A2 · Paper 4 practice · Ideal gases

Counted by the mole, moved by collisions.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Ideal gases: the mole, the equation of state, and the kinetic theory. Several questions deliberately carry one lesson’s idea into the next within the topic. Each is linked to the lessons it draws on; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Take R = 8.31 J mol⁻¹ K⁻¹, N₊ = 6.02 × 10²³ mol⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.
Keep these straight

Counting, and connecting.

01
Paper 4
[12 marks]

A cylinder contains 0.40 mol of an ideal gas, as shown in Fig. 1.1, at a temperature of 27 °C and a pressure of 1.5 × 10⁵ Pa.

ideal gas, 0.40 mol piston
Fig. 1.1

(a) State what is meant by one mole of a substance. [1]

(b) Calculate the number of molecules in the cylinder. [1]

(c) Calculate the volume occupied by the gas. [3]

(d) The gas is heated at constant volume until its temperature is 177 °C. Calculate the new pressure. [3]

(e) Show that the Boltzmann constant k = R / N₊ is about 1.38 × 10⁻²³ J K⁻¹, and state the equation of state written in terms of N and k. [2]

(f) State two assumptions about the molecules of an ideal gas. [2]

  • (a)One mole is the amount of substance that contains N₊ (6.02 × 10²³) particles ✓
  • (b)N = nN₊ = 0.40 × 6.02 × 10²³ = 2.4 × 10²³ molecules ✓
  • (c)pV = nRT, T = 300 K ✓; V = nRT / p = (0.40 × 8.31 × 300) / (1.5 × 10⁵) ✓ = 6.6 × 10⁻³ m³ ✓
  • (d)At constant volume p / T is constant ✓; T₂ = 450 K, so p₂ = p₁ × T₂ / T₁ = 1.5 × 10⁵ × 450 / 300 ✓ = 2.25 × 10⁵ Pa ✓
  • (e)k = R / N₊ = 8.31 / (6.02 × 10²³) = 1.38 × 10⁻²³ J K⁻¹ ✓; the equation of state is pV = NkT ✓
  • (f)Any two: the molecules have negligible volume compared with the container ✓; there are no intermolecular forces except during collisions; collisions are perfectly elastic; the motion is random ✓
02
Paper 4
[13 marks]

Fig. 2.1 represents the molecules of an ideal gas in random motion inside a container.

molecules in random motion
Fig. 2.1

(a) State three assumptions made in the kinetic theory of an ideal gas. [3]

(b) Kinetic theory gives pV = ⅓Nm⟨c²⟩. Use this together with pV = NkT to show that the mean translational kinetic energy of a molecule is ₃⁄₂ kT. [3]

(c) Calculate the mean translational kinetic energy of a molecule at a temperature of 300 K. [2]

(d) A nitrogen molecule has mass 4.7 × 10⁻²⁶ kg. Calculate its root-mean-square speed at 300 K. [3]

(e) State and explain how the root-mean-square speed changes if the thermodynamic temperature is doubled. [2]

  • (a)Any three: a large number of molecules in random motion ✓; molecular volume negligible compared with the container ✓; no intermolecular forces except during collisions, which are perfectly elastic ✓; time of collision negligible compared with time between collisions.
  • (b)Equate the two expressions for pV: ⅓Nm⟨c²⟩ = NkT ✓; so ⅓m⟨c²⟩ = kT, giving ½m⟨c²⟩ = ₃⁄₂ kT ✓, which is the mean translational kinetic energy of one molecule ✓
  • (c)½m⟨c²⟩ = ₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
  • (d)½m c_rms² = 6.2 × 10⁻²¹ J, so c_rms² = 2 × 6.2 × 10⁻²¹ / 4.7 × 10⁻²⁶ ✓ = 2.64 × 10⁵; c_rms = √(2.64 × 10⁵) ✓ = 5.1 × 10² m s⁻¹ ✓
  • (e)Since ½m c_rms² = ₃⁄₂ kT, c_rms ∝ √T ✓; doubling T multiplies the r.m.s. speed by √2 (about 1.41) ✓
03
Paper 4
[13 marks]

A small bubble of ideal gas rises from the bottom of a lake to the surface, as shown in Fig. 3.1. At the bottom its volume is 1.2 cm³, the pressure is 2.8 × 10⁵ Pa and the temperature is 8 °C. At the surface the pressure is 1.0 × 10⁵ Pa and the temperature is 18 °C.

surface, p₂ bottom, p₁ a gas bubble rising through water
Fig. 3.1

(a) Write down, in terms of p, V and T, the relationship that holds for a fixed mass of ideal gas between two states. [2]

(b) Calculate the volume of the bubble at the surface. [3]

(c) Explain, in molecular terms, why the pressure of the gas in the bubble is lower at the surface. [3]

(d) Calculate the amount, in mol, of gas in the bubble. [3]

(e) Suggest one reason the real bubble would not behave exactly as predicted. [2]

  • (a)For a fixed mass of ideal gas, pV / T is constant ✓, so p₁V₁ / T₁ = p₂V₂ / T₂ (T in kelvin) ✓
  • (b)T₁ = 281 K, T₂ = 291 K ✓; V₂ = p₁V₁T₂ / (T₁p₂) = (2.8 × 10⁵ × 1.2 × 291) / (281 × 1.0 × 10⁵) ✓ = 3.5 cm³ ✓
  • (c)As the bubble rises the surrounding water pressure falls, so the bubble expands ✓; with a larger volume the molecules are more spread out and strike unit area of the surface less often ✓, so the pressure they exert is lower ✓
  • (d)Using surface values, n = pV / RT = (1.0 × 10⁵ × 3.5 × 10⁻⁶) / (8.31 × 291) ✓ = 0.35 / 2418 ✓ = 1.4 × 10⁻⁴ mol ✓
  • (e)Any one: water vapour evaporates into the bubble, adding gas ✓; the temperature may not be uniform, or the gas is not perfectly ideal ✓
04
Paper 4
[10 marks]

A container holds 64 g of oxygen gas, which has a molar mass of 32 g mol⁻¹.

(a) Define the mole. [2]

(b) State the relation between the number of molecules N, the amount of substance n and the Avogadro constant. [1]

(c) Calculate the amount of substance, in moles. [2]

(d) Calculate the number of oxygen molecules. [2]

(e) Calculate the number of oxygen atoms. [1]

(f) State the value and unit of the Avogadro constant. [2]

  • (a)The amount of substance containing as many particles as there are atoms in 12 g of carbon-12, that is N₊ particles ✓✓
  • (b)N = nN₊ ✓
  • (c)n = 64/32 = 2.0 mol ✓✓
  • (d)N = nN₊ = 2.0 × 6.02 × 10²³ ✓ = 1.2 × 10²⁴ ✓
  • (e)Each O₂ molecule has 2 atoms, so 2.4 × 10²⁴ atoms ✓
  • (f)6.02 × 10²³ mol⁻¹ ✓✓
05
Paper 4
[10 marks]

A fixed mass of gas at a pressure of 1.0 × 10⁵ Pa occupies 250 cm³. It is compressed isothermally, as shown in Fig. 5.1.

p V pV = constant isothermal change (fixed T)
Fig. 5.1

(a) State Boyle’s law. [2]

(b) Calculate the pressure when the volume is reduced to 100 cm³. [2]

(c) Explain why the temperature must be constant for Boyle’s law to hold. [1]

(d) Describe the shape of the p-V graph for this change. [2]

(e) Calculate the volume at which the pressure would be 4.0 × 10⁵ Pa, at the same temperature. [2]

(f) State one condition on the gas for Boyle’s law to apply. [1]

  • (a)For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume (pV = constant) ✓✓
  • (b)p₂ = p₁V₁/V₂ = 1.0 × 10⁵ × 250/100 ✓ = 2.5 × 10⁵ Pa ✓
  • (c)From pV = nRT, pV is constant only if T (and n) are constant ✓
  • (d)A smooth curve (a rectangular hyperbola): p falls as V rises ✓✓
  • (e)V = p₁V₁/p = 1.0 × 10⁵ × 250 / 4.0 × 10⁵ ✓ = 62.5 cm³ ✓
  • (f)A fixed mass of ideal gas at constant temperature ✓
06
Paper 4
[10 marks]

A fixed mass of gas is kept at constant volume. At 27 °C its pressure is 1.2 × 10⁵ Pa.

(a) Convert 27 °C to kelvin. [1]

(b) State how the pressure depends on the thermodynamic temperature at constant volume. [2]

(c) Calculate the pressure at 127 °C. [2]

(d) Calculate the temperature, in °C, at which the pressure would be 2.0 × 10⁵ Pa. [3]

(e) Explain why the temperature must be in kelvin in these gas laws. [2]

  • (a)300 K ✓
  • (b)At constant volume the pressure is proportional to the thermodynamic temperature: p/T = constant ✓✓
  • (c)p₂ = p₁ × T₂/T₁ = 1.2 × 10⁵ × 400/300 ✓ = 1.6 × 10⁵ Pa ✓
  • (d)T = T₁ × p₂/p₁ = 300 × (2.0/1.2) = 500 K ✓✓ = 227 °C ✓
  • (e)The proportionality is measured from absolute zero; the Celsius zero is arbitrary, so p ∝ T fails in °C ✓✓
07
Paper 4
[11 marks]

A cylinder contains 0.16 kg of nitrogen gas (molar mass 28 g mol⁻¹) at 300 K and a pressure of 2.0 × 10⁵ Pa. Parts (a) and (b) use the mole; parts (c) and (d) use the equation of state.

(a) Calculate the amount of substance, in moles. [2]

(b) Calculate the number of nitrogen molecules. [2]

(c) Using pV = nRT, calculate the volume occupied by the gas. [3]

(d) The gas is heated at constant pressure to 450 K. Calculate the new volume. [2]

(e) State one assumption of the ideal-gas model used here. [2]

  • (a)n = 160/28 = 5.7 mol ✓✓
  • (b)N = nN₊ = 5.71 × 6.02 × 10²³ ✓ = 3.4 × 10²⁴ ✓
  • (c)V = nRT/p = 5.71 × 8.31 × 300 / (2.0 × 10⁵) ✓✓ = 7.1 × 10⁻² m³ ✓
  • (d)At constant p, V ∝ T: V₂ = 0.0712 × 450/300 ✓ = 1.1 × 10⁻¹ m³ ✓
  • (e)Any one: molecules have negligible volume; no intermolecular forces; collisions are elastic; motion is random ✓✓
08
Paper 4
[11 marks]

Fig. 8.1 represents the molecules of a gas in rapid random motion.

molecules in rapid random motion
Fig. 8.1

(a) Write the kinetic theory equation relating pressure to molecular motion. [1]

(b) State three assumptions of the kinetic theory of gases. [3]

(c) A gas has density 1.2 kg m⁻³ at a pressure of 1.0 × 10⁵ Pa. Calculate the root-mean-square speed of its molecules. [3]

(d) State what is meant by the root-mean-square speed. [2]

(e) State and explain how the r.m.s. speed changes if the kelvin temperature is doubled. [2]

  • (a)pV = ⅓Nm⟨c²⟩ ✓
  • (b)Any three: molecules are point-like (negligible volume); no forces between molecules except during collisions; collisions are perfectly elastic; motion is random; collision times are negligible ✓✓✓
  • (c)p = ⅓ρ⟨c²⟩, so ⟨c²⟩ = 3p/ρ = 3 × 10⁵/1.2 = 2.5 × 10⁵ ✓✓; c_rms = √(2.5 × 10⁵) = 5.0 × 10² m s⁻¹ ✓
  • (d)The square root of the mean of the squares of the molecular speeds ✓✓
  • (e)c_rms ∝ √T, so doubling T increases it by a factor of √2 (about 1.4) ✓✓
09
Paper 4
[11 marks]

This question links the equation of state with the kinetic theory. Part (b) connects pV = NkT with pV = ⅓Nm⟨c²⟩.

(a) Write the expression for the mean translational kinetic energy of a molecule in terms of temperature. [1]

(b) Show that this follows from pV = NkT and pV = ⅓Nm⟨c²⟩. [3]

(c) Calculate the mean translational kinetic energy of a molecule at 300 K. [2]

(d) An oxygen molecule has mass 5.3 × 10⁻²⁶ kg. Calculate its r.m.s. speed at 300 K. [3]

(e) State how the mean translational kinetic energy at a given temperature depends on the type of gas. [2]

  • (a)½m⟨c²⟩ = ₃⁄₂ kT ✓
  • (b)Equating: ⅓Nm⟨c²⟩ = NkT ✓; so m⟨c²⟩ = 3kT ✓; hence ½m⟨c²⟩ = ₃⁄₂ kT ✓
  • (c)₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
  • (d)½m c² = 6.21 × 10⁻²¹, so c² = 2 × 6.21 × 10⁻²¹ / 5.3 × 10⁻²⁶ ✓✓ = 2.34 × 10⁵; c_rms = 4.8 × 10² m s⁻¹ ✓
  • (e)It does not depend on the type of gas; it depends only on the temperature ✓✓
10
Paper 4
[12 marks]

A sealed flask of volume 2.0 × 10⁻³ m³ contains helium gas (molar mass 4.0 g mol⁻¹) at 1.5 × 10⁵ Pa and 300 K. This question uses all three lessons of the topic in turn.

(a) Using pV = nRT, calculate the amount of substance. [2]

(b) Calculate the number of helium atoms. [2]

(c) Calculate the mass of helium in the flask. [2]

(d) Calculate the mean translational kinetic energy of a helium atom. [2]

(e) The mass of a helium atom is the molar mass divided by N₊. Calculate the r.m.s. speed of the atoms. [3]

(f) State whether argon atoms (heavier) at the same temperature would have a greater or smaller r.m.s. speed. [1]

  • (a)n = pV/RT = 1.5 × 10⁵ × 2.0 × 10⁻³ / (8.31 × 300) ✓ = 0.12 mol ✓
  • (b)N = nN₊ = 0.120 × 6.02 × 10²³ ✓ = 7.2 × 10²² ✓
  • (c)mass = nM = 0.120 × 4.0 = 0.48 g = 4.8 × 10⁻⁴ kg ✓✓
  • (d)₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
  • (e)m = 4.0 × 10⁻³ / 6.02 × 10²³ = 6.6 × 10⁻²⁷ kg ✓; c² = 2 × 6.21 × 10⁻²¹ / 6.64 × 10⁻²⁷ = 1.87 × 10⁶ ✓; c_rms = 1.4 × 10³ m s⁻¹ ✓
  • (f)Smaller: the mean kinetic energy is the same, so a larger mass gives a smaller speed ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.