S1.5 Ideal gases

1. The particles lose a small amount of kinetic energy in every collision. — A student who thinks moving particles must gradually run down picks this. The model assumes that all collisions are elastic: kinetic energy passes between particles in a collision but the total is not lost, so a gas at constant temperature does not slow down.
2. The spaces between the particles are filled with air that separates them. — A student who cannot accept empty space between particles picks this. In the model the space between the particles is empty. Air is itself a gas made of particles, so it cannot fill the gaps between gas particles.
3. The particles repel one another, which keeps them spread through the container. — A student who pictures gas particles pushing each other apart picks this. The model assumes there are no intermolecular forces at all, attractive or repulsive; the gas spreads through the container because its particles move randomly.
4. The particles have a negligible volume compared with the volume of the gas. — This is one of the model's assumptions: the particles are treated as having no volume of their own compared with the space the gas occupies. The other key assumptions are that the particles are moving, exert no intermolecular forces, and collide elastically.

Syllabus statement S1.5.1 · Read this in Learn

1. At 150 K, as its particles have stopped moving and so no longer collide — A student who believes particles stop moving when cold picks this. The particles of a gas are always moving: at 150 K nitrogen is still a gas and exerts a pressure because its particles collide with the walls. They move more slowly, which is why the attractions between them matter more.
2. At 150 K, as its slower particles are affected more by the attractions between them — At low temperature the particles have less kinetic energy, so the intermolecular attractions, which the ideal model ignores, have a significant effect on their motion. Real gases deviate most at low temperature and high pressure.
3. At 600 K, as its faster particles collide more violently with each other — A student who thinks fast, energetic particles break the model picks this. A real gas behaves more ideally at high temperature, because the particles' kinetic energy is then large compared with the attractions between them. Deviation is greatest at low temperature.
4. At 600 K, as its particles expand when heated and take up more space — A student who thinks particles swell when heated picks this. Particles do not change size with temperature. At 600 K and the same pressure the gas occupies a larger volume and its particles move faster, so it behaves more ideally, not less.

Syllabus statement S1.5.2 · Read this in Learn

1. A straight line meeting the axis at −273 °C, where volume would be zero — Volume is proportional to absolute temperature. On a Celsius axis the line does not pass through the origin; extended backwards, it reaches zero volume at −273 °C (0 K), absolute zero. (A real gas would liquefy before this; the line is an extrapolation.)
2. A straight line meeting the axis at −273 °C, where the particles shrink to nothing — A student who thinks particles shrink as they cool picks this. The intercept is right but the reason is not: particles do not change size. The volume falls because slower particles hit the walls less hard and less often, so at constant pressure the gas occupies less space.
3. A straight line through the origin, since volume is proportional to temperature in °C — A student who treats 0 °C as zero temperature picks this. 0 °C is only the melting point of ice, and a gas at 0 °C still has a large volume. Volume is proportional to temperature in kelvin, so on a Celsius axis the line meets the axis at −273 °C.
4. A horizontal line, as the gas fills the same container at every temperature — A student who thinks a gas always has the volume of its container picks this. A gas does fill its container, but at constant pressure the container (for example a gas syringe) expands as the gas is heated, so the volume increases steadily with temperature.

Syllabus statement S1.5.3 · Read this in Learn

1. 9.41 × 10⁻⁵ m³ — A student who leaves the temperature in °C gets 0.0472 × 8.31 × 30 ÷ (1.25 × 10⁵) = 9.41 × 10⁻⁵ m³. PV = nRT needs absolute temperature: 30 °C = 303 K.
2. 9.51 × 10⁻¹ m³ — A student who substitutes the pressure in kPa gets 0.0472 × 8.31 × 303 ÷ 125 = 0.951 m³, 1000 times too large. With R in J K⁻¹ mol⁻¹ the pressure must be in Pa: 125 kPa = 1.25 × 10⁵ Pa.
3. 1.07 × 10⁻³ m³ — A student who uses the molar volume at STP gets 0.0472 mol × 22.7 dm³ mol⁻¹ = 1.07 dm³ = 1.07 × 10⁻³ m³. 22.7 dm³ mol⁻¹ applies only at 273 K and 100 kPa; at 303 K and 125 kPa use PV = nRT.
4. 9.51 × 10⁻⁴ m³ — T = 30 + 273 = 303 K and p = 125 kPa = 1.25 × 10⁵ Pa. V = nRT/p = 0.0472 × 8.31 × 303 ÷ (1.25 × 10⁵) = 9.51 × 10⁻⁴ m³.

Working T = 30 + 273 = 303 K; P = 125 kPa = 1.25 × 10⁵ Pa. V = nRT/P = (0.0472 mol × 8.31 J K⁻¹ mol⁻¹ × 303 K) ÷ (1.25 × 10⁵ Pa) = 118.8 J ÷ (1.25 × 10⁵ Pa) = 9.51 × 10⁻⁴ m³.

Syllabus statement S1.5.4 · Read this in Learn

1. Moving particles collide with the walls and exert a force on them. — In the model the particles move randomly and collide elastically with the walls. Each collision exerts a force on the wall; the pressure is the total force per unit area from these collisions, so it acts equally on every wall.
2. The particles repel one another, so they push outwards on the walls. — A student who pictures gas particles pushing each other apart picks this. The ideal gas model has no intermolecular forces, attractive or repulsive; the pressure comes from particles colliding with the walls as they move.
3. The weight of the particles presses on the container, mainly on its base. — A student who links pressure to weight, as for a solid resting on a surface, picks this. Gas pressure acts equally in all directions, including upwards on a lid; it comes from particle collisions with the walls, not from the weight of the gas.
4. The particles expand to fill the container and press against its walls. — A student who gives the particles the properties of the bulk gas picks this. Particles do not change size; a gas fills its container because its particles move freely through the empty space, and the pressure comes from their collisions with the walls.

Syllabus statement S1.5.1 · Read this in Learn

1. Helium, He, as its light atoms move the fastest and collide most violently — A student who thinks high particle speed makes a gas less ideal picks this. Helium is one of the most nearly ideal gases: its atoms are small and the London forces between them are extremely weak. Fast particles are less affected by attractions, not more.
2. Neon, Ne, as its atoms are the heaviest and so they occupy the most space — A student who equates a heavier particle with a larger one, and thinks size decides the deviation, picks this. Neon atoms (M = 20.18 g mol⁻¹) are heavier than ammonia molecules (17.04 g mol⁻¹), but the London forces between them are weak, so neon behaves almost ideally. The strength of the attractions between particles decides the deviation, not their mass.
3. Ammonia, NH₃, as hydrogen bonding attracts its molecules to one another — Ammonia molecules form hydrogen bonds with one another, so the attractions between them are much stronger than between the particles of the other three gases. The ideal gas model assumes no intermolecular forces, so the gas with the strongest attractions deviates most.
4. Methane, CH₄, as each molecule is held together by four C–H bonds — A student who counts the covalent bonds inside a molecule as the forces between molecules picks this. The four C–H bonds hold each methane molecule together but do not attract one methane molecule to another. Methane molecules attract one another only by London forces, which are much weaker than the hydrogen bonding between ammonia molecules.

Syllabus statement S1.5.2 · Read this in Learn

1. 22.7 dm³, as heavy xenon atoms move at the same speed as the helium atoms — A student who thinks all particles move at the same speed at a given temperature picks this. At the same temperature the particles have the same average kinetic energy, so the heavier xenon atoms move much more slowly than helium atoms. The xenon sample occupies 22.7 dm³ because volume depends only on the amount of gas, not because the speeds are equal.
2. 22.7 dm³, as volume depends only on the amount of gas at a given T and p — In an ideal gas the particles have negligible volume and exert no intermolecular forces, so at a given temperature and pressure the volume depends only on the amount of gas, not on the identity or mass of the particles. 1.00 mol of xenon at STP occupies the same 22.7 dm³ as 1.00 mol of helium.
3. More than 22.7 dm³, as its much larger atoms take up more of the space — A student who thinks bigger or heavier particles need more room picks this. The particles' own volume is negligible compared with the volume of the gas, so the size of the xenon atoms does not affect the molar volume: 1.00 mol of xenon occupies 22.7 dm³, the same as 1.00 mol of helium.
4. Less than 22.7 dm³, as a denser gas has its particles closer together — A student who explains density by particle spacing picks this: xenon is far denser than helium, so its atoms seem to be packed more closely. At the same temperature and pressure equal amounts of ideal gases occupy equal volumes, so the spacing is the same; xenon is denser because each of its atoms has a much larger mass.

Working Both samples are at the same temperature (273 K) and pressure (100 kPa) and contain the same amount of gas (1.00 mol). For an ideal gas V = nRT/p, which does not contain the molar mass, so V(Xe) = V(He) = 1.00 mol × 22.7 dm³ mol⁻¹ = 22.7 dm³.

Syllabus statement S1.5.3 · Read this in Learn

1. 1.59 × 10⁻³ mol — A student who leaves the temperature in °C gets 0.264 ÷ (8.31 × 20) = 1.59 × 10⁻³ mol. PV = nRT needs absolute temperature: 20 °C = 293 K.
2. 1.08 × 10⁻⁴ mol — T = 20 + 273 = 293 K and V = 2.20 cm³ = 2.20 × 10⁻⁶ m³. n = PV/RT = (1.20 × 10⁵ × 2.20 × 10⁻⁶) ÷ (8.31 × 293) = 0.264 ÷ 2435 = 1.08 × 10⁻⁴ mol.
3. 1.08 × 10⁻¹ mol — A student who converts cm³ to m³ by dividing by 1000, as for dm³, uses V = 2.20 × 10⁻³ m³ and gets 0.108 mol, 1000 times too large. 1 cm³ = 10⁻⁶ m³, so 2.20 cm³ = 2.20 × 10⁻⁶ m³.
4. 9.69 × 10⁻⁵ mol — A student who uses the molar volume at STP gets 2.20 × 10⁻³ dm³ ÷ 22.7 dm³ mol⁻¹ = 9.69 × 10⁻⁵ mol. 22.7 dm³ mol⁻¹ applies only at 273 K and 100 kPa; at 293 K and 1.20 × 10⁵ Pa use PV = nRT.

Working T = 20 + 273 = 293 K; V = 2.20 cm³ = 2.20 × 10⁻⁶ m³. n = PV/RT = (1.20 × 10⁵ Pa × 2.20 × 10⁻⁶ m³) ÷ (8.31 J K⁻¹ mol⁻¹ × 293 K) = 0.264 J ÷ 2435 J mol⁻¹ = 1.08 × 10⁻⁴ mol.

Syllabus statement S1.5.4 · Read this in Learn

1. 1.02 × 10² dm³ — A student who uses 16.00 g mol⁻¹, the molar mass of oxygen atoms, gets n = 72.0 ÷ 16.00 = 4.50 mol and V = 4.50 × 22.7 = 102 dm³. Oxygen gas consists of O₂ molecules, so M = 32.00 g mol⁻¹.
2. 1.63 × 10³ dm³ — A student who multiplies the mass by the molar volume gets 72.0 × 22.7 = 1634 dm³. The molar volume is the volume per mole, so the mass must first be converted to amount: n = 72.0 ÷ 32.00 = 2.25 mol.
3. 1.01 × 10¹ dm³ — A student who divides the molar volume by the amount gets 22.7 ÷ 2.25 = 10.1 dm³. The volume is the amount multiplied by the volume per mole: V = n × V_m = 2.25 × 22.7 = 51.1 dm³.
4. 5.11 × 10¹ dm³ — M(O₂) = 2 × 16.00 = 32.00 g mol⁻¹, so n = 72.0 ÷ 32.00 = 2.25 mol and V = n × V_m = 2.25 × 22.7 = 51.1 dm³ (5.11 × 10¹ dm³).

Working M(O₂) = 2 × 16.00 = 32.00 g mol⁻¹. n(O₂) = m/M = 72.0 g ÷ 32.00 g mol⁻¹ = 2.25 mol. V = n × V_m = 2.25 mol × 22.7 dm³ mol⁻¹ = 51.1 dm³ = 5.11 × 10¹ dm³.

Syllabus statement S1.5.3 · Read this in Learn

1. A straight line with a negative gradient, as p falls as 1/V rises — A student who thinks an inverse relationship is always a falling straight line picks this. p is inversely proportional to V, so p is directly proportional to 1/V: as 1/V rises p rises, giving a straight line through the origin with a positive gradient.
2. A curve that falls and levels off, as p and V are inversely related — A student who expects every inverse relationship to look like a falling curve picks this without reading the x-axis. The falling curve is the graph of p against V. Plotting against 1/V turns the inverse relationship into a direct one: a straight line through the origin.
3. A straight line through the origin, as p doubles whenever V is halved — At constant temperature p is inversely proportional to V, so p is directly proportional to 1/V: halving V doubles 1/V and doubles p. A graph of p against 1/V is therefore a straight line through the origin.
4. A horizontal line, as the particles' speed and hence p do not change — A student who thinks pressure depends only on how fast the particles move picks this. At constant temperature the average speed is unchanged, but in a smaller volume the particles hit the walls more often, so the pressure rises as V falls.

Syllabus statement S1.5.3 · Read this in Learn

1. It rises by about 7%, since its temperature in K does — At constant pressure V is proportional to absolute temperature. 20 °C = 293 K and 40 °C = 313 K, so V₂/V₁ = 313/293 = 1.07: the absolute temperature and the volume both rise by about 7%.
2. It doubles, since 20 °C to 40 °C doubles its temperature — A student who treats Celsius temperature as proportional to volume picks this. 0 °C is not absolute zero, so doubling the Celsius temperature does not double the absolute temperature: 293 K to 313 K is a rise of only about 7%.
3. It increases, since each gas particle expands when heated — A student who thinks particles swell when heated picks this. The volume does increase, but not because the particles grow: they move faster, so to keep the pressure constant the gas must occupy more space. The increase is a factor of 313/293.
4. It stays the same, since the gas still fills the same syringe — A student who thinks a gas always has the volume of its container picks this. The plunger of a gas syringe moves, so at constant pressure the gas expands as it is heated: V₂/V₁ = 313/293.

Syllabus statement S1.5.3 · Read this in Learn

1. 1.29 × 10⁶ Pa — A student who uses the Celsius temperatures gets 1.00 × 10⁵ × 4.00 × 87/27 = 1.29 × 10⁶ Pa. The combined gas law needs absolute temperatures: 300 K and 360 K.
2. 3.33 × 10⁵ Pa — A student who inverts the temperature ratio gets 1.00 × 10⁵ × 4.00 × 300/360 = 3.33 × 10⁵ Pa, as if heating lowered the pressure. From P₁V₁/T₁ = P₂V₂/T₂, P₂ = P₁ × (V₁/V₂) × (T₂/T₁).
3. 4.80 × 10⁵ Pa — T₁ = 300 K and T₂ = 360 K. P₂ = P₁ × (V₁/V₂) × (T₂/T₁) = 1.00 × 10⁵ × 4.00 × 1.20 = 4.80 × 10⁵ Pa. The compression and the heating both raise the pressure.
4. 4.00 × 10⁵ Pa — A student who applies only the pressure–volume relationship gets 1.00 × 10⁵ × 4.00 = 4.00 × 10⁵ Pa, ignoring the rise in temperature. Both variables changed, so the factor T₂/T₁ = 360/300 must be included.

Working T₁ = 27 + 273 = 300 K; T₂ = 87 + 273 = 360 K. From P₁V₁/T₁ = P₂V₂/T₂: P₂ = P₁ × (V₁/V₂) × (T₂/T₁) = 1.00 × 10⁵ Pa × (2.00 × 10⁻³ m³ ÷ 5.00 × 10⁻⁴ m³) × (360 K ÷ 300 K) = 1.00 × 10⁵ Pa × 4.00 × 1.20 = 4.80 × 10⁵ Pa.

Syllabus statement S1.5.4 · Read this in Learn

1. At 2.0 × 10⁷ Pa, as its particles collide more often and lose kinetic energy each time — A student who thinks every collision loses kinetic energy picks this. Collisions are elastic in the model and, for a real gas at 300 K, the average kinetic energy of the particles is set by the temperature, which is the same at both pressures. The deviation at high pressure comes from the particles' own volume and the attractions between them.
2. At 2.0 × 10⁷ Pa, as its particles are close enough for their volume and attractions to matter — At high pressure the particles are close together, so their own volume is no longer negligible compared with the volume of the gas, and the attractions between them become significant. The ideal gas model ignores both, so methane deviates more at 2.0 × 10⁷ Pa. Real gases deviate most at low temperature and high pressure.
3. At 1.0 × 10⁵ Pa, as its particles expand to fill the much larger volume of the gas — A student who thinks particles swell to fill their container picks this. Particles do not change size. At low pressure the particles are far apart, so their volume and the attractions between them are negligible and the gas behaves almost ideally.
4. Neither, as at 300 K its particles move at the same average speed at both pressures — A student who thinks deviation depends only on how fast the particles move (and so only on temperature) picks this. The average speed is indeed the same at both pressures, but at 2.0 × 10⁷ Pa the particles are much closer together, so their own volume and the attractions between them become significant.

Syllabus statement S1.5.2 · Read this in Learn

1. Attractions between the molecules pull them closer together, so the gas occupies a smaller volume than an ideal gas would. — The ideal model assumes no intermolecular forces. In a real gas the attractions between molecules reduce the volume (or the pressure on the walls) below the ideal value, so PV/nRT < 1 at A. At higher pressure (B) the molecules’ own volume dominates and the curve rises above the line.
2. The covalent bonds within each molecule are strong, and these bonds hold the neighbouring molecules close together. — Covalent bonds act within a molecule and have no effect on how molecules attract one another. The dip at A is caused by the intermolecular attractions (van der Waals forces) between molecules, which the ideal model ignores.
3. The molecules have a volume of their own, so the gas is squeezed into less space than the ideal model predicts. — The molecules’ own volume acts the other way: it makes the gas harder to compress, so PV/nRT rises above 1 (the region around B). The dip at A comes from the attractions between molecules, which draw them closer together.
4. The molecules lose kinetic energy in their frequent collisions, so they strike the walls with less force than ideal particles. — Collisions do not drain kinetic energy from a gas; at constant temperature the average kinetic energy is unchanged. The lower value of PV/nRT at A is due to attractions between the molecules, not to slower molecules.

Syllabus statement S1.5.2 · Read this in Learn

1. 272 °C — 227 × 48.0/40.0 treats volume as proportional to the Celsius temperature, which would put zero volume at 0 °C. The graph shows zero volume at −273 °C, so work in kelvin: 500 K × 48.0/40.0 = 600 K = 327 °C.
2. 144 °C — 500 × 40.0/48.0 = 417 K puts the volume ratio the wrong way round, giving a lower temperature for a larger volume. The graph shows volume rising with temperature: T = 500 K × 48.0/40.0 = 600 K = 327 °C.
3. 327 °C — The line reaches zero volume at −273 °C, so V is proportional to the kelvin temperature. V = 40.0 cm³ at 227 °C (500 K), so V = 48.0 cm³ at 500 × 48.0/40.0 = 600 K, which is 600 − 273 = 327 °C.
4. 600 °C — 600 is the temperature in kelvin. The question asks in °C, so subtract 273: 600 K = 327 °C.

Working From the graph, V = 40.0 cm³ at 227 °C and the line extrapolates to V = 0 at −273 °C, so V ∝ T/K. T₁ = 227 + 273 = 500 K. T₂ = T₁ × V₂/V₁ = 500 K × 48.0/40.0 = 600 K; t₂ = 600 − 273 = 327 °C.

Syllabus statement S1.5.3 · Read this in Learn