118 free, high-quality past-paper-style questions on Gas laws for IB Physics SL, each with a full mark scheme. Covers: Pressure; Moles, molecules & the Avogadro constant; The empirical gas laws; and more. For when you have finished the past papers, or want to practise one topic at a time.B.3 at Higher 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.575 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1B-1A-16PressurePaper 1AEasy1 mark
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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Step 1P = F/A = 250 / 0.020 = 1.25 × 10⁴ Pa.
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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.
A fixed mass of an ideal gas is compressed slowly at constant temperature to half of its original volume.
Which statement explains why the pressure of the gas increases?
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Step 1At constant temperature the mean kinetic energy, and so the mean speed, of the molecules is unchanged. In half the volume there are twice as many molecules per unit volume, so collisions with each unit area of wall per second double: the rate of change of momentum per unit area, the pressure, doubles.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThe temperature is constant, so the mean speed of the molecules does not change and nor does the average change in momentum per collision.
BThe ideal gas model assumes no intermolecular forces except during collisions. The pressure comes from collisions with the walls.
CCorrect: the number of molecules per unit volume doubles, so the collision rate per unit area of wall doubles while each collision is, on average, the same.
DCollisions between molecules conserve momentum and kinetic energy; they do not change the total momentum delivered to the walls. Pressure depends on collisions with the walls.
A delivery drone of mass m stands on level snow. It rests on four identical circular landing pads, each of diameter d, which share its weight equally. The gravitational field strength is g.
What is the pressure exerted on the snow by the pads?
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Step 1The force perpendicular to the snow is the weight mg. Each pad has area π(d/2)2 = πd2/4, so the four pads have a total area of πd2.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2P = F/A = mg/(πd2).
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the weight mg acts on a total area 4 × πd2/4 = πd2.
BThe mass has been used instead of the weight. Pressure is a force per unit area, so the force mg is needed.
COnly the area of one pad has been used: mg/(πd2/4). The weight is spread over all four pads.
DThe diameter has been used as the radius, giving a pad area of πd2 instead of πd2/4.
The kinetic model of an ideal gas is used to approximate the behaviour of real gases.
Which statement is not an assumption of the kinetic model of an ideal gas?
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Step 1The model assumes point-like molecules in random motion, elastic collisions and no intermolecular forces except during collisions. It does not require equal speeds: the molecules have a range of speeds, and only their mean square speed appears in P = ⅓ρv̄2.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis is an assumption: it is why the model works best at low density, where the molecules are far apart.
BThis is an assumption: no kinetic energy is lost in collisions, so the gas does not cool by itself.
CCorrect: the molecules move randomly with a spread of speeds; the model uses their mean square speed.
DThis is an assumption: it is why an ideal gas has no intermolecular potential energy.
7IBP2S-013-00-01The empirical gas lawsPaper 1AEasy1 mark
A nurse seals the nozzle of a plastic syringe containing air and pushes the plunger in very slowly to compress the air. Why must the plunger be moved slowly if the air is to obey Boyle's law?
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Boyle's law (pV = constant) applies only to a fixed amount of gas at constant temperature. Compressing slowly lets energy transfer to the surroundings, so the air stays at room temperature. The amount of gas is already fixed because the nozzle is sealed. The pressure must change during the compression. The collision rate with the plunger does increase, and this is what raises the pressure.
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Answer B
Answer: B
Common error: Rapid compression warms a gas. It then follows pV/T = constant rather than pV = constant.
8IBP2S-013-00-02The empirical gas lawsPaper 1AEasy1 mark
In an industrial drying oven, 0.50 m3 of air at 27 °C is heated to 327 °C. The pressure stays constant. What is the new volume of the air?
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At constant pressure, V ∝ T, where T is the temperature in kelvin. The temperatures are 300 K and 600 K, so the volume doubles: 0.50 × 600/300 = 1.0 m3. Option D comes from using Celsius temperatures. Option A treats V as inversely proportional to T. Option B wrongly assumes that constant pressure means constant volume.
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Answer C
Answer: C
Exam tip: Convert every temperature to kelvin before using any gas law.
9IBP2S-013-00-03Ideal & real gasesPaper 1AEasy1 mark
Under which conditions does a real gas behave least like an ideal gas?
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At high pressure, the molecules are close together, so their own volume is no longer negligible compared with the container. At low temperature, the molecules move slowly, so intermolecular attractions have a significant effect. Both conditions break assumptions of the ideal gas model. Low pressure combined with high temperature gives the most ideal behaviour.
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Answer D
Answer: D
Exam tip: Link each condition to a specific broken assumption: molecular volume (high p) and intermolecular forces (low T).
10IBP2S-013-00-04The empirical gas lawsPaper 1AEasy1 mark
The pressure p of the gas in a sealed, rigid pressure-cooker test vessel is measured at several temperatures θ in degrees Celsius. A graph of p against θ is plotted. Which statement describes the graph?
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At constant volume, p ∝ T, where T is in kelvin. Because T = θ + 273, the graph of p against θ is a straight line. Extrapolated to p = 0, it meets the axis at θ ≈ −273 °C, which is absolute zero. The line would pass through the origin only if the temperature were plotted in kelvin.
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Answer B
Answer: B
Common error: "Proportional" applies only to kelvin temperatures. A Celsius graph is linear but does not pass through the origin.
A hiker of weight 600 N stands still on snow wearing two snowshoes. Each snowshoe has a contact area of 0.20 m2 with the snow. What is the pressure exerted on the snow?
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The weight is spread over both snowshoes, so the total area is 2 × 0.20 = 0.40 m2. Pressure p = F/A = 600/0.40 = 1500 Pa. 3000 Pa uses only one snowshoe. 240 Pa comes from multiplying force by area. 6.7×10−4 Pa comes from dividing area by force.
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Answer A
Answer: A
Common error: Read whether the force acts on one surface or several. A person standing on two feet or two snowshoes spreads the force over the total contact area.
12IBP2S-013-03-02Moles, molecules & the Avogadro constantPaper 1AEasy1 mark
Which of the following expresses the pascal in SI base units?
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1 Pa = 1 N m−2, and 1 N = 1 kg m s−2. So 1 Pa = kg m s−2 × m−2 = kg m−1 s−2. kg m s−2 is the newton. kg m2 s−2 is the joule. kg m−2 s−2 comes from dividing the mass and the area separately rather than combining the powers of metres.
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Answer B
Answer: B
Exam tip: Pressure can also be written as J m⁻³ (energy per unit volume). This gives a quick check: kg m² s⁻² ÷ m³ = kg m⁻¹ s⁻².
13IBP2S-013-03-08The ideal gas equationPaper 1AEasy1 mark
Which change doubles the pressure of a fixed mass of ideal gas?
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From pV = NkT, at constant T and N the product pV is constant, so halving V doubles p. Doubling the Celsius temperature does not double the kelvin temperature. Doubling both V and T leaves p unchanged. Halving N halves p.
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Answer A
Answer: A
Common error: Gas-law proportionalities only hold for absolute temperature. For example, going from 20 °C to 40 °C increases T by only about 7%.
14IBP2S-013-01-01Internal energy of a gasPaper 1AEasy1 mark
A weather balloon is filled with helium (molar mass 4 g mol−1). A welding cylinder contains argon (molar mass 40 g mol−1). Both gases are at the same temperature. What is the ratio \(\frac{\text{rms speed of helium atoms}}{\text{rms speed of argon atoms}}\)?
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At the same temperature both gases have the same mean kinetic energy, \(\frac{1}{2}mc_{rms}^2=\frac{3}{2}kT\), so \(c_{rms}\propto\frac{1}{\sqrt{m}}\). The ratio is \(\sqrt{40/4}=\sqrt{10}\approx3.2\). Option 10 forgets the square root. The inverse ratio gives lighter atoms the lower speed. Option 1 confuses equal kinetic energy with equal speed.
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Answer A
Answer: A
Common error: Equal temperature means equal mean kinetic energy, not equal speed. Lighter particles must move faster to carry the same kinetic energy.
15IBP2S-013-01-02The empirical gas lawsPaper 1AEasy1 mark
During a test on an engine dynamometer, the air in a racing car tyre is heated from 17 °C to 307 °C. Which statement about the air molecules is correct?
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In kelvin the temperature rises from 290 K to 580 K, so it doubles. The mean kinetic energy is proportional to the absolute temperature, so it doubles. The rms speed increases by only \(\sqrt{2}\). A factor of 18 comes from wrongly dividing the Celsius values. A factor of \(\sqrt{2}\) applies to the speed, not the energy.
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Answer A
Answer: A
Exam tip: Convert to kelvin before forming any temperature ratio. Celsius ratios have no physical meaning.
16IBP2S-013-01-04Ideal & real gasesPaper 1AEasy1 mark
Which of the following is not an assumption of the kinetic model of an ideal gas?
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The kinetic model assumes the molecules move randomly with a range of speeds, described by a distribution. It does not assume they all share one speed. The other three statements are standard assumptions of the model.
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Answer A
Answer: A
Common error: The rms speed is an average over a spread of speeds. It does not mean every molecule moves at that speed.
17IBP2S-013-01-10Moles, molecules & the Avogadro constantPaper 1AEasy1 mark
Air in a patient's lungs is at body temperature, 37 °C. What is the mean translational kinetic energy of a gas molecule in the lungs?
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First convert the temperature: T = 310 K. Then \(\bar{E}_k=\frac{3}{2}kT=1.5\times1.38\times10^{-23}\times310=6.4\times10^{-21}\) J. Option 4.3×10−21 J is just kT. Option 7.7×10−22 J uses 37 instead of 310. Option 3.9×103 J uses R instead of k, which gives the energy per mole.
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Answer A
Answer: A
Exam tip: Use k for one molecule and R for one mole. A molecular energy should be around 10⁻²¹ J at everyday temperatures.
18IBP2S-013-02-01Moles, molecules & the Avogadro constantPaper 1AEasy1 mark
Two identical sealed steel cylinders in a laboratory store gas at the same pressure and the same temperature. One contains helium (molar mass 4 g mol−1). The other contains nitrogen (molar mass 28 g mol−1). Both gases behave ideally. Which statement about the number of gas molecules in the cylinders is correct?
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For an ideal gas \(pV = Nk_BT\), so \(N = \frac{pV}{k_BT}\). Pressure, volume and temperature are the same for both cylinders, so N is the same. The equation contains no molar mass or molecular structure. Helium is lighter, but this changes the mass of gas in the cylinder, not the number of molecules. The factor 7 is the ratio of the molar masses and applies to the masses of gas, not to the molecule numbers.
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Answer C
Answer: C
Common error: The ideal gas equation does not depend on the type of gas. Equal p, V and T always means equal numbers of molecules, whatever the molar mass.
19IBP2S-013-02-02Kinetic model & molecular speedsPaper 1AEasy1 mark
Carbon dioxide is used to carbonate drinks. Its molar mass is 44 g mol−1. The Avogadro constant is 6.0×1023 mol−1. What is the approximate mass of one carbon dioxide molecule?
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One mole has a mass of 44 g = 0.044 kg and contains NA molecules. The mass of one molecule is therefore \(m = \frac{0.044}{6.0\times10^{23}} \approx 7.3\times10^{-26}\) kg. The value 7.3×10−23 kg comes from leaving the molar mass in grams. The value 1.4×1022 comes from dividing NA by 44. The value 2.6×1025 comes from multiplying 44 by NA.
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Answer B
Answer: B
Exam tip: Convert the molar mass to kg mol⁻¹ before dividing by NA. A single molecule should have a tiny mass, of order 10⁻²⁶ kg.
20IBP2S-013-02-05Internal energy of a gasPaper 1AEasy1 mark
R is the molar gas constant and NA is the Avogadro constant. Which expression is equal to the Boltzmann constant kB?
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The ideal gas equation can be written as \(pV = nRT\) or as \(pV = Nk_BT\). Since \(N = nN_A\), it follows that \(nRT = nN_Ak_BT\), so \(k_B = \frac{R}{N_A}\). The quantity \(\frac{pV}{T}\) equals nR (or NkB) for a particular sample. It is not a constant.
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Answer C
Answer: C
Exam tip: Think of kB as the gas constant per molecule and R as the gas constant per mole.
21IBP2S-013-02-09Moles, molecules & the Avogadro constantPaper 1AEasy1 mark
Sample P is 2.0 g of hydrogen gas, H2 (molar mass 2.0 g mol−1). Sample Q is 32 g of oxygen gas, O2 (molar mass 32 g mol−1). Which statement is correct?
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Amount of substance is n = mass/molar mass. For P this is 2.0/2.0 = 1.0 mol. For Q it is 32/32 = 1.0 mol. Each sample therefore contains NA molecules. Both gases are diatomic, so they also contain the same number of atoms, 2NA. The factor 16 comes from comparing masses or molar masses, not amounts.
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Answer C
Answer: C
Common error: A larger mass does not mean more molecules. Always divide by the molar mass to find the amount in moles.
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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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.
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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.
24B-1A-39The empirical gas lawsPaper 1AMedium1 mark
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?
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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.
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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.
Step 3N = 2000 / 4.42 × 10⁻²¹ = 4.5 × 10²³ molecules, a little under one mole.
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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.
26B-1A-55The empirical gas lawsPaper 1AMedium1 mark
A fixed mass of an ideal gas is compressed slowly at a constant temperature T1. The experiment is repeated with the same gas at a constant higher temperature T2.
Which graph shows how the product PV varies with the pressure P at T1 (solid line) and at T2 (dashed line)?
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Step 1For a fixed mass of ideal gas PV = nRT. At constant temperature PV is constant, so each graph is a horizontal line whatever the pressure.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2PV is proportional to T, so the line for the higher temperature T2 is higher.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThese lines show PV proportional to P, which would need V to stay constant. At constant temperature V falls as P rises, keeping PV constant.
BThe lines are horizontal, but the higher temperature has been given the lower value of PV. PV = nRT increases with T.
CThese are the shapes of P–V isotherms (P ∝ 1/V). Plotted against P, the product PV does not change.
DCorrect: PV = nRT is constant along each isotherm and larger at the higher temperature.
27B-1A-57The ideal gas equationPaper 1AMedium1 mark
A rigid container holds an ideal gas at 47 °C. Half of the molecules of the gas leak out, and the temperature of the gas left in the container rises to 367 °C.
Which row gives the ratio (final value / initial value) for the pressure of the gas and for the average kinetic energy of a molecule?
PressureAverage kinetic energy of a molecule
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Step 1In kelvin the temperature doubles, from 320 K to 640 K.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2P = NkBT/V with V constant: P changes by ½ × 2 = 1. Ek = ³⁄₂kBT depends only on T, so it doubles.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe pressure is right, but the average kinetic energy has been taken as unchanged because the total internal energy of the gas is unchanged. The same energy is now shared among half as many molecules, so each has twice as much on average.
BCorrect: half the molecules at twice the absolute temperature exert the same pressure, while the average kinetic energy per molecule doubles with T.
CThe leak has been ignored for the pressure. Only half the molecules remain, which halves the pressure that the doubled temperature would otherwise give.
DCelsius temperatures have been used: 367/47 = 7.8 for the kinetic energy and ½ × 7.8 = 3.9 for the pressure. The gas laws and Ek = ³⁄₂kBT need the absolute temperature.
28B-1A-91Moles, molecules & the Avogadro constantPaper 1AMedium1 mark
Two identical sealed flasks are at the same temperature. One contains 1.0 g of helium (molar mass 4.0 g mol−1) and the other contains 1.0 g of nitrogen (molar mass 28 g mol−1). Both gases behave as ideal gases.
What is (pressure of the helium) / (pressure of the nitrogen)?
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Step 1The amounts of gas are n = m/M: 1.0/4.0 = 0.25 mol of helium and 1.0/28 = 0.036 mol of nitrogen, so, as N = nNA, the flask of helium contains seven times as many molecules.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2P = nRT/V with the same V and T, so the pressure ratio is (1.0/4.0)/(1.0/28) = 28/4.0 = 7.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe ratio has been inverted. The lighter gas has more molecules in the same mass, so it exerts the greater pressure.
BEqual masses have been taken to give equal pressures. Pressure depends on the number of molecules, not on their total mass.
C√7 is the ratio of the root mean square speeds of the molecules at the same temperature, not the ratio of the pressures.
DCorrect: P ∝ n at constant V and T, and n = m/M gives seven times as much helium.
29B-1A-92The empirical gas lawsPaper 1AMedium1 mark
A fixed mass of an ideal gas is in a cylinder closed by a freely moving piston, so that the pressure of the gas stays constant. The gas is heated and expands to a new equilibrium state at a higher temperature.
Compared with the initial state, what has happened to the mean speed of the molecules and to the number of collisions per second of the molecules with each unit area of the walls?
Mean speedCollisions per second per unit area
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Step 1A higher temperature means a larger mean kinetic energy, so the mean speed increases and each collision gives a larger change in momentum.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The pressure (rate of change of momentum per unit area) is unchanged, so there must be fewer collisions per second with each unit area: the gas has expanded and there are fewer molecules per unit volume.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis describes heating at constant volume, where the pressure rises. Here the gas expands, and the number of molecules per unit volume falls.
BCorrect: each collision delivers more momentum, so fewer collisions per unit area per second keep the pressure constant.
CThe pressure is constant but the temperature is not: the mean kinetic energy, and so the mean speed, increases with temperature.
DWith faster molecules an unchanged collision rate would give a larger rate of change of momentum, so the pressure would rise.
32B-1A-95Kinetic model & molecular speedsPaper 1AMedium1 mark
The kinetic theory of an ideal gas leads to the equation P = ⅓ρv̄2.
Which statement explains why the factor ⅓ appears in this equation?
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Step 1Only the velocity component perpendicular to a wall changes in a collision with it. Because the motion is random, the three perpendicular components have equal mean squares, so each is one third of v̄2.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AAt any instant about half of the molecules have a velocity component towards a given wall, and this fraction is already allowed for in the derivation.
BPressure is not shared out between walls: in equilibrium the gas exerts the same pressure on every wall.
CThe collisions are elastic in the model, so a molecule keeps all of its kinetic energy.
DCorrect: v̄2 = v̄x2 + v̄y2 + v̄z2 with three equal terms, so v̄x2 = v̄2/3.
33B-1A-96Internal energy of a gasPaper 1AMedium1 mark
Container X holds 3.0 mol of helium at 200 K. Container Y holds 1.0 mol of neon at 400 K. Both gases are monatomic and ideal.
What are the ratios (internal energy of X)/(internal energy of Y) and (average kinetic energy of an atom in X)/(average kinetic energy of an atom in Y)?
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The average kinetic energy of an atom, ³⁄₂ kBT, depends only on T: 200/400 = 0.50. The mass of the atoms does not matter.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe amount of gas has been ignored for the internal energy, so only the temperature ratio has been used.
BCorrect: U depends on both n and T, but the average kinetic energy of an atom depends on T alone.
CThe average kinetic energy has been given the same ratio as the internal energy. X has three times as many atoms, so each has, on average, 1.5/3 = 0.50 times as much.
DThe temperatures have been ignored: this treats both gases as if they were at the same temperature.
Which statement about the internal energy of the gas is correct?
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Step 1Internal energy = total random kinetic energy + total intermolecular potential energy. The ideal gas model has no intermolecular forces except in collisions, so its internal energy is ³⁄₂ NkBT, all kinetic. Attractive forces in a real gas give it intermolecular potential energy too.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: removing the intermolecular forces from the model removes the potential-energy term.
BU = ³⁄₂ NkBT depends only on N and T; compressing the gas at constant temperature does not change it.
CThe real gas also has intermolecular potential energy, which the ideal gas does not, so the internal energies differ.
DThe ideal gas model assumes there are no forces between the molecules except during collisions, so there is no intermolecular potential energy.
35IBP2S-013-00-05Moles, molecules & the Avogadro constantPaper 1AMedium1 mark
A sealed weather balloon rises through the atmosphere. During the ascent, the pressure of the gas inside falls to one quarter of its value at ground level. Its temperature falls from 290 K to 220 K. What is the ratio of the final volume to the volume at ground level?
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Using pV/T = constant: V2/V1 = (p1/p2)(T2/T1) = 4 × 220/290 ≈ 3.0. Option C ignores the temperature change. Option D inverts the temperature ratio. Option A inverts the whole result.
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Answer B
Answer: B
Exam tip: Check the direction of each effect. Falling pressure increases the volume, and falling temperature decreases it, so the answer must be a little less than 4.
36IBP2S-013-00-06Ideal & real gasesPaper 1AMedium1 mark
Which of the following is not an assumption of the kinetic model of an ideal gas?
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In the kinetic model, molecules have a range (distribution) of speeds. It is the mean kinetic energy that depends on temperature. Options A, B and C are all standard assumptions of the model.
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Answer D
Answer: D
Common error: The model assumes random motion with a distribution of speeds, not identical speeds.
37IBP2S-013-00-07Internal energy of a gasPaper 1AMedium1 mark
A fixed mass of helium is held in a cryogenic storage system. Its absolute pressure is doubled and its kelvin temperature is halved. Assume the helium behaves as an ideal gas throughout. By what factor does the density of the helium change?
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From pV = nRT with fixed n, V ∝ T/p. Density = mass/V, so for a fixed mass density ∝ p/T. Doubling p and halving T multiplies p/T by 2/(1/2) = 4, so the density becomes 4 times larger (D). Option A comes from finding the volume factor, (1/2)/2 = 1/4, and forgetting that density is inversely proportional to volume. Option B assumes the two changes cancel. Option C considers only the pressure change.
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Answer D
Answer: D
Exam tip: For a fixed mass of ideal gas, write density ∝ p/T and substitute both factors at once.
38IBP2S-013-00-08Ideal & real gasesPaper 1AMedium1 mark
In a high-pressure hydrogen fuel tank, the measured volume of the gas is greater than the volume predicted by the ideal gas equation for the same p, n and T. What is the best explanation?
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At very high pressure, the finite size of the molecules matters. The space available for their motion is less than the container volume, so a larger volume is needed than the ideal equation predicts. Attractive forces (option A) would make the volume smaller than predicted, not larger. Mean speed depends on temperature, not pressure. Inelastic collisions are not the cause.
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Answer B
Answer: B
Common error: Attractions make pV/(nRT) less than 1. Finite molecular volume makes it greater than 1.
39IBP2S-013-00-09The ideal gas equationPaper 1AMedium1 mark
A 12 L scuba cylinder contains air at a pressure of 200 bar. A diver breathes the air at a depth where the pressure is 3.0 bar. The temperature is constant. What volume would all the air in the cylinder occupy at 3.0 bar?
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Boyle's law: p1V1 = p2V2, so V2 = 12 × 200/3.0 = 800 L. Option A inverts the pressure ratio. Option B omits the cylinder volume. Option D is the volume at 1 bar, which is surface pressure, not the pressure at depth.
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Answer C
Answer: C
Exam tip: In Boyle's law any consistent pressure and volume units work, so bar and litres need no conversion.
40IBP2S-013-03-03Kinetic model & molecular speedsPaper 1AMedium1 mark
According to the kinetic model of an ideal gas, what is the origin of the pressure exerted by a gas on the walls of its container?
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When a molecule rebounds from a wall, its momentum changes. The rate of change of momentum of all the molecules striking the wall equals the force the wall exerts on them. By Newton's third law, the molecules exert an equal force on the wall, and this force per unit area is the pressure. An ideal gas has no intermolecular forces. Collisions between molecules do not push on the wall. Collisions with the wall are elastic, so no kinetic energy is lost.
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Answer C
Answer: C
Exam tip: A full explanation of gas pressure links three ideas: change of momentum at the wall, force as the rate of change of momentum, and pressure as force per unit area.
41IBP2S-013-03-04The empirical gas lawsPaper 1AMedium1 mark
The air in a racing-car tyre has an absolute pressure of 300 kPa at 7 °C. After several laps the air temperature reaches 35 °C. The volume of the tyre stays constant and no air escapes. What is the new absolute pressure?
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At constant volume, p/T is constant when T is in kelvin. T1 = 280 K and T2 = 308 K, so p2 = 300 × 308/280 = 330 kPa. 1500 kPa uses the Celsius ratio 35/7. 273 kPa inverts the temperature ratio. 300 kPa wrongly assumes that a constant volume means a constant pressure.
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Answer B
Answer: B
Common error: Always convert Celsius temperatures to kelvin before using any gas-law ratio.
42IBP2S-013-03-06Kinetic model & molecular speedsPaper 1AMedium1 mark
An ideal gas is held in a sealed container of fixed volume. The root mean square speed of its molecules is doubled. By what factor does the pressure change?
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From the kinetic model, p = ⅓ρ⟨c2⟩. The mass and volume are fixed, so the density ρ is constant. Doubling the rms speed multiplies the mean square speed ⟨c2⟩ by 4, so the pressure also increases by a factor of 4. A factor of 2 assumes pressure is proportional to speed. A factor of √2 confuses this with the effect of doubling the temperature. A factor of 16 squares the factor twice.
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Answer C
Answer: C
Common error: Do not mix up "doubling the rms speed" with "doubling the absolute temperature". Doubling T multiplies the rms speed by only √2, and it multiplies the pressure by 2.
43IBP2S-013-03-07Moles, molecules & the Avogadro constantPaper 1AMedium1 mark
Two identical rigid containers are at the same temperature. One contains 1.0 g of helium (molar mass 4 g mol−1). The other contains 1.0 g of nitrogen (molar mass 28 g mol−1). Both gases are ideal. What is (pressure of helium) ÷ (pressure of nitrogen)?
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At the same V and T, pressure is proportional to the number of moles, since p = nRT/V. Helium has 1.0/4 = 0.25 mol and nitrogen has 1.0/28 ≈ 0.036 mol, so the ratio is 28/4 = 7. A ratio of 1 assumes equal masses give equal pressures. A ratio of 1/7 assumes heavier molecules give a larger pressure. A ratio of √7 confuses the pressure ratio with the ratio of rms speeds.
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Answer B
Answer: B
Exam tip: Pressure depends on the number of particles, not on their mass. At the same temperature, lighter molecules move faster but strike the walls with smaller momentum, and these two effects cancel.
44IBP2S-013-03-10Kinetic model & molecular speedsPaper 1AMedium1 mark
A vacuum chamber used to test satellite components contains a small amount of residual gas at pressure p and absolute temperature T. k is the Boltzmann constant and R is the molar gas constant. Which expression gives the number of gas molecules per unit volume in the chamber?
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Start from pV = NkT and rearrange: N/V = p/(kT). The expression kT/p is the volume per molecule, which is the inverse of the required quantity. p/(RT) gives the number of moles per unit volume, not the number of molecules. pkT is dimensionally wrong.
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Answer A
Answer: A
Exam tip: Use k with numbers of molecules (N) and R with numbers of moles (n). The two constants are linked by R = NAk.
45IBP2S-013-01-03Internal energy of a gasPaper 1AMedium1 mark
A welding cylinder of internal volume 10 L contains argon, which can be treated as an ideal monatomic gas, at a pressure of 2.0×107 Pa. What is the internal energy of the argon?
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For an ideal monatomic gas \(U=\frac{3}{2}nRT=\frac{3}{2}pV\). Then \(U=1.5\times2.0\times10^{7}\times0.010=3.0\times10^{5}\) J. Option 2.0×105 J omits the factor 3/2. Option 1.0×105 J uses a factor 1/2. Option 3.0×108 J leaves the volume in litres.
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Answer A
Answer: A
Exam tip: For a monatomic ideal gas, \(\frac{3}{2}pV\) gives the internal energy directly, without finding n or T. Remember that 1 L = 10⁻³ m³.
46IBP2S-013-01-05Ideal & real gasesPaper 1AMedium1 mark
A fuel-cell test rig contains hydrogen (H2, molar mass 2 g mol−1) and oxygen (O2, molar mass 32 g mol−1) in thermal equilibrium. Which row gives the ratio of mean translational kinetic energy and the ratio of rms speed of hydrogen molecules to oxygen molecules?
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The gases are in thermal equilibrium, so they are at the same temperature. The mean translational kinetic energy \(\frac{3}{2}kT\) is therefore the same, giving a ratio of 1. Since \(c_{rms}\propto1/\sqrt{m}\), the speed ratio is \(\sqrt{32/2}=4\). A speed ratio of 16 forgets the square root. The other two rows wrongly assume kinetic energy depends on molecular mass.
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Answer A
Answer: A
Exam tip: Thermal equilibrium means equal temperature, and therefore equal mean translational kinetic energy whatever the gas.
47IBP2S-013-01-06Kinetic model & molecular speedsPaper 1AMedium1 mark
The air in a spacecraft cabin has a density of 1.2 kg m−3. The rms speed of its molecules is 500 m s−1. What is the pressure of the cabin air?
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Use \(p=\frac{1}{3}\rho c_{rms}^2\). Then \(p=\frac{1}{3}\times1.2\times500^2=1.0\times10^5\) Pa. Option 3.0×105 Pa omits the 1/3. Option 2.0×102 Pa does not square the speed. Option 1.5×105 Pa uses \(\frac{1}{2}\rho c^2\), confusing the pressure formula with kinetic energy.
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Answer A
Answer: A
Common error: The factor 1/3 arises because molecular motion is shared equally among three perpendicular directions. Do not replace it with the 1/2 from kinetic energy.
48IBP2S-013-01-07Internal energy of a gasPaper 1AMedium1 mark
The outlet of a bicycle pump is blocked. The air inside, treated as an ideal gas, is compressed so slowly that its temperature stays constant. What happens to the internal energy of the air?
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The internal energy of an ideal gas is the total random kinetic energy of its molecules. For a fixed amount of gas this depends only on temperature. The temperature is constant, so the internal energy is unchanged. Changes in pressure or volume alone do not change it. An ideal gas has no intermolecular potential energy, so option D is wrong.
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Answer C
Answer: C
Exam tip: For a fixed amount of ideal gas, ask only whether T changes. If T is constant, U is constant.
49IBP2S-013-01-08Kinetic model & molecular speedsPaper 1AMedium1 mark
A sealed, rigid aerosol can is left near a heater. The pressure of the gas inside increases by a factor of 4. Assuming ideal behaviour, by what factor does the rms speed of the gas molecules increase?
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At constant volume, \(p\propto T\), so T increases by a factor of 4. Since \(c_{rms}\propto\sqrt{T}\), the rms speed increases by \(\sqrt{4}=2\). A factor of 4 confuses speed with temperature. A factor of 16 squares instead of taking the root. Speeds must rise because the temperature rises.
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Answer A
Answer: A
Exam tip: Chain the proportionalities: p ∝ T at constant V, then crms ∝ √T.
A cyclist pumps air into a bicycle tyre. Assume the volume of the tyre stays constant. After pumping, the absolute pressure of the air in the tyre has doubled. Its absolute temperature has risen by 10%. What is \(\frac{\text{amount of air after pumping}}{\text{amount of air before pumping}}\)?
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From \(pV = nRT\) at constant V, \(n \propto \frac{p}{T}\). The ratio is therefore \(\frac{2}{1.1} \approx 1.8\). The answer 2.0 ignores the temperature rise. The answer 2.2 multiplies by 1.1 instead of dividing. The answer 0.55 is the reciprocal of the correct ratio.
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Answer D
Answer: D
Exam tip: Rearrange the equation so the unknown quantity is the subject, here \(n \propto p/T\), before putting in ratios. This stops you multiplying when you should divide.
51IBP2S-013-02-04The ideal gas equationPaper 1AMedium1 mark
A scuba diver breathes out a small bubble of air at a depth where the pressure is 3.0×105 Pa and the water temperature is 280 K. The bubble rises to the surface. At the surface the pressure is 1.0×105 Pa and the water temperature is 300 K. No air enters or leaves the bubble. What is \(\frac{\text{volume at surface}}{\text{volume at depth}}\)?
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The amount of air is fixed, so \(\frac{pV}{T}\) is constant. This gives \(\frac{V_2}{V_1} = \frac{p_1}{p_2}\times\frac{T_2}{T_1} = 3.0\times\frac{300}{280} \approx 3.2\). The answer 3.0 ignores the temperature change. The answer 2.8 inverts the temperature ratio. The answer 0.31 inverts the whole ratio.
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Answer A
Answer: A
Common error: Check your answer against the physics. Lower pressure and higher temperature both make the bubble bigger, so the ratio must be greater than 3.
A sealed medical syringe contains an ideal gas. The plunger is pushed in slowly, at constant temperature, until the gas occupies one third of its original volume. Which row describes the number of molecules per unit volume and the pressure of the gas?
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No gas escapes, so the same N molecules now occupy V/3. The number per unit volume, N/V, therefore triples. From \(p = \frac{N}{V}k_BT\) at constant T, the pressure also triples. The total number of molecules is unchanged, but the number per unit volume is not.
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Answer B
Answer: B
Common error: Do not confuse the total number of molecules N, which is fixed in a sealed container, with the number density N/V, which changes with volume.
53IBP2S-013-02-07Kinetic model & molecular speedsPaper 1AMedium1 mark
A weather balloon rises through the atmosphere. At its maximum altitude the air pressure is one quarter of the pressure at ground level. The air temperature there is 220 K, compared with 290 K at ground level. Treat air as an ideal gas. What is \(\frac{\text{density of air at maximum altitude}}{\text{density of air at ground level}}\)?
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From \(pV = nRT\), the density is \(\rho = \frac{nM}{V} = \frac{pM}{RT}\), so \(\rho \propto \frac{p}{T}\). The ratio is \(\frac{1}{4}\times\frac{290}{220} \approx 0.33\). The answer 0.25 ignores the temperature change. The answer 0.19 inverts the temperature ratio. The answer 3.0 inverts the whole ratio.
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Answer D
Answer: D
Exam tip: Remember that the density of an ideal gas is proportional to p/T. Colder air is denser at the same pressure, so the answer must be more than 0.25.
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?
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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.
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Step 3In degrees Celsius: 400 − 273 = 127 °C.
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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.
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?
Before the load is placed on the piston
New pressureNew volume
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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.
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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.
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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.
Cylinder X, of volume V, contains an ideal gas at pressure 3P0. Cylinder Y, of volume 2V, contains the same gas at pressure P0. Both are at the same temperature. A tap in the short tube joining the cylinders is opened. When the flow of gas has stopped, the gas is at its original temperature.
What is the final pressure of the gas?
Same temperature; the tap is opened
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Step 1At constant temperature the amount of gas is proportional to PV: total = 3P0V + P0(2V) = 5P0V.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2This gas now fills a volume 3V, so the final pressure is 5P0V/3V = 5P0/3.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: PV is conserved in total (same temperature, no gas lost), so P(3V) = 3P0V + 2P0V.
BThe two pressures have been averaged as if the cylinders had equal volumes. Y holds twice the volume, so its lower pressure counts for more.
CThe volumes of X and Y have been interchanged: (3P0 × 2V + P0 × V)/3V = 7P0/3.
DThe two pressures have been added. Each sample expands into the whole volume 3V, so each contributes less than its original pressure (P0 and 2P0/3).
59B-1A-99Internal energy of a gasPaper 1AHard1 mark
A rigid steel storage tank contains argon, a monatomic gas that behaves as an ideal gas. An open vent pipe keeps the argon in the tank at atmospheric pressure, which does not change. During the day the Sun warms the argon in the tank from 280 K to 320 K and some argon escapes through the vent.
What happens to the number of argon atoms in the tank and to the internal energy of the argon in the tank?
Number of atoms in the tankInternal energy of the argon in the tank
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Step 1P and V are fixed, so N = PV/(kBT) falls as T rises: atoms leave through the vent.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2U = ³⁄₂ NkBT = ³⁄₂ PV, which is constant: the fall in N exactly balances the rise in T.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe rise in temperature has been taken to dominate. With P and V fixed, NT stays constant, so the internal energy ³⁄₂ NkBT does not change.
BThe vent has been ignored and the gas treated as a fixed mass. With the pressure held constant in a rigid tank, warming must drive atoms out.
CThe loss of atoms has been taken to dominate. Each remaining atom has more kinetic energy, which exactly makes up for the atoms lost.
DCorrect: N ∝ 1/T at constant P and V, and U = ³⁄₂ PV stays the same.
60IBP2S-013-00-10Internal energy of a gasPaper 1AHard1 mark
Two hospital gas cylinders are at the same temperature. Cylinder X has volume V and contains oxygen at pressure 3p. Cylinder Y has volume 2V and contains oxygen at pressure p. The cylinders are joined by a valve of negligible volume, and the valve is opened. The temperature does not change. What is the final pressure?
Valve of negligible volume; same temperature
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The total amount of gas is conserved, and at constant T, n ∝ pV. The total pV is 3p × V + p × 2V = 5pV. This gas now fills a volume of 3V, so the final pressure is 5pV/3V = 5p/3. Option C takes a simple average of the two pressures. Option D adds the pressures. Option A divides the sum of the pressures by the number of volume units.
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Answer B
Answer: B
Exam tip: When gases mix at the same temperature, add the pV products, which are proportional to the number of moles.
61IBP2S-013-03-05The ideal gas equationPaper 1AHard1 mark
A sealed weather balloon rises through the atmosphere. The pressure of the gas inside falls to one quarter of its launch value, and its absolute temperature falls from 300 K to 225 K. The gas behaves ideally. By what factor does the volume of the balloon increase?
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For a fixed amount of gas, pV/T is constant. So V2/V1 = (p1/p2)(T2/T1) = 4 × 225/300 = 3.0. A factor of 4.0 ignores the cooling. A factor of 5.3 uses the temperature ratio upside down. A factor of 0.33 inverts the whole result.
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Answer A
Answer: A
Exam tip: When two variables change together, apply each change separately and check its effect. Lower pressure increases the volume and lower temperature decreases it, so the answer must lie below 4.
62IBP2S-013-03-09Ideal & real gasesPaper 1AHard1 mark
A container holds a mixture of hydrogen (H2) and oxygen (O2) gas in thermal equilibrium. The container holds equal numbers of molecules of each gas. Both gases are ideal. Which statement is correct?
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Each gas contributes a partial pressure p = NkT/V. N, T and V are the same for both gases, so their contributions are equal. The mean kinetic energy, 3/2 kT, depends only on temperature, so it is the same for both gases. The rms speed is proportional to 1/√m, so hydrogen's rms speed is √16 = 4 times larger, not 16 times. Heavier molecules move more slowly, so they do not exert more pressure.
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Answer C
Answer: C
Exam tip: At a given temperature, every ideal gas has the same mean kinetic energy per molecule. Differences in molecular mass appear only in the speeds.
63IBP2S-013-01-09Internal energy of a gasPaper 1AHard1 mark
Two storage tanks contain the same ideal monatomic gas. Tank X holds n moles at absolute temperature T. Tank Y holds 2n moles at absolute temperature T/2. Which row gives \(\frac{U_X}{U_Y}\) (ratio of internal energies) and \(\frac{c_X}{c_Y}\) (ratio of rms speeds)?
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Using \(U=\frac{3}{2}nRT\): \(U_X=\frac{3}{2}nRT\) and \(U_Y=\frac{3}{2}(2n)R\frac{T}{2}=\frac{3}{2}nRT\), so the ratio is 1. The rms speed depends only on T for a given gas. So \(c_X/c_Y=\sqrt{T/(T/2)}=\sqrt{2}\). A speed ratio of 2 forgets the square root. A U ratio of 2 ignores the amount of gas. Equal U does not imply equal speed.
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Answer A
Answer: A
Common error: Internal energy is an extensive quantity: it depends on the amount of gas. Mean kinetic energy per molecule and rms speed depend on T alone.
64IBP2S-013-02-08Internal energy of a gasPaper 1AHard1 mark
Container X has volume V and is kept at absolute temperature T. Container Y has volume 2V and is kept at absolute temperature 2T. The containers are joined by a narrow tube, so the gas pressure is the same in both. The gas is ideal. What is \(\frac{\text{number of molecules in X}}{\text{number of molecules in Y}}\)?
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For each container, \(N = \frac{pV}{k_BT}\), with the same p. This gives NX ∝ V/T and NY ∝ 2V/2T = V/T, so the ratio is 1 : 1. The answer 1 : 2 considers only the volumes. The answer 2 : 1 considers only the temperatures. The answer 1 : 4 multiplies the two effects instead of letting them cancel.
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Answer A
Answer: A
Exam tip: When the pressure is shared, write N ∝ V/T for each region. A larger volume and a higher temperature have opposite effects on N.
65IBP2S-013-02-10The ideal gas equationPaper 1AHard1 mark
A rigid industrial gas cylinder develops a slow leak. Over a day, the pressure of the gas falls from 6.0×105 Pa to 4.0×105 Pa. Over the same day, its temperature falls from 300 K to 250 K. What fraction of the original molecules has escaped?
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V is constant, so N ∝ p/T. Initially p/T = 6.0×105/300 = 2000 Pa K−1. Finally p/T = 4.0×105/250 = 1600 Pa K−1. The fraction remaining is 1600/2000 = 0.80, so the fraction escaped is 0.20. The answer 0.33 ignores the cooling, which accounts for part of the pressure drop. The answer 0.80 is the fraction remaining. The answer 0.44 uses an inverted temperature ratio.
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Answer B
Answer: B
Exam tip: Part of a pressure drop may be caused by cooling rather than by loss of gas. Compare p/T values, not pressures alone, to find the change in N.
66B-1B-06The empirical gas lawsPaper 1BEasy8 marks
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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Part (a)
the volume is Aℓ where the cross-sectional area A of the bore is constant, so V ∝ ℓ
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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
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OWTTE.
Part (c)
single straight line through the error bars, extended to ℓ = 0
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Part (d)
≈ −274 °C
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Accept −255 to −290 °C. Allow read-off from the candidate’s line.
Part (e)
1 / 68 × 100 = 1.5 %
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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
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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
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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»
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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)
67B-1B-27Kinetic model & molecular speedsPaper 1BEasy8 marks
A computer simulation models a monatomic ideal gas as a large number of identical atoms in a box. For each temperature T the simulation outputs the root mean square speed vrms of the atoms, where vrms² is the mean of the squares of the speeds. Because the simulation uses a limited number of atoms, each value of vrms has an uncertainty of ±1 %.
The graph shows vrms² against T for five of the six temperatures.
T / K
vrms / m s−1
vrms² / 105 m² s−2
100
353
1.25
200
494
2.44
300
611
400
706
4.98
500
783
6.13
600
866
7.50
Graph drawn to scale
(a)
Calculate the missing value of vrms² for T = 300 K and plot the point on the graph.
(1)
(b)
State the percentage uncertainty in each value of vrms².
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Show that the gradient of the line is 3kB/m, where m is the mass of one atom.
(1)
(e)
Determine m.
(2)
(f)
The gas modelled is one of helium (4.0 g mol−1), neon (20.2 g mol−1), argon (39.9 g mol−1) or krypton (83.8 g mol−1). Deduce which gas it is.
(1)
(g)
At 300 K the simulated pressure is 1.00 × 105 Pa. Calculate the density of the gas.
(1)
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Part (a)
611² = 3.73 «× 105 m² s−2» AND plotted correctly
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Accept 3.7.
Part (b)
±2 % «twice the percentage uncertainty in vrms»
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Part (c)
single straight line through all the error bars, passing through or very close to the origin
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Part (d)
½mvrms² = ³⁄₂kBT so vrms² = (3kB/m)T, a line through the origin of gradient 3kB/m
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Must see the average kinetic energy equated to ³⁄₂kBT.
Part (e)
gradient = 1250 m² s−2 K−1
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Accept 1210–1280. Power of ten from the axis needed.
m = 3 × 1.38 × 10−23/1250 = 3.3 × 10−26 kg
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Allow ECF.
Part (f)
M = mNA = 3.3 × 10−26 × 6.02 × 1023 = 19.9 g mol−1, so the gas is neon
Accept 0.80–0.82 kg m−3. Allow use of a value read from the line.
Answers: (a) 3.73 × 105 m² s−2 · (b) ±2 % · (e) 3.3 × 10−26 kg · (f) neon · (g) 0.80 kg m−3(the remaining parts are explanations — see the table above)
A computer simulation models N argon atoms moving randomly in a cubical box of side 40.0 nm. The temperature is set to 300 K. In each run the program adds up the impulse J delivered to one wall of the box by the atoms that collide with it during a time interval of 2.00 × 10−9 s. Because the number of collisions in this short time varies at random, each value of J has an uncertainty of ±3 %.
The table shows J and the pressure P on the wall. The graph shows P against N for five of the six runs.
N
J / 10−19 N s
P / 104 Pa
400
0.845
2.64
800
1.63
5.09
1200
2.51
1600
3.39
10.6
2000
4.05
12.7
2400
5.01
15.7
Graph drawn to scale
(a)
Outline why the impulse delivered to the wall is equal to the total change in momentum of the atoms that collide with it.
(1)
(b)
Show that the pressure on the wall for N = 1200 is about 7.8 × 104 Pa.
(1)
(c)
Plot this point on the graph and draw the line of best fit.
(2)
(d)
Determine the temperature of the atoms predicted by the gradient of your line, using PV = NkBT.
(3)
(e)
The uncertainty in the gradient is ±4 %. Comment on whether the simulation is consistent with the set temperature of 300 K.
(1)
(f)
Suggest one change to the simulation that would reduce the scatter of the values of J.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
by Newton’s third law the wall exerts an equal and opposite force on the atoms, so the impulse on the wall equals the total change in momentum of the atoms «2mv for each perpendicular rebound»
✓ 1
OWTTE.
Part (b)
P = J/(AΔt) = 2.51 × 10−19/((40.0 × 10−9)2 × 2.00 × 10−9) = 7.84 × 104 Pa
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (c)
point plotted correctly
✓ 1
Allow ECF from (b).
single straight line through all the error bars, passing through the origin
✓ 1
Part (d)
gradient ≈ 65 Pa «per atom»
✓ 1
Accept 63–67 Pa.
P = (kBT/V)N, so T = gradient × V/kB, with V = (40.0 × 10−9)3 = 6.40 × 10−23 m3
✓ 1
T ≈ 301 K
✓ 1
Accept 290–310 K. Allow ECF.
Part (e)
4 % of 301 K is about ±12 K, so 300 K lies within the range and the simulation is consistent
✓ 1
Allow ECF from (d). A comparison is needed.
Part (f)
use a longer time interval OR average J over several runs «so more collisions are included»
✓ 1
Answers: (b) 7.84 × 104 Pa · (d) 301 K (the remaining parts are explanations — see the table above)
A student uses sealed syringes to measure atmospheric pressure. For each syringe the plunger is pushed fully in, the nozzle is sealed and the plunger is pulled out slowly by a force sensor. The force F needed to keep the plunger moving steadily is recorded. The internal diameter d of each barrel is measured with vernier calipers to ±0.05 mm and each force has an uncertainty of ±0.5 N. The student assumes that the frictional force on the plunger is the same for every syringe.
The graph shows F against the cross-sectional area A of the barrel for five of the six syringes.
d / mm
A / 10−4 m2
F / N
10.4
0.849
11.2
12.0
1.131
13.5
14.6
19.4
19.2
2.895
32.0
21.6
3.664
39.3
26.2
5.391
56.9
Graph drawn to scale
(a)
Calculate A for d = 14.6 mm and its percentage uncertainty.
(2)
(b)
Plot this point on the graph and draw the line of best fit.
(1)
(c)
Determine the gradient of the line, including its unit.
(2)
(d)
Explain why the gradient is equal to the atmospheric pressure.
(1)
(e)
State the intercept on the F axis and outline what it represents.
(2)
(f)
A barometer in the laboratory reads 101.2 kPa. The uncertainty in the gradient is ±5 %. Comment on the student’s value.
(1)
(g)
A small volume of air stays trapped in the nozzle when the plunger is pushed in. Explain the effect of this air on the value of atmospheric pressure obtained.
70B-1B-05The empirical gas lawsPaper 1BMedium9 marks
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)
Show mark scheme
Marking point
Mark
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)
71B-1B-15The ideal gas equationPaper 1BMedium10 marks
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)
72B-1B-16Moles, molecules & the Avogadro constantPaper 1BMedium10 marks
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
Diagram not to scale
(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)
73B-1B-26The ideal gas equationPaper 1BMedium10 marks
A student determines the molar mass M of air. A steel cylinder of internal volume (3.00 ± 0.02) × 10−3 m³ is fitted with a valve and placed on a top-pan balance of resolution 0.01 g. Air is pumped into the cylinder in stages. After each stage the student waits for the cylinder to return to room temperature, 293 ± 1 K, and then records the gauge pressure p (the pressure above atmospheric pressure) and the mass m of the cylinder and its contents.
Each value of p has an uncertainty of ±0.10 × 105 Pa, shown by the error bars. The uncertainty in m is negligible.
p / 105 Pa
0.0
1.0
2.0
3.0
4.0
5.0
6.0
m / g
1846.52
1850.29
1853.42
1857.39
1860.61
1864.67
1867.86
Graph drawn to scale
(a)
The air in the cylinder warms while it is being pumped. Explain why the student waited for the cylinder to return to room temperature before reading the gauge.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Show that the gradient of the line is equal to MV/RT, where V is the volume of the cylinder and T is room temperature.
(1)
(d)
Determine M, using the gradient of your line.
(3)
(e)
The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in M and write down M with its uncertainty.
(2)
(f)
The accepted molar mass of air is 0.029 kg mol−1. Comment on the student's result.
(1)
(g)
State what is represented by the intercept of the line on the m axis.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 293 K is used in the calculation; while warm, the pressure of the same mass of air is higher than at 293 K, so the readings would not fit PV = nRT at 293 K «M would be too small»
✓ 1
OWTTE. Do not accept ‘for a fair test’.
Part (b)
single straight line through all the error bars
✓ 1
Part (c)
mass of air = nM = (p + Patm)VM/RT, so m = (MV/RT)p + constant «and the gradient is MV/RT»
✓ 1
Must see n = PV/RT with the total pressure, and the constant terms identified.
Part (d)
gradient = 3.57 g per 105 Pa from a large triangle
Value and uncertainty must have matching precision. Accept (0.029 ± 0.001) kg mol−1.
Part (f)
0.029 lies within the range 27.8–30.2 × 10−3 kg mol−1, so the result is consistent with the accepted value
✓ 1
Allow ECF from (e).
Part (g)
the mass of the empty cylinder plus the air it contains at atmospheric pressure «gauge pressure zero»
✓ 1
OWTTE. Do not accept ‘the mass of the cylinder’ alone.
Answers: (d) 3.57 g per 105 Pa; M = 2.90 × 10−2 kg mol−1 · (e) (29.0 ± 1.2) × 10−3 kg mol−1(the remaining parts are explanations — see the table above)
74B-1B-40The empirical gas lawsPaper 1BMedium12 marks
A student investigates how the pressure P of a fixed mass of air at constant volume depends on its temperature. A sealed glass flask of internal volume (2.50 ± 0.02) × 10−4 m3 is held completely under the water in a stirred water bath and joined by a short, narrow tube to a pressure sensor. After each change of bath temperature the student waits until the pressure reading is steady.
The bath temperature θ is read from a digital thermometer of resolution 0.1 °C. Each pressure reading has an uncertainty of ±0.5 kPa. The graph shows P against the kelvin temperature T, with T/K = θ/°C + 273.
θ / °C
P / kPa
4.8
96.3
18.6
100.6
31.9
105.5
46.2
110.1
59.7
115.2
73.5
119.7
87.1
124.8
Graph drawn to scale
(a)
Draw the line of best fit for the data.
(1)
(b)
Determine the gradient of the line, including its unit.
(2)
(c)
Show that the gradient is equal to nR/V, and hence determine the amount n of air in the flask.
(2)
(d)
The uncertainty in the gradient is ±2 %. Determine the absolute uncertainty in n and state n with its uncertainty.
(2)
(e)
The student’s hypothesis is that P is proportional to T. By calculating P/T and its uncertainty for the first and the last readings, determine whether these readings support the hypothesis.
(3)
(f)
The tube and the sensor contain a small volume of air that stays at room temperature, 20 °C. State and explain the effect of this on the value of n found in (c).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
single straight line through all the error bars «passing very close to every point»
✓ 1
Part (b)
gradient from a large triangle ≈ 0.346
✓ 1
Accept 0.336–0.356.
kPa K−1
✓ 1
Unit needed for MP2. Accept 345 Pa K−1 etc.
Part (c)
PV = nRT so P = (nR/V)T: a straight line through the origin of gradient nR/V
75B-1B-41Kinetic model & molecular speedsPaper 1BHard12 marks
A student uses a published table of the root mean square speeds v of the molecules of seven gases, all measured at the same temperature. The speeds were deduced from molecular-beam time-of-flight experiments and each has an uncertainty of ±2 %. M is the molar mass of the gas.
To test how v depends on M, the student plots log(v / m s−1) against log(M / g mol−1). The graph shows the points for six of the gases; the point for nitrogen has not been plotted.
Gas
M / g mol−1
v / m s−1
log(M / g mol−1)
log(v / m s−1)
He
4.00
1380
0.602
3.140
Ne
20.2
601
1.305
2.779
N2
28.0
520
Ar
39.9
431
1.601
2.634
CO2
44.0
418
1.643
2.621
Kr
83.8
296
1.923
2.471
Xe
131
240
2.117
2.380
Graph drawn to scale
(a)
Complete the table for nitrogen and plot the point on the graph.
(2)
(b)
Show that the uncertainty in each value of log(v / m s−1) is about ±0.009.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine the gradient of the line.
(2)
(e)
Kinetic theory predicts that v = √(3RT/M). Explain whether the graph supports this prediction.
(2)
(f)
The table does not state the temperature. Use the value of log(v / m s−1) on your line at log(M / g mol−1) = 1.00 to determine the temperature of the gases.
(3)
(g)
Outline one advantage of plotting log v against log M rather than v against M.
76B-1B-43The empirical gas lawsPaper 1BHard12 marks
A gas pycnometer measures the volume of a solid sample using helium. The sample chamber has an internal volume Vc = (100.0 ± 0.2) cm3 and the reference chamber has an internal volume Vr = (50.0 ± 0.2) cm3. A student places 85.42 g of dry sand in the sample chamber, evacuates the reference chamber and fills the sample chamber with helium to a pressure P1 = (200.0 ± 0.1) kPa. After waiting, the valve is opened and, after waiting again, the final pressure P2 is read. The temperature of the apparatus is the same at both readings.
The student repeats the procedure five times with the same P1.
Trial
1
2
3
4
5
P2 / kPa
115.0
115.2
114.9
115.3
115.1
Diagram NOT accurately drawn
(a)
Calculate the mean value of P2 and its absolute uncertainty.
(2)
(b)
The volume of helium in the sample chamber before the valve is opened is Vg. Show that Vg = P2Vr/(P1 − P2).
(2)
(c)
Calculate the volume Vs of the sand grains.
(1)
(d)
Determine the density of the sand grains and its absolute uncertainty. The uncertainty in the mass is negligible.
(3)
(e)
The sand is mainly quartz, of density 2.65 g cm−3. Comment on the student’s result.
(1)
(f)
In one trial the temperature of the apparatus rose slightly between the filling of the sample chamber and the reading of P2. State and explain the effect of this on the density of the sand calculated from that trial.
(2)
(g)
Suggest why helium, rather than air, is used in a pycnometer.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
mean = 115.1 kPa
✓ 1
uncertainty = ½ × range = ±0.2 kPa
✓ 1
Part (b)
the temperature is the same, so Boyle’s law applies to the fixed mass of helium: P1Vg = P2(Vg + Vr)
✓ 1
Must see the total final volume Vg + Vr.
Vg(P1 − P2) = P2Vr and rearrangement shown
✓ 1
Part (c)
Vg = 115.1 × 50.0/84.9 = 67.8 cm3, so Vs = 100.0 − 67.8 = 32.2 cm3
✓ 1
Allow ECF from (a).
Part (d)
ρ = 85.42/32.2 = 2.65 g cm−3
✓ 1
Allow ECF from (c).
fractional uncertainty in Vg = 0.2/115.1 + 0.3/84.9 + 0.2/50.0 = 0.93 %, so ΔVg = 0.63 cm3 and ΔVs = 0.83 cm3
✓ 1
Uncertainty of P1 − P2 = 0.1 + 0.2 = 0.3 kPa must be used. Allow ECF.
ρ = (2.65 ± 0.07) g cm−3
✓ 1
MP3 is for matching precision of value and uncertainty. Accept ±0.06 to ±0.08.
Part (e)
2.65 g cm−3 lies within the range of the result, so the result is consistent with the sand being quartz
✓ 1
Allow ECF from (d).
Part (f)
the pressure of a fixed amount of gas in a fixed volume increases with temperature, so P2 is too large and Vg «= P2Vr/(P1 − P2)» is too large
✓ 1
Vs = Vc − Vg is too small, so the calculated density is too large
✓ 1
MP2 only scores if MP1 scores.
Part (g)
helium atoms are very small, so they enter the narrow gaps and pores between the grains, AND helium behaves very nearly as an ideal gas «weak intermolecular forces, not adsorbed on surfaces»
✓ 1
Accept either point.
Answers: (a) (115.1 ± 0.2) kPa · (c) 32.2 cm3 · (d) (2.65 ± 0.07) g cm−3(the remaining parts are explanations — see the table above)
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)
Show mark scheme
Marking point
Mark
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)
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)
A laboratory vacuum chamber of volume 0.0400 m³ initially contains air at a pressure of 1.00 × 105 Pa and a temperature of 293 K. A simple piston pump removes air from the chamber.
In each stroke, the chamber is connected to the empty cylinder of the pump, of volume 4.00 × 10−3 m³, and the air spreads into both. The valve to the chamber is then closed and the air in the pump cylinder is expelled. The temperature of the air stays at 293 K throughout. Treat the air as an ideal gas.
Axes for (d)
(a)
Calculate the number of molecules of air in the chamber at the start.
(1)
(b)
Show that the pressure in the chamber after one stroke is about 9.1 × 104 Pa.
(1)
(c)
Determine whether 30 strokes are enough to reduce the pressure in the chamber to less than 5 % of its initial value.
(2)
(d)
Sketch, on the axes, a graph to show how the pressure in the chamber varies with the number of strokes, from 0 to 20 strokes.
(2)
(e)
Explain, in terms of molecules, why the pressure in the chamber decreases even though the temperature of the air does not change.
(2)
(f)
The chamber has a circular glass window of diameter 0.150 m. The atmospheric pressure outside is 1.00 × 105 Pa. Calculate the resultant force on the window when the pressure in the chamber is 2.0 × 103 Pa.
Must see full substitution OR answer to at least 3 s.f. Accept a Boyle’s law argument, P1V1 = P2V2.
Part (c)
each stroke multiplies the pressure by 0.0400/0.0440 = 1/1.1, so after 30 strokes P/P0 = (1/1.1)30 = 0.0573
✓ 1
= 5.7 %, which is more than 5 %, so 30 strokes are not enough
✓ 1
MP2 needs a comparison and a consistent conclusion. Do not accept a linear (constant decrease per stroke) model.
Part (d)
curve starting at 100 kPa and decreasing with decreasing gradient «falls by the same fraction each stroke»
✓ 1
Accept a set of points or a smooth curve.
passing through about 39 kPa at 10 strokes and about 15 kPa at 20 strokes
✓ 1
Accept ±3 kPa.
Part (e)
there are fewer molecules per unit volume, so there are fewer collisions with the walls per unit time
✓ 1
the temperature is the same, so the mean speed «and the change of momentum per collision» is unchanged; the rate of change of momentum per unit area «the pressure» is therefore smaller
✓ 1
OWTTE
Part (f)
area = π × 0.0752 = 1.77 × 10−2 m² and pressure difference = 9.8 × 104 Pa
✓ 1
F = 9.8 × 104 × 1.77 × 10−2 = 1.73 × 103 N «towards the inside of the chamber»
✓ 1
Award [2] for CNA. Award [1] max for 1.77 × 103 N (inside pressure ignored).
Answers: (a) 9.89 × 1023 · (b) 9.09 × 104 Pa · (c) 5.7 % — not enough · (f) 1.73 × 103 N (the remaining parts are explanations — see the table above)
Hot jam at 88 °C is poured into a glass jar and a metal lid is screwed on at once. The air in the small space under the lid (the headspace) is then at 88 °C and at atmospheric pressure, 1.01 × 105 Pa. The volume of the headspace is 2.4 × 10−5 m3 and does not change. The jar cools to a room temperature of 20 °C. Ignore the water vapour in the headspace and treat the air as an ideal gas.
The diagram shows a cross-section through the top of the sealed jar.
Diagram NOT accurately drawn
(a)
(i)
Show that the pressure of the air in the headspace at 20 °C is about 8.2 × 104 Pa.
(1)
(ii)
Explain, in terms of the molecules, why the pressure of the air in the headspace decreases as the jar cools.
(2)
(b)
Calculate the number of molecules of air in the headspace.
(1)
(c)
(i)
Draw, on the diagram, labelled arrows to show the forces that act on the lid of the cold jar. Ignore the weight of the lid.
(2)
(ii)
The inside diameter of the lid is 66 mm. Calculate the resultant force exerted on the lid by the air inside and outside the jar.
(2)
(d)
The lid has a small central button. The button is held down only if the pressure outside the jar exceeds the pressure inside by at least 1.2 × 104 Pa; otherwise it pops up, warning that the seal may have failed. Another jar is filled with jam that has cooled to 55 °C before the lid is put on. Determine whether the button of this jar is held down when the jar is at 20 °C.
(3)
(e)
Suggest why holding the lid of the cold jar under hot water makes the jar easier to open.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
P = 1.01 × 105 × 293/361 = 8.20 × 104 Pa
✓ 1
Must see full substitution OR answer to at least 3 s.f. Award [0] if Celsius temperatures are used.
Part (a)(ii)
the molecules move more slowly «lower mean kinetic energy», so each collision with the walls gives a smaller change in momentum
✓ 1
and there are fewer collisions per second with each unit area of the walls, so the rate of change of momentum per unit area «the pressure» is smaller
this is less than 1.2 × 104 Pa, so the button is not held down «it pops up»
✓ 1
MP3 needs a comparison and a consistent conclusion.
Part (e)
the air in the headspace warms, so its pressure increases «P ∝ T at constant volume»
✓ 1
the pressure difference «resultant force pushing the lid onto the rim» is smaller, so there is less friction to overcome when turning the lid «ALTERNATIVE for this mark: the metal lid expands more than the glass rim, so it grips the rim less tightly»
✓ 1
OWTTE.
Answers: (a)(i) 8.20 × 104 Pa · (b) 4.9 × 1020 · (c)(ii) 65 N · (d) 1.08 × 104 Pa, not held down (the remaining parts are explanations — see the table above)
A sealed snack bag contains 1.20 × 10−3 m3 of air at 1.01 × 105 Pa and 25 °C when it is loaded onto an aircraft. The bag is not tight, so the air in it is at the same pressure as the cabin until the bag reaches its maximum volume of 1.50 × 10−3 m3. At cruising height the cabin pressure is 7.6 × 104 Pa and the cabin temperature is 20 °C. Treat the air as an ideal gas.
Axes for (d)
(a)
Calculate the amount of air, in mol, in the bag.
(1)
(b)
Determine whether the bag reaches its maximum volume at cruising height.
(3)
(c)
(i)
Calculate the pressure of the air in the bag at cruising height.
(1)
(ii)
One flat side of the bag has an area of 0.020 m2. Calculate the resultant force exerted on this side by the air inside and outside the bag.
(2)
(d)
The change from the initial state S, shown on the axes, to the state at cruising height can be modelled as cooling at constant pressure from 25 °C to 20 °C, followed by expansion at a constant temperature of 20 °C until the bag reaches its maximum volume. Draw, on the axes, the change for this model. Label the final state C.
(2)
(e)
At cruising height a passenger opens the bag, lets its air come to cabin pressure and seals it again at its maximum volume. Calculate the volume of the air in the bag after landing, when it is at 1.01 × 105 Pa and 25 °C.
84IBP2S-013-03-12Moles, molecules & the Avogadro constantPaper 2Easy5 marks
A hospital oxygen cylinder has an internal volume of 10.0 litres. It contains oxygen at a pressure of 1.50×107 Pa and a temperature of 20 °C. Treat the oxygen as an ideal gas. The molar mass of oxygen (O2) is 32 g mol−1.
(a)
Calculate the amount of oxygen, in moles, in the cylinder.
(2)
(b)
Calculate the number of oxygen molecules in the cylinder.
(1)
(c)
Determine the mass of oxygen in the cylinder.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses V = 1.00×10−2 m3 and T = 293 K in n = pV/(RT).
✓ 1
n = (1.50×107 × 1.00×10−2)/(8.31 × 293) = 61.6 mol.
✓ 1
Part (b)
N = 61.6 × 6.02×1023 = 3.71×1025. ECF from (a).
✓ 1
Part (c)
Uses m = nM with M = 0.032 kg mol−1.
✓ 1
m = 61.6 × 0.032 = 1.97 kg. ECF from (a).
✓ 1
Exam tip: 1 litre = 10⁻³ m³. Convert the volume to m³ before substituting into pV = nRT.
85IBP2S-013-01-11Internal energy of a gasPaper 2Easy6 marks
A glass tube in an illuminated advertising sign contains neon, which can be treated as an ideal monatomic gas. The tube has an internal volume of 3.0×10−4 m3. The neon is at a pressure of 1.5×103 Pa and a temperature of 300 K.
(a)
State what is meant by the internal energy of an ideal gas.
(2)
(b)
Calculate the number of neon atoms in the tube.
(2)
(c)
Calculate the internal energy of the neon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
It is the total random kinetic energy of all the molecules/atoms . The intermolecular potential energy is zero for an ideal gas, so the internal energy is kinetic energy only .
✓ 2
Part (b)
Use N = pV/(kT) = (1.5×103 × 3.0×10−4)/(1.38×10−23 × 300) . N = 1.1×1020 .
✓ 2
Part (c)
Use U = (3/2)NkT or U = (3/2)pV . U = 1.5 × 1.5×103 × 3.0×10−4 = 0.68 J . ECF from (b) if U = (3/2)NkT is used.
✓ 2
Exam tip: In (c), U = (3/2)pV avoids carrying a rounded value of N forward. Either route earns full credit.
86IBP2S-013-02-11Moles, molecules & the Avogadro constantPaper 2Easy5 marks
A portable medical oxygen cylinder has an internal volume of 10.0 litres. When full, it contains oxygen at a pressure of 1.50×107 Pa and a temperature of 293 K. Treat the oxygen as an ideal gas. Molar mass of oxygen (O2) = 32 g mol−1; R = 8.31 J mol−1 K−1; NA = 6.02×1023 mol−1.
(a)
Calculate the amount of oxygen, in mol, in the full cylinder.
(2)
(b)
Determine the number of oxygen molecules in the cylinder.
(1)
(c)
Calculate the mass of oxygen in the cylinder.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Converts the volume to V = 1.00×10−2 m3 and uses n = pV/RT
✓ 1
n = (1.50×107 × 1.00×10−2)/(8.31 × 293) = 61.6 mol
✓ 1
Part (b)
N = nNA = 61.6 × 6.02×1023 = 3.71×1025
✓ 1
(Allow error carried forward from the amount calculated in the first part.)
—
Part (c)
Uses m = nM with M = 0.032 kg mol−1
✓ 1
m = 61.6 × 0.032 = 1.97 kg
✓ 1
(Allow error carried forward from the amount calculated in the first part.)
—
Common error: 1 litre = 1.00×10⁻³ m³, so 10.0 litres = 1.00×10⁻² m³. If you substitute the volume in litres, or the molar mass in g mol⁻¹, the answer is wrong by a factor of 1000.
88B-2-28The ideal gas equationPaper 2Medium12 marks
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)
In a hybrid airbag inflator, argon is stored in a steel cylinder of internal volume 3.00 × 10−4 m³ at a pressure of 2.40 × 107 Pa and a temperature of 20 °C. In a collision a valve opens and the argon flows into the airbag; a small charge heats the argon as it flows. When the bag is fully inflated its volume is 0.0600 m³ and the argon in it is at 360 K.
Argon is a monatomic gas. Treat it as an ideal gas unless told otherwise. Atmospheric pressure = 1.01 × 105 Pa.
(a)
(i)
Show that the cylinder contains about 3.0 mol of argon.
(1)
(ii)
Calculate the number of argon atoms in the cylinder.
(1)
(b)
Calculate the pressure of the argon in the fully inflated bag.
(2)
(c)
Calculate the increase in the internal energy of the argon between the cylinder and the inflated bag.
(2)
(d)
The face of the driver presses into the bag over a contact area of 0.040 m²; the rest of the head is surrounded by the air in the car. Estimate the resultant force on the head due to gas pressure at the instant described in (b).
(2)
(e)
The diameter of an argon atom is about 3.8 × 10−10 m. Determine, by estimating the mean separation of the atoms, whether the ideal gas model is suitable for the argon in the cylinder and for the argon in the inflated bag.
(3)
(f)
Explain, in terms of the atoms, why the pressure in the inflated bag is much smaller than the pressure in the cylinder, even though the temperature of the argon is higher.
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
N = nNA = 2.96 × 6.02 × 1023 = 1.78 × 1024
✓ 1
Part (b)
P = nRT/V = 2.96 × 8.31 × 360 / 0.0600
✓ 1
ALTERNATIVE: P1V1/T1 = P2V2/T2.
= 1.47 × 105 Pa
✓ 1
Award [2] for CNA.
Part (c)
ΔU = (3/2)nRΔT = 1.5 × 2.96 × 8.31 × (360 − 293)
✓ 1
= 2.47 × 103 J
✓ 1
Award [2] for CNA. Award [1] max for the internal energy at one temperature only (e.g. 1.33 × 104 J).
Part (d)
resultant pressure on the head = 1.474 × 105 − 1.01 × 105 = 4.64 × 104 Pa
✓ 1
Allow ECF from (b).
F = ΔPA = 4.64 × 104 × 0.040 = 1.86 × 103 N
✓ 1
Accept 1.8 × 103 to 1.9 × 103 N. Award [1] max for 5.9 × 103 N (atmospheric pressure not subtracted).
Part (e)
cylinder: separation ≈ (V/N)1/3 = (3.0 × 10−4/1.78 × 1024)1/3 = 5.5 × 10−10 m, about 1.5 atomic diameters
✓ 1
bag: separation ≈ 3.2 × 10−9 m, about 8.5 atomic diameters
✓ 1
in the cylinder the atoms are so close that their volume and the forces between them are significant, so the model is not suitable; in the bag it is a good approximation
✓ 1
MP3 needs both conclusions consistent with the candidate’s values.
Part (f)
the number of atoms per unit volume in the bag is 200 times smaller, so there are far fewer collisions with each unit area of the wall per second
✓ 1
the higher temperature increases the mean speed only by a factor of about √(360/293) = 1.11, a small increase in the momentum change per collision
✓ 1
OWTTE
Answers: (a)(i) 2.96 mol · (a)(ii) 1.78 × 1024 · (b) 1.47 × 105 Pa · (c) 2.47 × 103 J · (d) 1.9 × 103 N (the remaining parts are explanations — see the table above)
90B-2-74Moles, molecules & the Avogadro constantPaper 2Medium12 marks
At the summit of a high mountain the air pressure is 3.4 × 104 Pa and the temperature is −20 °C. At sea level the air pressure is 1.01 × 105 Pa and the temperature is 15 °C. In each breath a climber takes in 5.0 × 10−4 m3 of outside air, measured at the outside pressure and temperature. 21 % of the molecules in air are oxygen molecules. Treat air as an ideal gas.
(a)
(i)
Calculate the amount of air, in mol, in one breath at the summit.
(1)
(ii)
Calculate the number of oxygen molecules in one breath at the summit.
(1)
(b)
In the lungs the air from one breath warms to 37 °C at constant pressure. Calculate the volume of this air in the lungs.
(2)
(c)
At sea level the climber takes 12 breaths per minute. At the summit the climber takes 28 breaths per minute, each of the same volume as before. Determine whether the climber takes in at least as many oxygen molecules per minute at the summit as at sea level.
(3)
(d)
Explain, in terms of molecules, why the air pressure at the summit is lower than at sea level.
(2)
(e)
Calculate (root mean square speed of the air molecules at the summit) / (root mean square speed of the air molecules at sea level).
(2)
(f)
Outline why the ideal gas model is a good approximation for the air at the summit.
Allow ECF from (a)(i). Accept the comparison in numbers of oxygen molecules.
the summit value is smaller, so the climber does not take in as many oxygen molecules per minute
✓ 1
MP3 needs a comparison of calculated values and a consistent conclusion.
Part (d)
there are fewer molecules per unit volume «about 0.38 of the sea-level value», so fewer collisions with each unit area per second
✓ 1
the lower temperature means the molecules are slower, so each collision gives a smaller change in momentum «and collisions are less frequent»
✓ 1
OWTTE.
Part (e)
³⁄₂ kBT = ½ mv̄2, so vrms ∝ √T
✓ 1
Accept vrms = √(3RT/M).
ratio = √(253/288) = 0.937
✓ 1
Award [1] max for 0.878 «square root omitted».
Part (f)
the pressure/density is low «and the temperature far above the boiling points of nitrogen and oxygen», so the molecules are far apart and intermolecular forces and the volume of the molecules are negligible
✓ 1
OWTTE. A condition and its molecular consequence are both needed.
Answers: (a)(i) 8.1 × 10−3 mol · (a)(ii) 1.0 × 1021 · (b) 6.1 × 10−4 m3 · (c) 0.253 mol min−1 vs 0.226 mol min−1 · (e) 0.937 (the remaining parts are explanations — see the table above)
91B-2-75Kinetic model & molecular speedsPaper 2Medium13 marks
A small satellite of total mass 4.0 kg is moved by a cold-gas thruster. Its tank has an internal volume of 4.0 × 10−4 m3 and contains nitrogen (molar mass 0.028 kg mol−1) at a pressure of 1.5 × 107 Pa and a temperature of 290 K. When the thruster fires, nitrogen leaves the nozzle at a constant rate of 1.2 × 10−4 kg s−1 with a speed of 700 m s−1 relative to the satellite. The gas in the tank stays at 290 K. Treat nitrogen as an ideal gas.
(a)
(i)
Show that the tank contains about 2.5 mol of nitrogen.
(1)
(ii)
Calculate the mass of nitrogen in the tank.
(1)
(iii)
Calculate the root mean square speed of the nitrogen molecules in the tank.
(2)
(b)
Explain, with reference to momentum, how the gas leaving the nozzle exerts a force on the satellite.
(2)
(c)
Calculate the force exerted on the satellite by the thruster.
(1)
(d)
The tank is a cylinder. Each of its flat ends has an internal area of 2.8 × 10−3 m2.
(i)
Calculate the force exerted by the nitrogen on one end of the full tank.
(1)
(ii)
The force in (d)(i) is about 5 × 105 times the force in (c). Explain why the nitrogen in the closed tank exerts no resultant force on the satellite when the thruster is not firing.
(2)
(e)
The satellite is initially at rest relative to a nearby space station. The thruster then fires in one direction for 5.0 minutes. Determine the speed of the satellite relative to the station at the end of this firing.
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
m = 2.49 × 0.028 = 7.0 × 10−2 kg
✓ 1
Accept 0.070 kg.
Part (a)(iii)
ρ = m/V = 6.97 × 10−2/4.0 × 10−4 = 174 kg m−3
✓ 1
ALTERNATIVE: vrms = √(3RT/M) for both marks.
vrms = √(3P/ρ) = √(3 × 1.5 × 107/174) = 508 m s−1
✓ 1
Allow ECF from (a)(ii). Award [1] max for 293 m s−1 «factor 3 omitted».
Part (b)
the gas is given momentum away from the satellite «backwards»: the satellite exerts a force on the gas equal to its rate of change of momentum
✓ 1
by Newton’s third law the gas exerts an equal and opposite force «forwards» on the satellite
✓ 1
Part (c)
F = Δp/Δt = 1.2 × 10−4 × 700 = 0.084 N
✓ 1
Part (d)(i)
F = PA = 1.5 × 107 × 2.8 × 10−3 = 4.2 × 104 N
✓ 1
Part (d)(ii)
the pressure is the same on every part of the wall, so the force on each end «and on each part of the curved wall» is balanced by an equal and opposite force on the opposite wall
✓ 1
OWTTE.
the molecules move randomly in all directions, so the total momentum they give to the walls each second is zero in every direction «the gas and the tank form a closed system with no external force»
✓ 1
ALTERNATIVE for MP2: when the nozzle is open there is no wall opposite the nozzle, so the forces are no longer balanced.
Part (e)
mass of nitrogen released = 1.2 × 10−4 × 300 = 3.6 × 10−2 kg
✓ 1
ALTERNATIVE: impulse = 0.084 × 300 = 25 N s.
conservation of momentum: 4.0v = 3.6 × 10−2 × 700 «the change in the mass of the satellite is small»
✓ 1
Allow ECF from (c).
v = 6.3 m s−1
✓ 1
Accept 6.3–6.4 m s−1. Award [3] for CNA.
Answers: (a)(i) 2.49 mol · (a)(ii) 7.0 × 10−2 kg · (a)(iii) 508 m s−1 · (c) 0.084 N · (d)(i) 4.2 × 104 N · (e) 6.3 m s−1(the remaining parts are explanations — see the table above)
A constant-volume gas thermometer consists of a glass bulb of volume 1.20 × 10−4 m3 filled with helium, joined by a capillary tube of negligible volume to a pressure sensor. Helium is a monatomic gas. The thermometer is calibrated by placing the bulb in melting ice at 273 K, where the pressure is 40.0 kPa, and then in steam at 373 K, where the pressure is 54.7 kPa. The graph shows these two points.
Graph drawn to scale
(a)
(i)
Draw, on the graph, the straight line through the two calibration points, extended to P = 0.
(1)
(ii)
State what your line suggests about the relationship between P and T.
(1)
(b)
The bulb is placed in a flask of liquid oxygen. The pressure is 13.2 kPa. Calculate the temperature of the liquid oxygen in °C.
(2)
(c)
(i)
Calculate the number of helium atoms in the bulb.
(1)
(ii)
Calculate the internal energy of the helium when the bulb is in melting ice.
(1)
(d)
The thermometer is emptied and refilled with nitrogen to the same pressure at 273 K. Nitrogen boils at 77 K. Explain why helium is a better choice than nitrogen for measuring the temperature of liquid oxygen.
(2)
(e)
In a real thermometer the capillary tube contains a small volume of helium that stays at room temperature. State and explain how this affects the temperature measured for the liquid oxygen.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
straight line through both points that reaches P = 0 at «or very close to» T = 0
✓ 1
Part (a)(ii)
P is «directly» proportional to T «P/T constant at constant volume»
at 90 K nitrogen is close to the temperature at which it liquefies, so its molecules move slowly and the intermolecular «attractive» forces are significant; nitrogen deviates noticeably from ideal-gas behaviour
✓ 1
so its pressure is no longer proportional to T «it is lower than the ideal value»; helium liquefies only at a much lower temperature «about 4 K» and remains nearly ideal
✓ 1
OWTTE.
Part (e)
the measured temperature is too high
✓ 1
No mark for the effect without a reason.
some of the gas does not cool, so the pressure falls by a smaller factor than the temperature of the bulb «and T is calculated from the pressure»
✓ 1
OWTTE.
Answers: (b) 90.1 K = −183 °C · (c)(i) 1.27 × 1021 · (c)(ii) 7.2 J (the remaining parts are explanations — see the table above)
93B-2-78The empirical gas lawsPaper 2Medium13 marks
A student models a temperature-operated actuator. Helium is trapped in a horizontal cylinder by a piston of cross-sectional area 2.0 × 10−3 m2 that moves without friction. A spring of spring constant 1.0 × 103 N m−1 joins the piston to a fixed support, and the other side of the piston is in contact with the air of the room, at a pressure of 1.0 × 105 Pa. In the initial state A the spring has its natural length, and the helium is at 290 K and has a volume of 3.0 × 10−4 m3. The helium is then heated slowly, so that the piston moves out and compresses the spring. The diagram shows the apparatus during the heating. Helium is a monatomic gas; treat it as an ideal gas.
Diagram NOT accurately drawnAxes for (d)(ii)
(a)
Calculate the amount of helium, in mol, in the cylinder.
(1)
(b)
Draw, on the diagram, labelled arrows to show the horizontal forces that act on the piston during the heating, when it is at rest.
(2)
(c)
The piston has moved out by 2.0 cm.
(i)
Show that the pressure of the helium is about 1.1 × 105 Pa.
(1)
(ii)
Determine the temperature of the helium.
(2)
(d)
(i)
Deduce that, as the helium is heated, the graph of the pressure P of the helium against its volume V is a straight line.
(2)
(ii)
Draw, on the axes, the change from state A to the state in (c). Label the final state B.
(1)
(e)
Calculate the increase in the internal energy of the helium from A to B.
(2)
(f)
Explain, in terms of molecules, why the pressure of the helium increases although its volume also increases.
arrow away from the helium «outwards» labelled force from the helium/gas pressure AND two arrows towards the helium labelled force from the air/atmosphere and force from the spring
✓ 1
All arrows must act on the piston. Do not credit a weight or normal force for MP1 or MP2.
the outward arrow is clearly longer than each inward arrow «equal to their sum: the piston is in equilibrium»
✓ 1
MP2 only scores if MP1 scores.
Part (c)(i)
PA = P0A + kx, so P = 1.0 × 105 + 1.0 × 103 × 0.020/2.0 × 10−3 = 1.10 × 105 Pa
✓ 1
Must see the force balance with the atmospheric pressure included AND full substitution.
ALTERNATIVE: ΔU = ³⁄₂nRΔT with the values from (a) and (c)(ii).
ΔU = 11 J
✓ 1
Allow ECF. Award [2] for CNA. Award [1] max for 56 J «internal energy in state B, not the increase».
Part (f)
the temperature rises, so the molecules move faster: each collision with the walls/piston gives a larger change in momentum, and collisions with each unit area are more frequent
✓ 1
OWTTE.
the volume increases only a little «by 13 %, against 25 % for the kelvin temperature», so the fall in the number of molecules per unit volume is outweighed and the rate of change of momentum per unit area «the pressure» increases
✓ 1
OWTTE. A comparison of the two effects is needed.
Answers: (a) 1.24 × 10−2 mol · (c)(i) 1.10 × 105 Pa · (c)(ii) 362 K · (e) 11 J (the remaining parts are explanations — see the table above)
94B-2-80Internal energy of a gasPaper 2Medium12 marks
Two rigid tanks, X and Y, contain helium, a monatomic gas that behaves as an ideal gas. The tanks are joined by a pipe of negligible volume with a closed valve. The tanks and the pipe are thermally insulated from the surroundings. The diagram shows the volume, pressure and temperature of the helium in each tank.
Diagram NOT accurately drawn
(a)
(i)
Calculate the amount of helium, in mol, in tank X.
(1)
(ii)
Show that the internal energy of the helium in tank X is about 1.2 × 104 J.
(1)
(iii)
Calculate the internal energy of the helium in tank Y.
(1)
(b)
State, with a reason, in which tank the atoms have the greater average kinetic energy.
(1)
(c)
The valve is opened and the helium in the two tanks mixes. Eventually the helium is all at the same temperature.
(i)
Explain why the total internal energy of the helium does not change.
(2)
(ii)
Determine the final temperature of the helium.
(3)
(iii)
Show that the final pressure of the helium is about 2.2 × 105 Pa.
(1)
(iv)
Deduce that the final pressure would be the same for any initial temperatures of the helium in X and Y, provided that the initial pressures and volumes are unchanged.
95IBP2S-013-00-12Kinetic model & molecular speedsPaper 2Medium9 marks
Before a race, a racing-car tyre is inflated to an absolute pressure of 2.4 × 105 Pa at a temperature of 18 °C. The volume of air in the tyre is 0.030 m3 and may be assumed constant.
(a)
Calculate the amount, in mol, of air in the tyre.
(2)
(b)
After several laps, the temperature of the air in the tyre is 95 °C. Calculate the new pressure.
(2)
(c)
Explain, in terms of molecules, why the pressure increases as the temperature rises.
(3)
(d)
The team wants the pressure at 95 °C to be 2.8 × 105 Pa. Determine the amount of air that must be released from the tyre.
(2)
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Part (a)
Uses n = pV/(RT) with T = 291 K . n = (2.4 × 105 × 0.030)/(8.31 × 291) = 2.98 mol .
✓ 2
Part (b)
p2 = p1T2/T1, using kelvin temperatures 368 K and 291 K . p2 = 2.4 × 105 × 368/291 = 3.0 × 105 Pa .
✓ 2
Part (c)
The mean kinetic energy, and so the mean speed, of the molecules increases . Molecules hit the tyre walls more often . There is a greater change in momentum per collision, so the force on the walls (per unit area) increases .
✓ 3
Part (d)
n required = (2.8 × 105 × 0.030)/(8.31 × 368) = 2.75 mol . Amount released = 2.98 − 2.75 = 0.23 mol . ECF from (a).
✓ 2
Common error: Tyre gauges read pressure above atmospheric pressure. The gas laws need absolute pressure, which is the value given here.
96IBP2S-013-00-13The ideal gas equationPaper 2Medium8 marks
An industrial gas, X, is stored in steel cylinders. The table gives the measured value of the compression factor Z = pV/(nRT) for X at 300 K and different pressures.
State the value of Z for an ideal gas at all pressures.
(1)
(b)
Outline why Z is less than 1 at moderate pressures such as 10 MPa.
(2)
(c)
Explain why Z is greater than 1 at 40 MPa.
(2)
(d)
A cylinder of volume 0.050 m3 holds X at 20 MPa and 300 K. Using the table, determine the actual amount of X in the cylinder.
(3)
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Part (a)
Z = 1 .
✓ 1
Part (b)
Molecules are close enough for attractive intermolecular forces to be significant . These pull molecules away from the walls, reducing the rate of momentum change at the walls, so the pressure is lower than ideal and pV < nRT .
✓ 2
Part (c)
Molecules are very close together, so their own volume is a significant fraction of the container volume and repulsive forces dominate . The free volume available for motion is less than V, so for the same n, V and T the pressure is higher than ideal (or at the same p the volume is larger than ideal), giving pV > nRT .
✓ 2
Part (d)
Ideal amount: n = pV/(RT) = (20 × 106 × 0.050)/(8.31 × 300) = 401 mol . Uses Z = 0.97 at 20 MPa, so n = pV/(ZRT) . n = 401/0.97 ≈ 414 mol .
✓ 3
Exam tip: When Z is less than 1, the gas is more compressible than an ideal gas. A real cylinder then holds more moles than the ideal gas equation predicts.
A football has a constant internal volume of 5.5×10−3 m3. Before inflation, it contains air at a pressure of 1.0×105 Pa and a temperature of 290 K. The ball must be inflated to a pressure of 1.9×105 Pa. Each stroke of a hand pump pushes 1.5×10−4 m3 of atmospheric air (1.0×105 Pa, 290 K) into the ball. Assume the temperature stays at 290 K during inflation and that air behaves as an ideal gas.
(a)
Calculate the amount, in moles, of air in the ball before inflation.
(2)
(b)
Determine the number of pump strokes needed.
(3)
(c)
The ball is used on a cold day, and the air inside cools to 275 K. No air escapes. Calculate the new pressure in the ball.
(2)
(d)
Explain, in terms of the motion of the air molecules, why the pressure in the ball falls when it cools.
(2)
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Notes
Part (a)
n = pV/(RT) = (1.0×105 × 5.5×10−3)/(8.31 × 290).
✓ 1
n = 0.23 mol.
✓ 1
Part (b)
Recognises that the extra gas needed corresponds to Δp × V = 0.9×105 × 5.5×10−3 = 495 J, or Δn = 0.205 mol.
✓ 1
Each stroke delivers pV = 1.0×105 × 1.5×10−4 = 15 J, or 6.2×10−3 mol.
✓ 1
Number of strokes = 495/15 = 33.
✓ 1
Part (c)
At constant volume, p2 = p1T2/T1 = 1.9×105 × 275/290.
✓ 1
p2 = 1.8×105 Pa.
✓ 1
Part (d)
A lower temperature means a lower mean kinetic energy and a lower mean speed of the molecules.
✓ 1
So the molecules make fewer collisions with the wall per second, with a smaller change of momentum per collision, so the force per unit area is smaller.
✓ 1
Exam tip: When the temperature is constant, you can compare amounts of gas using the product pV. This avoids rounding errors from calculating moles at every step.
98IBP2S-013-03-14The empirical gas lawsPaper 2Medium6 marks
A student traps a fixed mass of air in a rigid flask of volume 2.0×10−4 m3. The flask is placed in a water bath, and the student records the pressure of the air at several temperatures.
θ / °C
p / kPa
0
101.0
20
108.4
40
115.8
60
123.2
80
130.6
Data from the table
(a)
Show that the rate of change of pressure with temperature is about 0.37 kPa K−1.
(2)
(b)
Determine, using the data, a value for absolute zero in °C.
(2)
(c)
Determine the amount of air, in moles, in the flask.
(2)
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Part (a)
Uses a wide interval, e.g. Δp = 130.6 − 101.0 = 29.6 kPa over Δθ = 80 K.
✓ 1
Gradient = 29.6/80 = 0.37 kPa K−1.
✓ 1
Part (b)
Extrapolates to p = 0: θ = 0 − 101.0/0.37.
✓ 1
θ = −273 °C. ECF from (a).
✓ 1
Part (c)
At constant V, the gradient Δp/ΔT = nR/V, so n = gradient × V/R (gradient = 370 Pa K−1).
✓ 1
n = 370 × 2.0×10−4/8.31 = 8.9×10−3 mol. ECF from (a).
✓ 1
Exam tip: A change of 1 °C equals a change of 1 K. A gradient found from Celsius data can therefore be used directly in kelvin-based equations, but convert kPa to Pa first.
99IBP2S-013-03-15Kinetic model & molecular speedsPaper 2Medium8 marks
A lander on Mars measures an atmospheric pressure of 610 Pa and an atmospheric density of 0.015 kg m−3. The Martian atmosphere is mostly carbon dioxide, with molar mass 0.044 kg mol−1. Treat the atmosphere as an ideal gas.
(a)
State two assumptions of the kinetic model of an ideal gas.
(2)
(b)
Calculate the root mean square speed of the gas molecules.
(2)
(c)
Determine the temperature of the Martian atmosphere at the landing site.
(2)
(d)
Calculate the mean translational kinetic energy of a molecule in this atmosphere.
(2)
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Part (a)
Accepted assumptions: molecules are point particles (their volume is negligible compared with the container volume); no intermolecular forces except during collisions; collisions are elastic; time spent in a collision is negligible compared with the time between collisions; molecules move randomly; there is a very large number of molecules.
—
First valid assumption from the list.
✓ 1
Second, different valid assumption from the list.
✓ 1
Part (b)
Uses p = ⅓ρ⟨c2⟩, so crms = √(3p/ρ) = √(3 × 610/0.015).
✓ 1
crms = 349 m s−1; 350 m s−1 is also accepted.
✓ 1
Part (c)
Uses T = pM/(ρR), or T = M⟨c2⟩/(3R).
✓ 1
T = 610 × 0.044/(0.015 × 8.31) ≈ 215 K. A value of 216 K from a rounded crms is accepted; ECF from (b).
✓ 1
Part (d)
Ek = (3/2)kT = 1.5 × 1.38×10−23 × 215.
✓ 1
Ek = 4.5×10−21 J; ECF from (c).
✓ 1
Common error: In p = ⅓ρ⟨c²⟩, ρ is the density of the gas, not the mass of one molecule. Check that the units give m² s⁻² before taking the square root.
100IBP2S-013-01-12Kinetic model & molecular speedsPaper 2Medium6 marks
A computer simulation models a pure gas as an ideal gas. It outputs the rms speed crms of the molecules at four temperatures T: T / K: 200, 300, 400, 500 crms / m s−1: 395, 483, 558, 624
(a)
Show that the data are consistent with crms2 being proportional to T.
(2)
(b)
Determine the molar mass of the gas.
(3)
(c)
Suggest the identity of the gas.
(1)
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Part (a)
Calculates crms2/T for at least three rows, e.g. 780, 778, 778, 779 m2 s−2 K−1 . States that the ratio is constant to within about 0.5%, so crms2 ∝ T .
✓ 2
Part (b)
Uses crms2 = 3RT/M . Rearranges to M = 3R ÷ (c2/T), e.g. 3 × 8.31/779, or uses one row, e.g. 3 × 8.31 × 500/6242 . M = 0.032 kg mol−1 . Allow ECF from the ratio found in (a).
✓ 3
Part (c)
Names oxygen, O2, because it is consistent with a molar mass of 32 g mol−1 . Accept any other gas or vapour with a molar mass of about 32 g mol−1, e.g. methanol vapour or silane. Allow ECF from (b).
✓ 1
Exam tip: In a 'show that' question with data, calculate the ratio for several rows and state explicitly how constant it is. Checking one row is not enough.
101IBP2S-013-01-13Moles, molecules & the Avogadro constantPaper 2Medium9 marks
A match football has an internal volume of 5.0×10−3 m3. It is inflated to an absolute pressure of 1.8×105 Pa at 20 °C. Treat the air as an ideal gas of molar mass 0.029 kg mol−1, and assume the volume of the ball stays constant.
(a)
Calculate the amount of air, in moles, inside the ball.
(2)
(b)
Calculate the rms speed of the air molecules at 20 °C.
(2)
(c)
The ball is left on a sunny pitch and the air inside warms to 40 °C. Determine the new pressure in the ball.
(2)
(d)
Explain, in terms of the kinetic model, why the pressure increases when the air warms.
(3)
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Part (a)
Use n = pV/(RT) with T = 293 K . n = (1.8×105 × 5.0×10−3)/(8.31 × 293) = 0.37 mol .
✓ 2
Part (b)
Use crms = √(3RT/M) = √(3 × 8.31 × 293/0.029) . crms = 5.0×102 m s−1 .
✓ 2
Part (c)
Use p/T = constant, with temperatures in kelvin: p2 = 1.8×105 × 313/293 . p2 = 1.9×105 Pa .
✓ 2
Part (d)
A higher temperature means a greater mean kinetic energy, so the molecules have a greater (rms) speed . Molecules collide with the wall more frequently . Each collision involves a greater change of momentum, so the force on the wall per unit area increases .
✓ 3
Common error: Gauge pressure and Celsius temperature both give wrong answers in gas-law calculations. Use absolute pressure and kelvin throughout.
102IBP2S-013-01-15Internal energy of a gasPaper 2Medium9 marks
A satellite's ion thruster uses xenon as propellant. Xenon can be treated as an ideal monatomic gas of molar mass 0.131 kg mol−1. Before launch, the rigid tank of volume 0.050 m3 contains 5.0 kg of xenon at 290 K.
(a)
Calculate the pressure of the xenon in the tank.
(3)
(b)
Calculate the internal energy of the xenon.
(2)
(c)
In orbit, before any xenon is used, the tank cools to 250 K. Determine the change in internal energy of the xenon.
(2)
(d)
Explain why the internal energy of an ideal gas contains no potential energy term.
(2)
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Notes
Part (a)
n = 5.0/0.131 = 38.2 mol . Use p = nRT/V . p = 38.2 × 8.31 × 290/0.050 = 1.8×106 Pa .
✓ 3
Part (b)
Use U = (3/2)nRT . U = 1.5 × 38.2 × 8.31 × 290 = 1.4×105 J . ECF from (a) for n.
✓ 2
Part (c)
Use ΔU = (3/2)nRΔT = 1.5 × 38.2 × 8.31 × (250 − 290) . ΔU = −1.9×104 J, i.e. a decrease of 1.9×104 J . ECF from (a)/(b).
✓ 2
Part (d)
In an ideal gas there are no intermolecular forces, except during collisions . So no work is done changing molecular separations and the intermolecular potential energy is zero (or constant); the internal energy is purely random kinetic energy .
✓ 2
Exam tip: State the sign of ΔU and say clearly whether it is an increase or a decrease. A bare magnitude may not earn the final mark.
103IBP2S-013-02-12The empirical gas lawsPaper 2Medium7 marks
An engineer tests a sealed pneumatic syringe used in a robot gripper. The syringe contains a fixed mass of air at 295 K. The plunger is moved slowly, and the volume V of the air is recorded for each pressure p.
p / kPa: 100, 125, 160, 200 V / cm3: 48.0, 38.4, 30.0, 24.0
kB = 1.38×10−23 J K−1
(a)
Show that the data are consistent with the air behaving as an ideal gas at constant temperature.
(2)
(b)
Determine the number of air molecules in the syringe.
(2)
(c)
Calculate the volume of the air when the pressure is 240 kPa at the same temperature.
(1)
(d)
Outline why the plunger must be moved slowly during the test.
(2)
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Notes
Part (a)
Calculates pV for at least three rows (or all four) in consistent units, e.g. 100×103 × 48.0×10−6 = 4.80 J
✓ 1
States that pV is the same (4.80 J) for every row, so p ∝ 1/V, as expected for an ideal gas at constant T
✓ 1
Part (b)
N = pV/(kBT) = 4.80/(1.38×10−23 × 295)
✓ 1
N = 1.18×1021 (allow error carried forward from the value of pV found in the first part)
✓ 1
Part (c)
V = 4.80/(2.40×105) = 2.00×10−5 m3 = 20.0 cm3 (allow error carried forward from the value of pV found in the first part)
✓ 1
Part (d)
Compressing the gas quickly does work on it and raises its temperature
✓ 1
Moving the plunger slowly allows energy to transfer thermally to the surroundings, so the temperature stays constant at 295 K
✓ 1
Exam tip: To show a proportionality from a table, calculate the product or ratio for several rows and state clearly that it is constant.
104IBP2S-013-02-13Kinetic model & molecular speedsPaper 2Medium8 marks
Near the surface of Mars, the atmosphere is mostly carbon dioxide. Its typical pressure is 610 Pa and its temperature is 210 K. Treat the atmosphere as an ideal gas of molar mass 44 g mol−1. R = 8.31 J mol−1 K−1; kB = 1.38×10−23 J K−1; NA = 6.02×1023 mol−1
(a)
Calculate the number of gas molecules per cubic metre near the Martian surface.
(2)
(b)
Determine the density of the Martian atmosphere near the surface.
(2)
(c)
An instrument box on a Mars rover was sealed on Earth with air at 1.01×105 Pa and 293 K. The box is rigid. Calculate the pressure of the air inside the box when it is at 210 K on Mars.
(2)
(d)
The lid of the box has an area of 0.040 m2. Determine the resultant force on the lid due to the gas pressures inside and outside the box.
(2)
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Part (a)
N/V = p/(kBT)
✓ 1
= 610/(1.38×10−23 × 210) = 2.10×1023 m−3
✓ 1
Part (b)
ρ = pM/(RT), or ρ = (N/V) × (M/NA)
✓ 1
ρ = 610 × 0.044/(8.31 × 210) = 1.54×10−2 kg m−3 (allow error carried forward from the number per cubic metre if the second method is used)
✓ 1
Part (c)
For fixed n and V, p/T is constant
✓ 1
p = 1.01×105 × 210/293 = 7.24×104 Pa
✓ 1
Part (d)
Pressure difference = 7.24×104 − 610 = 7.18×104 Pa
✓ 1
F = Δp × A = 7.18×104 × 0.040 = 2.87×103 N, directed outwards (allow error carried forward from the pressure calculated for the box at 210 K)
✓ 1
Exam tip: A resultant force due to gas pressure needs the pressure difference across the surface. Do not use the pressure on one side only.
A football is inflated indoors to an absolute pressure of 1.80×105 Pa at 20 °C. It is then taken onto a pitch where the temperature is −5 °C. The volume of the ball stays constant at 5.6×10−3 m3. Treat the air as an ideal gas. R = 8.31 J mol−1 K−1
(a)
Calculate the pressure of the air in the ball when it has cooled to −5 °C.
(2)
(b)
Explain, in terms of molecules, why the pressure falls as the ball cools.
(2)
(c)
Determine the amount of air, in mol, that must be pumped into the ball at −5 °C to restore the pressure to 1.80×105 Pa.
(2)
(d)
Each stroke of a hand pump delivers 1.2×10−4 m3 of outside air, measured at 1.01×105 Pa and −5 °C. Estimate the number of strokes needed.
(2)
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Notes
Part (a)
Uses absolute temperatures 293 K and 268 K with p/T constant
✓ 1
p = 1.80×105 × 268/293 = 1.65×105 Pa
✓ 1
Part (b)
A lower temperature means the molecules have a lower mean kinetic energy, so on average they move more slowly
✓ 1
Collisions with the wall are less frequent and each one produces a smaller change of momentum, so the force per unit area is smaller
✓ 1
Part (c)
Uses Δn = ΔpV/(RT) with T = 268 K and Δp = 1.80×105 − (answer to (a)), about 1.5×104 Pa
✓ 1
Δn = 1.54×104 × 5.6×10−3/(8.31 × 268) ≈ 0.039 mol. Accept 0.038–0.039 mol; using the rounded 1.65×105 Pa gives 0.038 mol (ECF from (a))
Number of strokes = 0.039/5.4×10−3 ≈ 7.1, so about 7 strokes. Accept 7, or 8 when justified as the whole number needed to reach the full pressure (ECF from (c))
✓ 1
Common error: Using −5 and 20 directly in p/T gives a meaningless negative pressure. The gas laws need temperatures in kelvin.
106IBP2S-013-02-16Ideal & real gasesPaper 2Medium6 marks
A semiconductor fabrication plant uses a stainless-steel vacuum chamber of volume 0.050 m3. The chamber is pumped down to an ultra-high vacuum. The residual gas is at a pressure of 1.0×10−7 Pa and a temperature of 295 K. kB = 1.38×10−23 J K−1
(a)
Calculate the number of gas molecules remaining in the chamber.
(2)
(b)
Estimate the mean separation of the gas molecules in the chamber.
(2)
(c)
Outline why the residual gas is expected to behave very closely as an ideal gas.
(2)
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Part (a)
N = pV/(kBT) = 1.0×10−7 × 0.050/(1.38×10−23 × 295)
✓ 1
N = 1.2×1012
✓ 1
Part (b)
Volume per molecule = V/N = 0.050/1.23×1012 ≈ 4.1×10−14 m3. Allow 4.2×10−14 m3 if N is rounded to 1.2×1012, and allow error carried forward from the first part.
✓ 1
Separation ≈ (4.1×10−14)1/3 ≈ 3.4×10−5 m. Accept 3×10−5 m to 3.5×10−5 m.
✓ 1
Part (c)
The molecular separation (≈10−5 m) is very much larger than the size of a molecule (≈10−10 m), so the volume of the molecules themselves is negligible.
✓ 1
The molecules spend almost all their time far apart, so intermolecular forces are negligible except during collisions.
✓ 1
Exam tip: To estimate molecular spacing, find the volume available to each molecule (V/N) and take its cube root. Quote the answer to one or two significant figures.
107B-2-08Kinetic model & molecular speedsPaper 2Hard14 marks
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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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)
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)
109B-2-36Kinetic model & molecular speedsPaper 2Hard14 marks
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.
Cross-section of the cubical box: one molecule bouncing between two opposite walls
(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)
110B-2-45The ideal gas equationPaper 2Hard15 marks
A zero-pressure scientific balloon is launched containing 2.40 × 103 m³ of helium at a pressure of 1.01 × 105 Pa and a temperature of 288 K. The envelope is open at its base, so the helium is always at the same pressure as the surrounding air, and the volume of the helium increases freely as the balloon rises until the envelope is full. The volume of the full envelope is 2.00 × 105 m³.
Assume that the helium is always at the temperature of the surrounding air and that the helium and the air behave as ideal gases. Molar mass of air = 0.029 kg mol−1; molar mass of helium = 0.0040 kg mol−1. The table gives data for the atmosphere.
altitude / km
pressure / Pa
temperature / K
0
1.01 × 105
288
10
2.65 × 104
223
20
5.53 × 103
217
30
1.20 × 103
227
(a)
Calculate the amount of helium in the balloon.
(1)
(b)
Determine whether the envelope is full when the balloon reaches an altitude of 30 km.
(2)
(c)
Show that the buoyant force on the balloon at launch is about 2.9 × 104 N. Ignore the volume of the envelope and the payload.
(2)
(d)
The total mass of the envelope and the payload is 2200 kg. Calculate the resultant force on the balloon at launch.
(2)
(e)
Deduce that, while the volume of the helium is increasing freely, the buoyant force on the balloon does not change as the balloon rises.
(4)
(f)
Explain why the balloon stops rising at some altitude after the envelope has become full.
(3)
(g)
Suggest one reason why, in practice, the helium is not always at the temperature of the surrounding air.
Must see full substitution OR answer to at least 3 s.f.
Part (d)
mass of helium = 1.01 × 105 × 0.0040 = 405 kg, so weight = (2200 + 405) × 9.81 = 2.56 × 104 N
✓ 1
resultant force = 2.88 × 104 − 2.56 × 104 = 3.3 × 103 N upwards
✓ 1
Award [1] max for 7.2 × 103 N (mass of helium omitted). Allow ECF from (c).
Part (e)
density of the surrounding air ρ = PMair/(RT)
✓ 1
volume of helium V = nHeRT/P, with the same P and T as the air
✓ 1
so ρV = nHeMair, which does not depend on P or T
✓ 1
so Fb = ρVg = nHeMairg is constant «assuming g is constant»
✓ 1
Award [2] max for a numerical check at one other altitude only.
Part (f)
once the envelope is full, the volume of the balloon can no longer increase
✓ 1
the density of the air decreases as the altitude increases
✓ 1
so the buoyant force decreases until it equals the weight and the resultant force is zero
✓ 1
OWTTE
Part (g)
the envelope and gas absorb solar radiation and warm up / at night the envelope and gas radiate energy to space and cool below the temperature of the air
✓ 1
Accept any one.
Answers: (a) 1.01 × 105 mol · (b) 1.59 × 105 m³ — not yet full · (c) 2.88 × 104 N · (d) 3.3 × 103 N (the remaining parts are explanations — see the table above)
A hydrogen fuel-cell car stores hydrogen gas (molar mass 2.0 × 10−3 kg mol−1) in a tank of internal volume 0.125 m3. When the tank is full the hydrogen is at a pressure of 7.0 × 107 Pa and a temperature of 15 °C. Under these conditions the measured density of hydrogen is 40 kg m−3.
(a)
(i)
Show that the ideal gas model predicts a density of about 58 kg m−3 for the hydrogen in the full tank.
(1)
(ii)
Calculate the percentage by which the ideal gas model overestimates the mass of hydrogen in the tank.
(1)
(iii)
Explain, with reference to the molecules, why the ideal gas model is not a good approximation for the hydrogen in the full tank.
(2)
(b)
(i)
Calculate the mass of hydrogen in the full tank.
(1)
(ii)
Calculate the number of hydrogen molecules in the full tank.
(1)
(c)
The energy released per kilogram of hydrogen used in the fuel cell is 1.2 × 108 J kg−1, and 50 % of this energy is used to overcome the resistive forces on the car. On a motorway the average resistive force is 520 N. Determine whether the car can travel 500 km on one full tank.
(3)
(d)
During fast refuelling the hydrogen warms. Refuelling stops when the pressure reaches 7.0 × 107 Pa, at which time the hydrogen is at 85 °C.
(i)
Using the ideal gas model, calculate the pressure in the tank when the hydrogen has cooled to 15 °C.
(2)
(ii)
Suggest why the answer to (d)(i) is only an estimate.
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
(58.5 − 40)/40 × 100 = 46 %
✓ 1
Accept 45–47 %.
Part (a)(iii)
at this very high density the molecules are very close together, so the volume of the molecules is not negligible compared with the volume of the gas
✓ 1
and «repulsive» forces act between the molecules when they are this close, so the gas takes up more space than the model predicts «lower density at this pressure»
this is more than 500 km, so the car can travel 500 km
✓ 1
MP3 needs a comparison and a consistent conclusion. Award [2] max if the ideal-gas mass is used.
Part (d)(i)
P/T constant: P = 7.0 × 107 × 288/358
✓ 1
Kelvin temperatures needed.
= 5.6 × 107 Pa
✓ 1
Award [2] for CNA.
Part (d)(ii)
hydrogen is not an ideal gas at these pressures «see (a)», so P is not exactly proportional to T
✓ 1
OWTTE. Accept: the tank volume may change slightly with pressure/temperature.
Answers: (a)(i) 58.5 kg m−3 · (a)(ii) 46 % · (b)(i) 5.0 kg · (b)(ii) 1.5 × 1027 · (c) 580 km · (d)(i) 5.6 × 107 Pa (the remaining parts are explanations — see the table above)
In a simple model, the core of the Sun is made of fully ionised hydrogen: equal numbers of protons and electrons moving freely. The density of the core is 1.5 × 105 kg m−3 and its temperature is 1.5 × 107 K. The mass of an electron is negligible compared with the mass of a proton, 1.67 × 10−27 kg. In the model, the protons and the electrons together behave as an ideal gas of separate particles.
(a)
(i)
Show that the number of protons per unit volume in the core is about 9 × 1031 m−3.
(1)
(ii)
Determine the pressure in the core.
(2)
(b)
Calculate the root mean square speed of the protons in the core.
(2)
(c)
(i)
Show that the mean separation of the particles in the core is about 2 × 10−11 m.
(1)
(ii)
The density of the core is about 150 times that of water. Suggest why the ideal gas model can still be used for the core. The diameter of a proton is about 2 × 10−15 m.
(2)
(d)
Using P = ⅓ρv̄2, explain why the electrons contribute about half of the pressure in the core although they make up a negligible fraction of its density.
(2)
(e)
Outline why the core of the Sun does not collapse under its own gravitational attraction.
volume per particle = 1/1.8 × 1032 m3, so separation ≈ (1/1.8 × 1032)1/3 = 1.77 × 10−11 m
✓ 1
Must see full substitution OR answer to at least 3 s.f. Accept 2.2 × 10−11 m from the protons alone.
Part (c)(ii)
the particles are protons and electrons, about 104 times smaller than their separation, so the volume of the particles is negligible «the high density is due to the mass, not the size, of the particles»
✓ 1
the temperature is so high that the particles move very fast and the forces between them have little effect except during close encounters «collisions»
✓ 1
OWTTE.
Part (d)
electrons and protons are at the same temperature, so they have the same mean kinetic energy ½ mv̄2 = ³⁄₂ kBT: the electrons have a far larger mean square speed
✓ 1
in P = ⅓ρv̄2 the pressure of each kind of particle depends on ρv̄2 = (N/V) × mv̄2, which is the same for electrons and protons because their numbers are equal
✓ 1
OWTTE.
Part (e)
the pressure in the core «gas pressure and radiation pressure» produces an outward force on the layers above
✓ 1
this balances the inward gravitational force, so the core is in equilibrium
✓ 1
Answers: (a)(i) 9.0 × 1031 m−3 · (a)(ii) 3.7 × 1016 Pa · (b) 6.1 × 105 m s−1 · (c)(i) 1.8 × 10−11 m (the remaining parts are explanations — see the table above)
113IBP2S-013-00-15The empirical gas lawsPaper 2Hard8 marks
In a diesel engine cylinder, 500 cm3 of air at 1.0 × 105 Pa and 300 K is compressed rapidly to 25 cm3. The pressure rises to 5.0 × 106 Pa. Assume that the air behaves as an ideal gas and that no air leaks out.
(a)
Calculate the temperature of the air after compression.
(2)
(b)
Explain why Boyle's law cannot be used to predict the final pressure.
(2)
(c)
Calculate the number of air molecules in the cylinder.
(2)
(d)
Determine the ratio of the root mean square speed of the molecules after compression to that before compression.
Boyle's law requires a constant temperature . The compression is rapid, so work done on the gas increases its internal energy and there is no time for energy to transfer to the surroundings, so the temperature rises .
✓ 2
Part (c)
N = pV/(kT) = (1.0 × 105 × 5.0 × 10−4)/(1.38 × 10−23 × 300) . N = 1.2 × 1022 . Using the final values (5.0 × 106 Pa, 2.5 × 10−5 m3 and the temperature from the first part) is equally acceptable.
✓ 2
Part (d)
vrms ∝ √T because ½m(vrms)2 = (3/2)kT . Ratio = √(750/300) = 1.58 . Allow follow-through from the temperature calculated in the first part.
✓ 2
Exam tip: The combined law pV/T = constant holds for any change of a fixed amount of ideal gas, even a rapid one, if the final p, V and T are used.
114IBP2S-013-00-16The ideal gas equationPaper 2Hard7 marks
A diver at a depth of 25 m in a freshwater lake releases an air bubble of radius 4.0 mm. The water at this depth is at 8 °C. The bubble rises to the surface, where the water is at 20 °C and the atmospheric pressure is 1.01 × 105 Pa. The density of the water is 1000 kg m−3. Assume that the air in the bubble is always at the temperature of the surrounding water and that no air dissolves.
Diagram not to scale
(a)
Calculate the pressure of the air in the bubble at a depth of 25 m.
(2)
(b)
Determine the radius of the bubble when it reaches the surface.
(3)
(c)
A scuba diver breathes compressed air at this depth. Explain why the diver must not hold their breath while ascending to the surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
p = patm + ρgh = 1.01 × 105 + 1000 × 9.8 × 25 . p = 3.46 × 105 Pa .
✓ 2
Part (b)
Use pV/T = constant, so V2/V1 = (p1/p2)(T2/T1) = (3.46/1.01)(293/281), with temperatures in kelvin . V2/V1 = 3.57 . r2 = 4.0 × 3.571/3 = 6.1 mm . Allow error carried forward from the pressure found in the first part.
✓ 3
Part (c)
As the diver rises, the external pressure falls, so the air trapped in the lungs expands. The air stays at constant body temperature, so Boyle's law applies . Its volume would increase by about p1/p2 = 3.46/1.01 ≈ 3.4 times, which over-expands the lungs and can rupture or damage them . Accept any reasoning that the trapped air would expand several times over and damage the lungs. The factor is 3.4 rather than the 3.57 found for the bubble because body temperature does not change.
✓ 2
Common error: Volume scales with the cube of the radius. Take the cube root of the volume ratio, and do not multiply the radius by the volume ratio. Always convert temperatures to kelvin before using the gas laws.
115IBP2S-013-03-16The empirical gas lawsPaper 2Hard9 marks
A vertical cylinder used in an engineering test rig is sealed at its upper end by a frictionless piston. The piston has mass 12 kg and cross-sectional area 4.0×10−3 m2. The air above the piston is at atmospheric pressure, 1.01×105 Pa. The trapped gas forms a column of height 0.250 m at 290 K, with the piston in equilibrium. The gas is ideal.
Gas at 290 K with the piston in equilibrium
(a)
Show that the pressure of the trapped gas is about 1.3×105 Pa.
(2)
(b)
The gas is slowly heated to 360 K, and the piston rises freely. Calculate the distance moved by the piston.
(2)
(c)
The piston is then clamped in its new position, and the gas cools back to 290 K. Calculate the new pressure of the gas.
(2)
(d)
Determine the magnitude and direction of the force the clamp must exert on the piston to hold it in place after cooling.
The pressure is constant, so the height is proportional to T: h2 = 0.250 × 360/290 = 0.310 m.
✓ 1
Distance moved = 0.310 − 0.250 = 0.060 m.
✓ 1
Part (c)
The volume is constant, so p2 = 1.30×105 × 290/360.
✓ 1
p2 = 1.05×105 Pa. ECF from (a).
✓ 1
Part (d)
Recognises that the clamp force balances the fall in the gas force: F = (p − p2)A, or F = p0A + mg − p2A.
✓ 1
F = (1.30×105 − 1.05×105) × 4.0×10−3 ≈ 1.0×102 N. ECF from (c).
✓ 1
The force is upward, because the reduced gas pressure can no longer support the piston.
✓ 1
Exam tip: Draw a free-body diagram of the piston in your head first. Gas pressure pushes up, while atmospheric pressure and the piston's weight push down. Any imbalance must be supplied by the clamp.
116IBP2S-013-01-14Ideal & real gasesPaper 2Hard8 marks
In the exosphere, the outermost layer of Earth's atmosphere, the gas temperature can reach 1500 K. The escape speed from Earth is 11.2 km s−1. The mass of a hydrogen atom is 1.67×10−27 kg. The mass of an oxygen molecule is 5.3×10−26 kg.
(a)
Show that the mean translational kinetic energy of a hydrogen atom at 1500 K is about 3×10−20 J.
(2)
(b)
Calculate the rms speed of hydrogen atoms at this temperature.
(2)
(c)
The rms speed of hydrogen atoms is less than the escape speed. Outline why hydrogen is nevertheless gradually lost from the atmosphere.
(2)
(d)
Determine the temperature at which the rms speed of oxygen molecules would equal the escape speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses E = (3/2)kT = 1.5 × 1.38×10−23 × 1500 . E = 3.1×10−20 J, i.e. about 3×10−20 J .
✓ 2
Part (b)
Uses crms = √(2E/m) or √(3kT/m), e.g. √(2 × 3.1×10−20/1.67×10−27) . crms = 6.1×103 m s−1 . Accept 6.0×103 m s−1 if the rounded value 3×10−20 J is used; allow ECF from (a).
✓ 2
Part (c)
The atoms have a distribution (range) of speeds, so some move much faster than the rms speed . The small fraction with speeds above 11.2 km s−1 (moving outwards) escapes; the speed distribution is continually re-established (by collisions in the denser gas lower down / energy input to the gas), so a small fraction keeps escaping over time .
✓ 2
Part (d)
Rearranges (1/2)mc2 = (3/2)kT to give T = mc2/(3k) = 5.3×10−26 × (1.12×104)2/(3 × 1.38×10−23) . T = 1.6×105 K .
✓ 2
Exam tip: For 'outline' questions about escape or evaporation, mention the speed distribution. The rms value alone never explains why some particles escape.
117IBP2S-013-01-16Internal energy of a gasPaper 2Hard8 marks
In a cryogenics laboratory, two rigid, thermally insulated tanks are joined by a closed valve. Both contain helium, which can be treated as an ideal monatomic gas. Tank A: volume 0.020 m3, 2.0 mol at 300 K. Tank B: volume 0.030 m3, 3.0 mol at 400 K. The valve is opened and the gas comes to equilibrium.
The valve is closed initially; diagram not to scale
(a)
Outline why the total internal energy of the helium is the same before and after the valve is opened.
(2)
(b)
Show that the final temperature of the helium is 360 K.
(2)
(c)
Calculate the ratio \(\frac{\text{final rms speed}}{\text{initial rms speed in tank A}}\) for the helium atoms.
(2)
(d)
Determine the final pressure of the helium.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The tanks are insulated, so no thermal energy is transferred to or from the surroundings . The tanks are rigid, so the gas does no work on the surroundings (no external volume change); therefore U is conserved .
crms ∝ √T, so the ratio = √(360/300) . Ratio = 1.10 . ECF from (b).
✓ 2
Part (d)
Use p = ntotalRTf/Vtotal = 5.0 × 8.31 × 360/0.050 . p = 3.0×105 Pa . ECF from (b).
✓ 2
Common error: Averaging the two temperatures (giving 350 K) ignores the different amounts of gas. Weight each temperature by its number of moles, because internal energy is what is conserved.
118IBP2S-013-02-15The ideal gas equationPaper 2Hard7 marks
A hot-air balloon has an envelope of fixed volume 2800 m3. The envelope is open at the bottom, so the air inside is always at atmospheric pressure, 1.01×105 Pa. The outside air is at 285 K. The total mass of the envelope, basket, burner and passengers is 650 kg; ignore the volume they occupy. Treat air as an ideal gas of molar mass 29 g mol−1. R = 8.31 J mol−1 K−1
(a)
Explain why the mass of air inside the envelope decreases as the burner heats it.
(2)
(b)
Show that the mass of outside air that would fill the envelope is about 3.5×103 kg.
(2)
(c)
Determine the minimum temperature of the air inside the envelope for the balloon to lift off.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The pressure (atmospheric) and volume of the air inside stay constant because the envelope is open
✓ 1
From pV = nRT, n ∝ 1/T (or density ∝ 1/T), so as T rises some air (molecules) leaves through the opening
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