A2 · Paper 4 practice · Thermodynamics

Heat in, work out, books balanced.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Thermodynamics: internal energy and the first law. With only two lessons in the topic, most questions link them, taking a process and tracking both the energy transfers and the internal energy. The sign convention is ΔU = q + W, where W is the work done on the gas. Each question is linked to its lessons; 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 where needed, and use the first law ΔU = q + W, with W the work done on the gas.
Keep these straight

Bookkeeping with energy.

01
Paper 4
[12 marks]

Fig. 1.1 represents the molecules of a fixed mass of ideal gas in random motion.

ideal gas: internal energy is the total molecular kinetic energy
Fig. 1.1

(a) State what is meant by the internal energy of a system. [2]

(b) Explain why, for an ideal gas, the internal energy is entirely kinetic. [2]

(c) The temperature of the gas is raised. State and explain the effect on its internal energy. [2]

(d) State whether the internal energy of this gas depends on its pressure and volume separately or only on its temperature, and justify your answer briefly. [2]

(e) The gas is 0.50 mol of a monatomic ideal gas at 300 K. Using U = ₃⁄₂ nRT, calculate its internal energy. [3]

(f) The gas is now compressed to half its volume at constant temperature. State the change in its internal energy. [1]

  • (a)Internal energy is the sum of the random distribution of kinetic and potential energies ✓ of all the molecules of the system ✓
  • (b)An ideal gas has no intermolecular forces ✓, so there is no molecular potential energy; the internal energy is therefore all kinetic ✓
  • (c)The internal energy increases ✓; a higher temperature means a greater mean molecular kinetic energy (₃⁄₂ kT per molecule) ✓
  • (d)It depends only on the temperature ✓; since U is all kinetic and the mean KE depends only on T, the same T gives the same U whatever the p and V ✓
  • (e)U = ₃⁄₂ nRT = 1.5 × 0.50 × 8.31 × 300 ✓✓ = 1.9 × 10³ J ✓
  • (f)Zero (the temperature is unchanged, so U is unchanged) ✓
02
Paper 4
[13 marks]

A gas expands at a constant pressure of 1.5 × 10⁵ Pa, its volume increasing from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³. Fig. 2.1 shows this change on a p–V diagram.

p V V₁ V₂ work = pΔV constant pressure
Fig. 2.1

(a) Show that the work done by a gas expanding at constant pressure is W = pΔV. [2]

(b) Calculate the work done by the gas in this expansion. [2]

(c) State what the work done is represented by on the p–V diagram. [1]

(d) During the expansion, 900 J of thermal energy is supplied to the gas. Use the first law to find the change in its internal energy. [3]

(e) State and explain whether the temperature of the gas rises, falls or stays the same. [2]

(f) State the sign of the work done on the gas if instead it were compressed at constant pressure. [1]

(g) The gas instead expands by the same amount with no thermal energy supplied. State and explain what happens to its temperature. [2]

  • (a)At constant pressure the force on the piston is F = pA ✓; if it moves a distance d, W = Fd = pA d = p(Ad) = pΔV ✓
  • (b)W = pΔV = 1.5 × 10⁵ × (5.0 − 2.0) × 10⁻³ ✓ = 1.5 × 10⁵ × 3.0 × 10⁻³ = 450 J ✓
  • (c)The area under the line (between the line and the V axis) ✓
  • (d)ΔU = q + W, with W the work done on the gas = −pΔV = −450 J ✓; ΔU = 900 + (−450) ✓ = +450 J ✓
  • (e)The temperature rises ✓; ΔU is positive, so the internal energy increases, and for an ideal gas a higher internal energy means a higher temperature ✓
  • (f)Positive (work is done on the gas) ✓
  • (g)q = 0, so ΔU = W = −450 J ✓; the internal energy falls, so the temperature falls ✓
03
Paper 4
[13 marks]

A fixed mass of ideal gas is taken through several separate processes, each starting from the state S shown in Fig. 3.1.

p V S constant V constant p isothermal three processes from the same starting state S
Fig. 3.1

(a) State the first law of thermodynamics in the form ΔU = q + W, defining each symbol and its sign convention. [3]

(b) The gas is heated at constant volume, gaining 600 J of thermal energy. State the work done on the gas and the change in internal energy. [2]

(c) In a separate isothermal expansion the gas does 400 J of work. State the change in internal energy and the heat transferred. [2]

(d) In a separate adiabatic compression, 250 J of work is done on the gas. State the change in internal energy and what happens to the temperature. [2]

(e) Using the first law, explain why compressing a gas rapidly raises its temperature. [2]

(f) Explain why ΔU = 0 for any process that returns the gas to its original state. [2]

  • (a)ΔU is the increase in internal energy ✓; q is the thermal energy supplied to the gas (positive when heat is added) ✓; W is the work done on the gas (positive when the gas is compressed, negative when it expands) ✓
  • (b)Constant volume means ΔV = 0, so W = 0 ✓; ΔU = q + 0 = +600 J ✓
  • (c)Isothermal means T constant, so ΔU = 0 ✓; then 0 = q + W with W = −400 J, so q = +400 J (heat is absorbed) ✓
  • (d)q = 0 and W = +250 J, so ΔU = +250 J ✓; the internal energy rises, so the temperature rises ✓
  • (e)A rapid compression gives little time for heat to leave, so q ≈ 0 ✓; work is done on the gas (W positive), so ΔU = W is positive and the temperature rises ✓
  • (f)Internal energy is a function of state ✓; returning to the same state means the same internal energy, so the net change ΔU is zero ✓
04
Paper 4
[10 marks]

This question is about the meaning of internal energy.

(a) Explain what is meant by the internal energy of a system. [2]

(b) State why the motion of the molecules is described as random. [1]

(c) For an ideal gas, state and explain which type of molecular energy contributes to the internal energy. [2]

(d) Hence explain why the internal energy of an ideal gas depends only on its temperature. [2]

(e) State what happens to the internal energy of an ideal gas if its temperature is held constant while p and V change. [1]

(f) State what happens to the internal energy of a fixed mass of ideal gas if its kelvin temperature is doubled. [2]

  • (a)The sum of the random distribution of the kinetic and potential energies of the molecules of the system ✓✓
  • (b)The molecules move in all directions with a wide range of speeds ✓
  • (c)Only kinetic energy ✓: an ideal gas is assumed to have no intermolecular forces, so no molecular potential energy ✓
  • (d)The internal energy is the total molecular kinetic energy, and the mean kinetic energy is proportional to temperature ✓✓
  • (e)It is unchanged (ΔU = 0) ✓
  • (f)U is proportional to T, so the internal energy also doubles ✓✓
05
Paper 4
[10 marks]

A real gas is compared with an ideal gas.

(a) State the two contributions to the internal energy of a substance. [2]

(b) Explain why the internal energy of an ideal gas has no potential-energy contribution. [2]

(c) A fixed mass of ideal gas is heated from 300 K to 600 K. State the ratio of its final to initial internal energy, and explain. [3]

(d) Explain why two samples of the same ideal gas at the same temperature but different volumes have the same internal energy per mole. [2]

(e) State whether the internal energy depends on the current state of the gas or on the path taken to reach it. [1]

  • (a)The random kinetic energy and the potential energy of the molecules ✓✓
  • (b)An ideal gas is assumed to have no intermolecular forces, so there is no molecular potential energy ✓✓
  • (c)U ∝ T, so the ratio is 600/300 = 2 ✓✓; the internal energy depends only on temperature ✓
  • (d)Internal energy of an ideal gas depends only on temperature, not on volume or pressure ✓✓
  • (e)Only on the current state: internal energy is a function of state ✓
06
Paper 4
[10 marks]

A gas at a constant pressure of 1.5 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³, as shown in Fig. 6.1.

p V V₁ V₂ W = pΔV (area) constant pressure
Fig. 6.1

(a) Write the expression for the work done by a gas at constant pressure. [1]

(b) Calculate the work done by the gas during the expansion. [2]

(c) State the sign of the work done on the gas, and explain. [2]

(d) State what the work done corresponds to on the p-V diagram. [1]

(e) The gas is then compressed back to 2.0 × 10⁻³ m³ at a constant pressure of 2.5 × 10⁵ Pa. Calculate the work done on the gas. [2]

(f) State the net work done on the gas over the two stages. [2]

  • (a)W = pΔV ✓
  • (b)W = 1.5 × 10⁵ × (5.0 − 2.0) × 10⁻³ ✓ = 4.5 × 10² J ✓
  • (c)Negative ✓: the gas expands and does work on the surroundings, so work done on the gas is −450 J ✓
  • (d)The area under the line ✓
  • (e)W on gas = pΔV = 2.5 × 10⁵ × 3.0 × 10⁻³ ✓ = +7.5 × 10² J ✓
  • (f)Net = −450 + 750 = +3.0 × 10² J done on the gas ✓✓
07
Paper 4
[11 marks]

0.050 mol of an ideal monatomic gas is heated at constant volume; 250 J of thermal energy is supplied. Parts (a) to (c) use the first law; parts (d) and (e) use internal energy.

(a) State the first law of thermodynamics, defining each symbol. [2]

(b) State the work done on the gas at constant volume, and explain. [2]

(c) Hence state the increase in internal energy. [1]

(d) Explain, in terms of the molecules, what this increase in internal energy represents. [2]

(e) State and explain what happens to the temperature of the gas. [2]

(f) If instead the gas were allowed to expand as it was heated, state whether the temperature rise would be larger or smaller for the same heat supplied. [2]

  • (a)ΔU = q + W ✓: q is the heat supplied to the gas, W is the work done on the gas ✓
  • (b)W = 0 ✓: the volume is constant, so no work is done ✓
  • (c)ΔU = q = +250 J ✓
  • (d)An increase in the mean kinetic energy of the molecules (faster random motion) ✓✓
  • (e)The temperature rises ✓, because the internal energy (and mean kinetic energy) of an ideal gas is proportional to T ✓
  • (f)Smaller ✓: some energy leaves as work done by the gas, so less is left to raise the internal energy ✓
08
Paper 4
[11 marks]

An ideal gas expands isothermally. This question links internal energy with the first law.

(a) State what happens to the internal energy of the gas, and explain. [2]

(b) Apply the first law to this process. [2]

(c) State the direction of heat flow as the gas expands isothermally. [2]

(d) The gas does 600 J of work on the surroundings. State the work done on the gas and the heat supplied. [2]

(e) Describe the isothermal expansion on a p-V diagram and state what the area under it represents. [2]

(f) Explain why the temperature is unchanged even though the gas absorbs heat. [1]

  • (a)It is unchanged ✓: the internal energy of an ideal gas depends only on temperature, which is constant ✓
  • (b)ΔU = 0, so q = −W ✓: the heat supplied equals the work done by the gas ✓
  • (c)Heat flows into the gas ✓✓
  • (d)Work done on the gas = −600 J ✓; q = +600 J ✓
  • (e)A curve (pV = constant) ✓; the area under it is the work done by the gas ✓
  • (f)All the absorbed heat leaves again as work, so the internal energy and hence the temperature stay the same ✓
09
Paper 4
[11 marks]

A gas is compressed rapidly so that no heat enters or leaves it (an adiabatic change).

(a) State the value of q for an adiabatic process. [1]

(b) Apply the first law to an adiabatic change. [2]

(c) The gas is compressed, so work is done on it. State and explain what happens to its internal energy and temperature. [3]

(d) 300 J of work is done on the gas. State the change in internal energy. [1]

(e) Explain, in molecular terms, why an adiabatic compression raises the temperature. [2]

(f) State and explain what happens to the temperature during an adiabatic expansion. [2]

  • (a)q = 0 ✓
  • (b)ΔU = W ✓: the change in internal energy equals the work done on the gas ✓
  • (c)Work is done on the gas (W positive), so ΔU is positive ✓: the internal energy rises ✓ and the temperature rises (U ∝ T) ✓
  • (d)ΔU = +300 J ✓
  • (e)Molecules rebound from the inward-moving piston with greater speed, gaining kinetic energy ✓✓
  • (f)The temperature falls ✓: with q = 0 the gas does work as it expands, so ΔU is negative ✓
10
Paper 4
[12 marks]

A fixed mass of ideal gas is taken through a cycle and returns to its original state, as suggested by Fig. 10.1. This question applies the first law to the steps and uses the fact that internal energy is a function of state.

p V start cyclic process: gas returns to its original state
Fig. 10.1

(a) In the first step (constant volume), 200 J of heat is supplied. State the work done on the gas and the change in internal energy. [2]

(b) In the second step (constant-pressure expansion), 500 J of heat is supplied and the gas does 180 J of work. Calculate the change in internal energy. [2]

(c) After the complete cycle the gas returns to its starting state. State the total change in internal energy over the cycle, and explain. [3]

(d) Over the whole cycle the net work done on the gas is +120 J. Calculate the net heat transferred, and state its direction. [3]

(e) Explain why the internal energy at the end of the second step is greater than at the start of the cycle. [2]

  • (a)W = 0 (constant volume) ✓; ΔU = q = +200 J ✓
  • (b)ΔU = q + W = 500 + (−180) ✓ = +320 J ✓
  • (c)ΔU over the cycle = 0 ✓: internal energy is a function of state, and the gas returns to the same temperature ✓✓
  • (d)Over the cycle 0 = q_net + W_net, so q_net = −120 J ✓✓: 120 J of heat is released by the gas ✓
  • (e)The temperature is higher there, and the internal energy of an ideal gas depends only on temperature ✓✓

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