41IBP2H-014-01-09The ideal gas equationPaper 2Easy5 marks
A racing-car tyre is filled with air at an absolute pressure of 2.9 × 105 Pa when the air temperature is 7 °C. After several laps the air in the tyre reaches 35 °C. Assume the volume of the tyre stays constant and no air escapes.
(a)
State the gas law that applies to the air in the tyre.
(1)
(b)
Calculate the absolute pressure of the air in the tyre at 35 °C.
(2)
(c)
Outline, using the kinetic model, why the pressure increases.
(2)
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Notes
Part (a)
The pressure law (accept Gay-Lussac's law): for a fixed mass of gas at constant volume, the pressure is proportional to the absolute temperature (p/T = constant).
✓ 1
Part (b)
Convert to kelvin: T1 = 280 K and T2 = 308 K (accept 280.15 K and 308.15 K).
42IBP3H-B3-20r12-09The kinetic model of an ideal gasPaper 2Easy5 marks
A theatre smoke machine contains a tank of volume 2.5×10−3 m3 holding 0.80 mol of nitrogen gas at 295 K. Treat the nitrogen as an ideal gas. Use R = 8.31 J mol−1 K−1 and NA = 6.02×1023 mol−1.
(a)
State two assumptions of the kinetic model of an ideal gas.
(2)
(b)
Calculate the pressure of the nitrogen in the tank. Give your answer to 3 significant figures.
(2)
(c)
Calculate the number of nitrogen molecules in the tank. Give your answer to 3 significant figures.
(1)
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Notes
Part (a)
The volume of the molecules is negligible compared with the volume of the container
✓ 1
There are no intermolecular forces except during collisions
✓ 1
Accept other valid assumptions, for example elastic collisions, random motion or Newton's laws applying.
—
Part (b)
Uses pV = nRT
✓ 1
p = (0.80 × 8.31 × 295)/(2.5×10−3) = 7.84×105 Pa
✓ 1
Part (c)
N = 0.80 × 6.02×1023 = 4.82×1023
✓ 1
Exam tip: Keep the volume in m³ when using R = 8.31 J mol⁻¹ K⁻¹ to get the pressure in pascals.
43IBP3H-B3-20r12-10Boyle's law with a gas syringePaper 2Easy6 marks
A volcanologist collects 450 cm3 of gas from a fumarole in a syringe. The gas is at a pressure of 1.00×105 Pa and a temperature of 290 K. The gas is first compressed slowly to a pressure of 2.50×105 Pa at constant temperature. It is then heated at constant pressure until its temperature reaches 350 K.
(a)
State Boyle's law.
(2)
(b)
Calculate the volume of the sample after the compression.
(2)
(c)
Calculate the volume of the sample after it is then heated to 350 K at constant pressure. Use the volume from (b). Give your answer to 3 significant figures.
(2)
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Part (a)
For a fixed mass of gas at constant temperature
✓ 1
the pressure is inversely proportional to the volume, i.e. pV is constant
✓ 1
Part (b)
p1V1 = p2V2
✓ 1
V2 = 450 × (1.00×105)/(2.50×105) = 180 cm3
✓ 1
Part (c)
V ∝ T at constant pressure, so V3 = V2 × 350/290
✓ 1
V3 = 180 × 350/290 = 217.2, so 217 cm3
✓ 1
Exam tip: State the condition (fixed mass, constant temperature or pressure) whenever you quote an empirical gas law.
44B-2-12Kinetic model of an atmospherePaper 2Medium9 marks
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)
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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)
45B-2-24The ideal gas equationPaper 2Medium10 marks
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)
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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)
46B-2-37Pressure and the kinetic modelPaper 2Medium12 marks
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)
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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)
47IBP2H-014-00-10The ideal gas equationPaper 2Medium8 marks
A portable medical oxygen cylinder has an internal volume of 1.00 ×10−2 m3. When full, it contains oxygen at a pressure of 1.50 ×107 Pa and a temperature of 20 °C. The molar mass of oxygen is 32 g mol−1. Assume the oxygen behaves as an ideal gas and that its temperature remains at 20 °C throughout.
(a)
Calculate the mass of oxygen in the full cylinder.
(3)
(b)
A patient breathes the oxygen at a rate of 2.0 litres per minute, the volume being measured at atmospheric pressure 1.01 ×105 Pa and 20 °C. The cylinder must be replaced when its pressure falls to 1.0 ×106 Pa. Determine the time, in minutes, for which the cylinder can be used. (1 litre = 1.0 ×10−3 m3)
(3)
(c)
Suggest two reasons why oxygen at 1.50 ×107 Pa may not behave as an ideal gas.
(2)
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Notes
Part (a)
n = pV/(RT) = (1.50 ×107 × 1.00 ×10−2)/(8.31 × 293) ; n = 61.6 mol ; mass = 61.6 × 0.032 = 1.97 kg
✓ 3
Part (b)
Either method earns the first mark. Method 1: usable gas has pV = (1.50 ×107 − 1.0 ×106) × 1.00 ×10−2 = 1.40 ×105 J. Method 2: amount removed = 61.6 − 4.1 = 57.5 mol (ECF from (a)) ; volume at atmospheric pressure = 1.40 ×105/1.01 ×105 = 1.39 m3 ≈ 1390 litres (1386 L unrounded) ; time = 1386/2.0 ≈ 690 min (693 min, about 11.5 h). Accept 690–695 min
✓ 3
Part (c)
Any two valid reasons, for example: the molecules are close together, so their own volume is not negligible compared with the cylinder volume ; intermolecular forces are no longer negligible because the molecules are close together for a significant fraction of the time
✓ 2
Exam tip: At constant temperature the quantity pV is proportional to the amount of gas. Comparing pV before and after avoids calculating the number of moles twice.
48IBP2H-014-00-13Pressure and the kinetic modelPaper 2Medium7 marks
A basketball has a fixed internal volume of 7.1 ×10−3 m3. Initially it contains air at 1.0 ×105 Pa and 20 °C. It is pumped up to an absolute pressure of 1.6 ×105 Pa, the temperature remaining at 20 °C. Each stroke of the pump delivers 3.0 ×10−4 m3 of air, measured at 1.0 ×105 Pa and 20 °C. Treat air as an ideal gas.
(a)
Calculate the number of molecules in the ball at 1.6 ×105 Pa.
(2)
(b)
Determine the minimum number of pump strokes needed.
(3)
(c)
The ball is taken to an outdoor court where the temperature is 2.0 °C. Calculate the new pressure of the air in the ball.
(2)
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Part (a)
N = pV/(kT) = (1.6 ×105 × 7.1 ×10−3)/(1.38 ×10−23 × 293) ; N = 2.8 ×1023
49IBP2H-014-00-14Pressure and molecular speedPaper 2Medium7 marks
A lander on Mars measures the atmospheric pressure near the surface as 610 Pa and the density of the atmosphere as 0.0154 kg m−3. The atmosphere may be treated as an ideal gas of carbon dioxide, molar mass 0.044 kg mol−1.
(a)
Explain, using the kinetic model, how the molecules of a gas exert a pressure on a surface.
(3)
(b)
Calculate the root mean square speed of the carbon dioxide molecules.
(2)
(c)
Determine the temperature of the atmosphere near the lander.
(2)
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Part (a)
Molecules collide elastically with the surface and rebound, so each undergoes a change of momentum ; the surface exerts a force on each molecule to change its momentum, so by Newton's third law the molecules exert an equal and opposite force on the surface ; the very large number of collisions per second gives a steady average force, and pressure is this force per unit area
✓ 3
Part (b)
p = ⅓ρc2 so crms = √(3p/ρ) ; crms = √(3 × 610/0.0154) = 345 m s−1
✓ 2
Part (c)
½mc2 = (3/2)kT with m = M/NA, giving T = Mc2/(3R) ; T = 0.044 × 3452/(3 × 8.31) = 210 K . ECF from (b)
✓ 2
Exam tip: Dividing k by the molecular mass is the same as dividing R by the molar mass, so T = Mc²/(3R) avoids calculating the mass of a single molecule.
50IBP2H-014-00-16Molecular speedsPaper 2Medium8 marks
A meteorological sounding balloon is filled with helium at ground level, where the pressure is 1.0 ×105 Pa and the temperature is 288 K. The balloon's volume at launch is 4.0 m3. At a certain altitude the pressure is 2.5 ×104 Pa and the temperature is 223 K. Assume the helium is an ideal gas at the same pressure and temperature as its surroundings and that none escapes.
(a)
Calculate the amount of helium, in mol, in the balloon.
(2)
(b)
Calculate the volume of the balloon at the altitude.
(2)
(c)
Outline why a helium atom inside the balloon has the same mean translational kinetic energy as a nitrogen molecule in the surrounding air, but a greater root mean square speed.
(2)
(d)
Calculate the mean translational kinetic energy of a helium atom at the altitude.
(2)
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Part (a)
n = pV/(RT) = (1.0 ×105 × 4.0)/(8.31 × 288) ; n = 167 mol (≈170 mol)
✓ 2
Part (b)
pV/T constant (or V = nRT/p using the answer to (a)): V2 = V1 × (p1/p2) × (T2/T1) = 4.0 × 4.0 × 223/288 ; V2 = 12.4 m3 (12 m3 to 2 s.f.)
✓ 2
Part (c)
Mean translational kinetic energy depends only on temperature, (3/2)kT, and both gases are at the same temperature ; helium atoms have a smaller mass, so since ½mcrms2 is the same, crms must be larger
✓ 2
Part (d)
E = (3/2)kT ; E = 1.5 × 1.38 ×10−23 × 223 = 4.6 ×10−21 J
✓ 2
Exam tip: In the combined gas law, check each ratio makes physical sense: lower pressure should make the volume larger, while lower temperature should make it smaller.
51IBP2H-014-01-10The ideal gas equationPaper 2Medium7 marks
A scuba diver at a depth of 20 m in sea water fills her lungs to a volume of 6.0 L with air from her regulator. The air is supplied at the surrounding water pressure. The density of sea water is 1030 kg m−3 and atmospheric pressure is 1.01 × 105 Pa. The air in her lungs is at 37 °C.
(a)
Calculate the total pressure at a depth of 20 m.
(2)
(b)
She rises to the surface without breathing out. Calculate the volume the air in her lungs would occupy at the surface if her lungs could expand freely.
(2)
(c)
State one assumption you made when calculating the volume at the surface.
(1)
(d)
Determine the amount, in moles, of air in her lungs at depth.
(2)
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Part (a)
Pressure due to the water: ρgh = 1030 × 9.8 × 20 = 2.02 × 105 Pa (accept g = 9.81).
Allow error carried forward from the total pressure calculated in the first part.
—
Exam tip: The pressure at depth is atmospheric pressure plus ρgh. Leaving out the atmospheric term is a common way to lose the first mark. Always convert litres to m³ and °C to K before using pV = nRT.
52IBP2H-014-01-11The kinetic model of an ideal gasPaper 2Medium8 marks
A spacecraft carries a rigid helium tank of volume 0.050 m3. Before launch the helium is at a pressure of 2.0 × 107 Pa and a temperature of 290 K. Treat the helium as an ideal monatomic gas.
(a)
Calculate the amount of helium in the tank.
(2)
(b)
Show that the internal energy of the helium is about 1.5 MJ.
(2)
(c)
Explain why the internal energy of an ideal gas depends only on its temperature and the amount of gas.
(2)
(d)
In orbit, sunlight warms the tank to 320 K. Calculate the increase in the internal energy of the helium.
(2)
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Part (a)
n = pV/(RT) = 2.0 × 107 × 0.050/(8.31 × 290).
✓ 1
n = 415 mol (accept 410–420 mol).
✓ 1
Part (b)
U = (3/2)nRT, or equivalently U = (3/2)pV.
✓ 1
U = 1.5 × 2.0 × 107 × 0.050 = 1.5 × 106 J.
✓ 1
Part (c)
In an ideal gas there are no intermolecular forces except during collisions, so the intermolecular potential energy is zero.
✓ 1
The internal energy is therefore the total random kinetic energy of the molecules, N × (3/2)kT, which depends only on N and T.
✓ 1
Part (d)
ΔU = (3/2)nRΔT with ΔT = 30 K, or ΔU = 1.5 × 106 × 30/290.
53IBP2H-014-01-12Changes of state on a p–V diagramPaper 2Medium8 marks
A fixed mass of ideal monatomic gas in a cylinder with a piston starts in state A: pressure 1.0 × 105 Pa, volume 2.0 × 10−3 m3, temperature 300 K. The gas expands at constant pressure to state B, with volume 6.0 × 10−3 m3. It is then cooled at constant volume to state C, with pressure 0.50 × 105 Pa.
(a)
Calculate the temperature of the gas in state B.
(2)
(b)
Calculate the temperature of the gas in state C.
(2)
(c)
The processes A → B → C are plotted on a p–V diagram, with p on the vertical axis and V on the horizontal axis. Describe the shape, position and direction of each part of this graph.
(2)
(d)
Determine the change in internal energy of the gas between state A and state C.
(2)
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Part (a)
At constant pressure V/T is constant, so TB = 300 × 6.0/2.0.
✓ 1
TB = 900 K.
✓ 1
Part (b)
At constant volume p/T is constant, so TC = TB × 0.50/1.0 = 900 × 0.50/1.0.
✓ 1
TC = 450 K.
✓ 1
Allow a follow-through from the value of TB calculated in the previous part.
—
Part (c)
A → B: a horizontal straight line at p = 1.0 × 105 Pa, from V = 2.0 × 10−3 m3 to V = 6.0 × 10−3 m3, with the process directed towards increasing V (to the right).
✓ 1
B → C: a vertical straight line at V = 6.0 × 10−3 m3, from p = 1.0 × 105 Pa down to p = 0.50 × 105 Pa, with the process directed towards decreasing p (downwards).
✓ 1
Part (d)
Uses ΔU = (3/2)(pCVC − pAVA) = 1.5 × (300 − 200), or ΔU = (3/2)nRΔT with n ≈ 0.080 mol and ΔT = 150 K.
✓ 1
ΔU = +150 J, which is an increase.
✓ 1
Allow a follow-through from the value of TC calculated earlier.
—
Model sketch
Exam tip: Internal energy depends only on the state of the gas, not on the path taken. Compare pV, or T, at the start and end points only.
54IBP2H-014-01-15Molecular speedsPaper 2Medium8 marks
In a microchip fabrication chamber, nitrogen gas (molar mass 28 g mol−1) is kept at a pressure of 1.0 Pa and a temperature of 300 K. Treat the gas as ideal.
(a)
Calculate the number of nitrogen molecules per cubic metre in the chamber.
(2)
(b)
Calculate the rms speed of the nitrogen molecules.
(2)
(c)
Show that your answers to (a) and (b) are consistent with a pressure of 1.0 Pa.
(2)
(d)
Estimate the average separation of the molecules. Comment on whether treating the gas as ideal is justified, given that a nitrogen molecule has a diameter of about 3 × 10−10 m.
This is about 500 times the molecular diameter, so the volume of the molecules and the forces between them are negligible. The ideal gas model is justified.
✓ 1
ECF from (a).
—
Exam tip: To estimate the spacing between particles, take the cube root of the volume per particle. This links the number density directly to the ideal-gas assumptions.
55IBP3H-B3-20r12-11Boyle's law with a gas syringePaper 2Medium7 marks
A sealed shipping container of volume 33 m3 holds air at 1.01×105 Pa and 293 K. Treat the air as an ideal gas. Use R = 8.31 J mol−1 K−1. The container volume does not change.
(a)
Calculate the amount of air in the container, in mol. Give your answer to 3 significant figures.
(2)
(b)
The temperature of the air rises to 318 K with no air escaping. Calculate the new pressure. Give your answer to 3 significant figures.
(2)
(c)
A relief valve opens whenever the pressure reaches 1.05×105 Pa and releases air until the pressure falls to that value, with the temperature staying at 318 K. Calculate the amount of air that escapes, in mol. Use the unrounded value of n from (a) and give your answer to 3 significant figures.
(3)
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Part (a)
n = pV/RT
✓ 1
n = (1.01×105 × 33)/(8.31 × 293) = 1368.9, so 1.37×103 mol
56IBP3H-B3-20r12-13Kinetic model of an ideal gasPaper 2Medium7 marks
Gas leaves a volcanic vent at pressure 1.8×105 Pa. The density of the gas is 0.65 kg m−3.
(a)
Use the equation p = (1/3)ρ⟨c2⟩ to determine the r.m.s. speed of the gas molecules. Give your answer to 3 significant figures.
(2)
(b)
The gas is carbon dioxide, with molar mass 44 g mol−1. Determine the temperature of the gas. Use NA = 6.02×1023 mol−1 and kB = 1.38×10−23 J K−1. Use the unrounded value of ⟨c2⟩ and give your answer to 3 significant figures.
(3)
(c)
The gas cools to 900 K at constant volume with no change in the amount of gas. Calculate the new pressure. Use the unrounded value of the temperature from (b) and give your answer to 3 significant figures.
(2)
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Part (a)
⟨c2⟩ = 3p/ρ = 3 × 1.8×105/0.65 = 8.308e+05 m2 s−2
✓ 1
crms = √⟨c2⟩ = 911.5, so 911 m s−1
✓ 1
Part (b)
Mass of one molecule m = 0.044/6.02×1023 = 7.31e-26 kg
✓ 1
Uses (1/2)m⟨c2⟩ = (3/2)kBT
✓ 1
T = 7.31e-26 × 8.308e+05/(3 × 1.38×10−23) = 1467, so 1.47×103 K
✓ 1
Part (c)
At constant volume p ∝ T
✓ 1
p = 1.8×105 × 900/1467 = 1.10×105 Pa
✓ 1
Common error: Take the square root of ⟨c²⟩ to get the r.m.s. speed, but use ⟨c²⟩ itself in energy equations.
57IBP3H-B3-20r12-14The empirical gas lawsPaper 2Medium7 marks
A sealed tank of fixed volume on a cargo ship contains a fixed amount of gas. The pressure p is measured at four temperatures T: 15 °C: 93.7 kPa 35 °C: 100.1 kPa 55 °C: 106.6 kPa 75 °C: 113.2 kPa
Data from the question
(a)
Calculate the value of p/T, with T in kelvin (take T/K = T/°C + 273), for each of the four readings. Give each value to 3 significant figures in kPa K−1.
(2)
(b)
Explain what your answers to (a) show about the relationship between p and T.
(1)
(c)
Use the readings at 15 °C and 75 °C to estimate the temperature in °C at which a straight line through them reaches zero pressure. Give your answer to 3 significant figures.
(2)
(d)
The volume of the gas is 1.50×10−3 m3. Calculate the amount of gas, in mol, using the reading at 15 °C. Use R = 8.31 J mol−1 K−1 and give your answer to 3 significant figures.
(2)
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Part (a)
Converts to 288 K, 308 K, 328 K and 348 K
✓ 1
p/T = 0.325, 0.325, 0.325, 0.325 kPa K−1, i.e. 0.325 for all four
✓ 1
Part (b)
The ratio is constant, so p is directly proportional to the kelvin temperature at constant volume and amount of gas
✓ 1
Part (c)
Gradient = (113.2 − 93.7)/60 = 0.325 kPa K−1
✓ 1
Zero pressure at 15 − 93.7/0.325 = -273.3, so −273 °C
✓ 1
Part (d)
n = pV/RT with p = 9.37×104 Pa and T = 288 K
✓ 1
n = (9.37×104 × 1.50×10−3)/(8.31 × 288) = 0.0587, so 5.87×10−2 mol
✓ 1
Exam tip: Calculating a ratio such as p/T for every reading is a quick test for direct proportionality; the ratio must stay constant within the uncertainty.
58IBP3H-B3-20r12-16The ideal gas equationPaper 2Medium6 marks
Two stage-effect air tanks, A and B, are connected by a closed valve. Tank A has volume 2.0×10−3 m3 and contains air at 3.0×105 Pa. Tank B has volume 6.0×10−3 m3 and contains air at 1.0×105 Pa. Both are at 300 K. Treat the air as an ideal gas and the volume of the valve as negligible. Use R = 8.31 J mol−1 K−1.
Diagram not to scale
(a)
Calculate the total amount of gas in the two tanks. Give your answer to 3 significant figures.
(2)
(b)
Calculate the pressure after the valve is opened and the temperature is still 300 K.
(2)
(c)
The gas is then heated to 380 K at constant volume. Calculate the new pressure. Give your answer to 3 significant figures.
(2)
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Part (a)
Uses n = pV/RT for each tank
✓ 1
n = (3.0×105 × 2.0×10−3 + 1.0×105 × 6.0×10−3)/(8.31 × 300) = 0.4813, so 0.481 mol
✓ 1
Part (b)
Final volume 8.0×10−3 m3
✓ 1
p = nRT/V = 0.4813 × 8.31 × 300/8.0×10−3 = 1.5×105 Pa
✓ 1
Part (c)
p ∝ T at constant volume
✓ 1
p = 1.5×105 × 380/300 = 1.90×105 Pa
✓ 1
Exam tip: When tanks are connected, the total amount of gas is conserved: add the amounts n = pV/RT, not the pressures.
59B-2-25Kinetic model of an ideal gasPaper 2Hard9 marks
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.
Diagram not to scale
(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)
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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)
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)
61IBP2H-014-00-11Boyle's law with a gas syringePaper 2Hard9 marks
A student uses a gas syringe connected by a short length of tubing to a pressure sensor to investigate Boyle's law. The volume V read from the syringe scale does not include the gas in the tubing. The temperature of the gas is constant. The results are shown below.
V / cm3: 50.0, 40.0, 30.0, 20.0, 10.0 p / kPa: 100, 124, 163, 236, 433
Diagram not to scale
(a)
Outline why the plunger should be moved slowly during the experiment.
(2)
(b)
Show that the product pV, using the syringe readings, is not constant.
(2)
(c)
The student suggests that a fixed extra volume V0 of gas in the tubing explains the data, so that p(V + V0) is constant. Determine V0, using the first and last rows of data.
(3)
(d)
Outline how a graph could be used to determine V0 from all the data.
(2)
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Part (a)
Compressing the gas does work on it, which tends to raise its temperature ; moving slowly allows time for energy to be transferred to the surroundings, so the temperature stays constant
✓ 2
Part (b)
pV calculated for at least two rows, e.g. 5000 kPa cm3 at 50.0 cm3 and 4330 kPa cm3 at 10.0 cm3 (all values: 5000, 4960, 4890, 4720, 4330) ; the values fall steadily as V decreases, by about 13 % overall, so the trend is systematic and pV is not constant
Plot V against 1/p, which gives a straight line because V = C/p − V0. Alternatively, plot 1/p against V, which gives a straight line because 1/p = (V + V0)/C ; for V against 1/p, the intercept on the V axis is −V0. For 1/p against V, the intercept on the V axis is −V0, or V0 = intercept ÷ gradient
✓ 2
Exam tip: A steady trend in pV that grows at small volumes points to a hidden constant volume, not to random error. The effect is largest when the syringe volume is smallest.
62IBP2H-014-00-12Kinetic model of an atmospherePaper 2Hard9 marks
The escape speed from the Earth is 11.2 km s−1. Molar masses: nitrogen N2 28 g mol−1, hydrogen H2 2.0 g mol−1.
(a)
Show that the mass of one nitrogen molecule is about 4.7 ×10−26 kg.
(2)
(b)
Calculate the root mean square speed of nitrogen molecules at 300 K.
(2)
(c)
Calculate the temperature at which the root mean square speed of hydrogen molecules would equal the escape speed.
(3)
(d)
The upper atmosphere is at about 1000 K, far below the temperature in (c). Explain why hydrogen has nevertheless escaped from the Earth's atmosphere while nitrogen has not.
(2)
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Part (a)
m = M/NA = 0.028/6.02 ×1023 ; m = 4.65 ×10−26 kg ≈ 4.7 ×10−26 kg
✓ 2
Part (b)
½mc2 = (3/2)kT so crms = √(3kT/m) ; crms = √(3 × 1.38 ×10−23 × 300/4.65 ×10−26) = 517 m s−1 (accept 510–520 m s−1)
✓ 2
Part (c)
m(H2) = 0.0020/6.02 ×1023 = 3.32 ×10−27 kg ; T = mc2/(3k) ; T = 3.32 ×10−27 × (1.12 ×104)2/(3 × 1.38 ×10−23) = 1.0 ×104 K
✓ 3
Part (d)
Molecules have a distribution of speeds, so a small fraction have speeds several times the rms value and can exceed the escape speed ; hydrogen's rms speed is √14 ≈ 3.7 times that of nitrogen at the same temperature, so a far larger fraction of hydrogen molecules reach escape speed, and over long times the hydrogen is lost
✓ 2
Common error: Remember to convert molar mass from g mol⁻¹ to kg mol⁻¹ before dividing by the Avogadro constant; forgetting gives a molecular mass 1000 times too big.
Outline why Z falls below 1 as the pressure increases up to about 10 MPa.
(2)
(c)
Explain why Z exceeds 1 at 30 MPa.
(2)
(d)
A tank of volume 0.080 m3 holds the gas at 20 MPa and 300 K. Determine the amount of gas actually in the tank.
(3)
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Part (a)
Z = 1 at all pressures
✓ 1
Part (b)
Molecules are closer together so attractive intermolecular forces become significant ; molecules approaching a wall are pulled back by neighbours, reducing the force on the wall, so p (and pV) is less than the ideal value nRT
✓ 2
Part (c)
The volume of the molecules themselves becomes a significant fraction of the tank volume ; the free space available is less than V, so molecules hit the walls more often and the pressure is greater than the ideal prediction, giving pV > nRT
✓ 2
Part (d)
Reads Z = 0.90 at 20 MPa ; n = pV/(ZRT) ; n = (2.0 ×107 × 0.080)/(0.90 × 8.31 × 300) = 710 mol (713 mol)
✓ 3
Common error: Multiplying by Z instead of dividing gives 578 mol. Since Z < 1 means pV is smaller than nRT, the true amount must be larger than the ideal estimate of 642 mol.
64IBP2H-014-01-13Molecular speedsPaper 2Hard8 marks
An acoustics laboratory measured the speed of sound in four monatomic gases, all at 300 K.
Gas
Molar mass / g mol−1
Speed of sound / m s−1
Helium
4.0
1020
Neon
20.2
454
Argon
39.9
323
Krypton
83.8
223
(a)
Calculate the rms speed of helium atoms at 300 K.
(2)
(b)
Show, using helium and one other gas from the table, that the ratio speed of sound ÷ rms speed is the same for these gases.
(2)
(c)
Explain, in terms of molecular motion, why the speed of sound in a gas is related to the rms speed of its molecules.
(2)
(d)
Xenon is monatomic with a molar mass of 131.3 g mol−1. Predict the speed of sound in xenon at 300 K.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
crms = √(3RT/M) = √(3 × 8.31 × 300 / 4.0 × 10−3).
✓ 1
crms = 1.37 × 103 m s−1.
✓ 1
Part (b)
Helium: 1020/1367 = 0.746.
✓ 1
Second gas with its rms speed calculated and the ratio compared. For example, argon: crms = √(3 × 8.31 × 300/0.0399) = 433 m s−1, giving 323/433 = 0.746. Neon gives 454/609 = 0.746 and krypton gives 223/299 = 0.746. The ratios agree at about 0.75.
✓ 1
Allow ECF from (a).
—
Part (c)
A sound wave is passed through the gas by molecules moving and colliding with their neighbours, transferring momentum and energy. So the disturbance travels at a rate set by how fast the molecules move.
✓ 1
Any one of the following for the second mark.
✓ 1
• The mean kinetic energy of the molecules is proportional to T, so the molecular speeds scale as √(T/M). The wave speed therefore scales in the same way, and the ratio is constant.
—
• Faster molecules carry the disturbance to their neighbours sooner, so the wave speed is proportional to the typical molecular speed.
—
• The wave speed is a fixed fraction of crms for gases of the same type, here all monatomic.
—
Part (d)
At constant T, v ∝ 1/√M, so v = 323 × √(39.9/131.3). Alternatively, v = 0.746 × √(3RT/M) with M = 0.1313 kg mol−1.
✓ 1
v = 178 m s−1 (accept 177–179 m s−1).
✓ 1
Allow ECF from (b).
—
Common error: Convert molar masses from g mol⁻¹ to kg mol⁻¹ before using √(3RT/M). Otherwise the speed comes out √1000 ≈ 32 times too small.
65IBP2H-014-01-14Kinetic model of an atmospherePaper 2Hard9 marks
A simple model of the atmosphere of Mars uses a uniform temperature of 210 K. The escape speed from Mars is 5.0 km s−1. Planetary scientists use a rule of thumb: a gas is kept over geological timescales only if the rms speed of its molecules is less than one-sixth of the escape speed.
(a)
Calculate the mean translational kinetic energy of a gas molecule at 210 K.
(2)
(b)
Calculate the rms speed of hydrogen molecules (molar mass 2.0 g mol−1) at 210 K.
(2)
(c)
Determine, using the rule of thumb, the minimum molar mass of a gas that Mars can keep. Hence deduce whether Mars can keep helium (4.0 g mol−1) and water vapour (18 g mol−1).
(3)
(d)
Explain why some molecules of a gas can escape even when the rms speed is well below the escape speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Ek = (3/2)kT = 1.5 × 1.38 × 10−23 × 210.
✓ 1
Ek = 4.3 × 10−21 J.
✓ 1
Part (b)
Either crms = √(3RT/M) = √(3 × 8.31 × 210 / 2.0 × 10−3), or use ½mc2 = Ek with m = 2.0 × 10−3 / 6.02 × 1023 kg.
✓ 1
crms = 1.6 × 103 m s−1.
✓ 1
Allow error carried forward from the kinetic energy found in the first part.
—
Part (c)
Maximum allowed rms speed = 5000/6 = 833 m s−1.
✓ 1
M = 3RT/c2 = 3 × 8.31 × 210 / 8332 = 7.5 × 10−3 kg mol−1 = 7.5 g mol−1.
✓ 1
Helium escapes because 4.0 < 7.5. Water vapour is kept because 18 > 7.5.
✓ 1
Allow error carried forward for the conclusion.
—
Part (d)
The molecules have a distribution of speeds, and a small fraction move much faster than the rms speed.
✓ 1
Fast molecules high in the atmosphere, where collisions are rare, can reach escape speed and leave, so over long times the gas is steadily lost.
✓ 1
Exam tip: At a given temperature, the mean kinetic energy is the same for every gas. Only the molecular mass decides how fast the molecules move.
66IBP2H-014-01-16The empirical gas lawsPaper 2Hard9 marks
A student uses a constant-volume gas thermometer. It consists of a glass bulb of volume 250 cm3 containing a fixed mass of dry air. She records the pressure p of the air at five Celsius temperatures θ: θ = 10 °C, p = 103.6 kPa θ = 30 °C, p = 110.9 kPa θ = 50 °C, p = 118.2 kPa θ = 70 °C, p = 125.5 kPa θ = 90 °C, p = 132.9 kPa Take R = 8.31 J mol−1 K−1.
Data from the question
(a)
Determine, using the data, a value for absolute zero in °C.
(3)
(b)
Determine the amount of air in the bulb.
(2)
(c)
Explain why the pressure varies linearly with the Celsius temperature.
(2)
(d)
The glass bulb actually expands slightly as it is heated. State and explain the effect of this on the value of absolute zero determined from these data.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Gradient of p against θ = (132.9 − 103.6)/(90 − 10) = 0.366 kPa °C−1. Any valid pair of points or a best-fit line is accepted.
✓ 1
Extrapolate to p = 0: θ0 = 10 − 103.6/0.366.
✓ 1
θ0 = −273 °C (accept −272 °C to −274 °C).
✓ 1
Part (b)
Use n = pV/(RT) with T in kelvin, e.g. n = 103.6 × 103 × 250 × 10−6/(8.31 × 283). Alternatively, n = (gradient × V)/R with the gradient in Pa K−1.
✓ 1
n = 0.011 mol (0.0110 mol).
✓ 1
Part (c)
With n and V constant, pV = nRT gives p = (nR/V)T, so p ∝ T in kelvin. Also accept the kinetic argument: a higher T means faster molecules, so the collisions with the walls are harder and more frequent.
✓ 1
T = θ + 273, so p = (nR/V)(θ + 273). This is a linear function of θ, a straight line that is not through the origin and reaches p = 0 at θ = −273 °C.
✓ 1
Part (d)
The volume increases with temperature, so the pressures measured at higher temperatures are slightly lower than they would be at a truly constant volume. The gradient of p against θ is therefore smaller.
✓ 1
With a smaller gradient, the line extrapolates to p = 0 further below 0 °C, so the value found is lower (more negative) than the true value.
✓ 1
Common error: Using T = 10 instead of 283 K in pV = nRT gives an amount of gas about 28 times too large. Only the intercept extrapolation is done on the Celsius scale; every gas-law calculation needs kelvin.
67IBP3H-B3-20r12-15The ideal gas equationPaper 2Hard8 marks
A gas bubble forms at a depth of 32 m in an acidic crater lake. At that depth the water temperature is 280 K and the bubble has volume 0.52 cm3. The density of the lake water is 1050 kg m−3, the atmospheric pressure is 1.01×105 Pa and the surface temperature is 303 K. The bubble rises to the surface without gas entering or leaving it. Assume the gas inside the bubble is ideal and its pressure equals the pressure of the surrounding water.
(a)
Calculate the pressure of the water at a depth of 32 m. Use g = 9.8 m s−2 and give your answer to 3 significant figures.
(2)
(b)
Determine the volume of the bubble at the surface. Use the unrounded value of the pressure from (a) and give your answer in cm3 to 3 significant figures.
(3)
(c)
At a shallower depth h2 the temperature of the water is 285 K, and the bubble volume there is 1.5 times smaller than its volume at the surface (surface temperature 303 K). Determine h2. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
p = patm + ρgh
✓ 1
p = 1.01×105 + 1050 × 9.8 × 32 = 4.30×105 Pa
✓ 1
Part (b)
Uses p1V1/T1 = p2V2/T2 with temperatures in kelvin
h2 = (142500 − 1.01×105)/(1050 × 9.8) = 4.03, so 4.03 m
✓ 1
Common error: The pressure on a bubble below the surface is the atmospheric pressure plus the pressure due to the water column, not the water column alone.
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