IB DP Physics: First law of thermodynamics, gas processes and work done by a gas (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions follow the IB Physics guide, first assessment 2025, topic B.4 (first half: first law, work, internal energy and gas processes). All of B.4 is additional higher level, so every question is HL only. Entropy, the second law and heat engines are in a separate unit. Where needed use R = 8.31 J mol⁻¹ K⁻¹. All gases are monatomic ideal gases.

Revise first with the study guide and the revision notes. The course page is IB DP Physics and the printable list is the IB DP Physics checklist. For a quick check of your level, try the free 10-minute diagnostics.

Questions

1. In a closed cylinder, 120 J of heat enters a gas and a piston does 45 J of work on it. Find ΔU. [2]

2. (a) State what is meant by a closed system and by an isolated system. [2] (b) State the change in internal energy of an isolated sample of ideal gas, and give a reason. [1]

3. A gas at a constant pressure of 2.4 × 10⁵ Pa expands from 1.5 × 10⁻³ m³ to 4.0 × 10⁻³ m³ while absorbing 1.50 kJ of heat.

(a) Calculate the work done by the gas. [2] (b) Calculate the change in internal energy. [2]

4. A rigid container holds 3.0 mol of gas. The gas is heated from 280 K to 350 K. Determine the change in internal energy, the work done by the gas and the heat supplied. [3]

5. 0.25 mol of gas at 310 K expands isothermally while absorbing 540 J of heat.

(a) State the change in internal energy and the work done by the gas. [2] (b) Explain, in terms of the particles, why the pressure of the gas falls. [2]

6. A gas changes from 3.0 × 10⁵ Pa and 2.0 × 10⁻³ m³ to 1.0 × 10⁵ Pa and 5.0 × 10⁻³ m³ along a straight line on a P-V diagram.

(a) Calculate the work done by the gas. [2] (b) Calculate the change in internal energy. [2] (c) Determine the heat transferred to the gas. [1]

7. A gas at 1.2 × 10⁵ Pa and 1.0 × 10⁻³ m³ expands adiabatically to 3.0 × 10⁻³ m³.

(a) Calculate the final pressure. [2] (b) Calculate the work done by the gas. [3] (c) State whether the temperature of the gas rises or falls. [1]

8. A gas in an insulated cylinder is compressed quickly by a piston.

(a) State the heat transferred to the gas. [1] (b) State the sign of the work W and of the change in internal energy under the Clausius convention. [2] (c) Explain, using the particle model, why the temperature rises. [2]

9. 0.40 mol of gas starts at A with T = 300 K and V = 5.0 × 10⁻³ m³. It is heated at constant volume to B, where the pressure has doubled. It then expands at constant pressure to C, where the volume has doubled.

(a) Determine the temperatures at B and at C. [2] (b) Calculate the total change in internal energy from A to C. [2] (c) Calculate the total work done by the gas. [2] (d) Calculate the total heat supplied. [2]

10. 0.050 mol of gas at 290 K and 1.0 × 10⁵ Pa is compressed adiabatically to one eighth of its volume.

(a) Calculate the final pressure. [2] (b) Show that the final temperature is 1160 K (to 3 s.f.). [2] (c) Calculate the change in internal energy. [2] (d) State the work done by the gas. [1] (e) Explain why the gas ends up hotter even though no heat enters it. [1]

11. A gas is taken from state 1 (3.0 × 10⁵ Pa, 2.0 × 10⁻³ m³) to state 2 (3.0 × 10⁵ Pa, 4.0 × 10⁻³ m³), then to state 3 (1.5 × 10⁵ Pa, 4.0 × 10⁻³ m³).

(a) Calculate the work done by the gas from 1 to 2. [2] (b) Calculate the change in internal energy and the heat supplied from 1 to 2. [3] (c) For the change from 2 to 3, state the work done and determine ΔU and Q. [3] (d) Show that the change from 1 to 3 has ΔU = 0, and state what this tells you about the temperatures of states 1 and 3. [1] (e) The gas could instead go from 1 to 3 by a different route. State whether ΔU, Q and W would each be the same or could differ. [1]

Answers

1. W = −45 J because work is done on the gas [1]. ΔU = Q − W = 120 − (−45) = 165 J [1] [2] Examiner insight: A bare 75 J (from subtracting the wrong way round) shows a sign slip; writing W = −45 J explicitly makes your intention clear.

2. (a) Closed: matter cannot cross the boundary, but energy can cross it as heat or work [1]. Isolated: the boundary lets through neither matter nor energy [1] (b) ΔU = 0, because Q = 0 and W = 0 for an isolated system [1] [3] Examiner insight: Answers that say a closed system “cannot exchange anything” lose the idea that energy still crosses its boundary.

3. (a) W = PΔV = 2.4 × 10⁵ × (4.0 − 1.5) × 10⁻³ [1] = 600 J [1] (b) ΔU = Q − W = 1500 − 600 [1] = 900 J [1] [4] Examiner insight: Convert kJ to J before subtracting; mixing 1.50 kJ with 600 J is a common slip that loses the final answer.

4. Volume is fixed, so W = 0 [1]. ΔU = 3/2 nRΔT = 1.5 × 3.0 × 8.31 × 70 [1] = 2.62 kJ, and Q = ΔU + W = 2.62 kJ [1] [3] Examiner insight: Justify W = 0 by saying the volume does not change; stating the value alone gives no reasoning to credit.

5. (a) ΔU = 0 as the temperature is constant [1]; so W = Q = 540 J [1] (b) Temperature is constant, so the mean kinetic energy and speed of the particles are unchanged [1]. The volume is larger, so collisions with a given area of wall are less frequent and the pressure is lower [1] [4] Examiner insight: Explanations that mention only “the gas has more room” miss the link to collision rate and are not enough on their own for an explain question.

6. (a) Area under the line is a trapezoid: average pressure = 2.0 × 10⁵ Pa [1]; W = 2.0 × 10⁵ × 3.0 × 10⁻³ = 600 J [1] (b) P₁V₁ = 600 J and P₂V₂ = 500 J [1]; ΔU = 3/2 × (500 − 600) = −150 J [1] (c) Q = ΔU + W = −150 + 600 = 450 J [1] [5] Examiner insight: ΔU can be found from the end points without n or T; candidates who try to find a temperature first often introduce rounding errors.

7. (a) P₂ = P₁(V₁/V₂)^(5/3) = 1.2 × 10⁵ × (1/3)^(5/3) [1] = 1.92 × 10⁴ Pa [1] (b) Q = 0 so W = −ΔU [1]; ΔU = 3/2 × (1.923 × 10⁴ × 3.0 × 10⁻³ − 1.2 × 10⁵ × 1.0 × 10⁻³) = 3/2 × (57.7 − 120) = −93.5 J [1]; W = +93.5 J [1] (c) The temperature falls, since ΔU is negative [1] [6] Examiner insight: Keep unrounded intermediate values for the pressure; rounding to 1.9 × 10⁴ Pa before using it changes the third significant figure of the work.

8. (a) Q = 0 [1] (b) W is negative (work is done on the gas) [1]; ΔU is positive [1] (c) The piston does work on the gas and no heat escapes, so the random kinetic energy of the particles increases [1]; temperature measures the mean kinetic energy so it rises [1] [5] Examiner insight: Mention both parts of the chain: energy transferred by work, then the effect on particle kinetic energy. “Compressing makes it hot” alone is a restatement, not an explanation.

9. (a) At constant volume P ∝ T, so T_B = 2 × 300 = 600 K [1]. At constant pressure V ∝ T, so T_C = 2 × 600 = 1200 K [1] (b) ΔU = 3/2 nRΔT = 1.5 × 0.40 × 8.31 × 900 [1] = 4.49 kJ [1] (c) A to B: W = 0. P_B = 2nRT_A/V_A = 3.99 × 10⁵ Pa [1]; B to C: W = 3.99 × 10⁵ × 5.0 × 10⁻³ = 1.99 kJ [1] (d) Q = ΔU + W = 4.49 + 1.99 [1] = 6.48 kJ [1] [8] Examiner insight: Work is only done on the isobaric leg. Adding a PΔV term for the vertical leg gives a wrong total work, and the heat total then inherits the error.

10. (a) P₂ = P₁ × 8^(5/3) [1] = 1.0 × 10⁵ × 32 = 3.2 × 10⁶ Pa [1] (b) T₂/T₁ = (P₂V₂)/(P₁V₁) = 32 × (1/8) = 4 [1]; T₂ = 4 × 290 = 1160 K [1] (c) ΔU = 3/2 nRΔT = 1.5 × 0.050 × 8.31 × 870 [1] = 542 J [1] (d) W = −ΔU = −542 J [1] (e) Work is done on the gas, which raises its internal energy, and Q = 0 means none of it is lost as heat [1] [8] Examiner insight: On a “show that” question, state the intermediate ratio before the final value; going straight to 1160 K with no working does not demonstrate the result.

11. (a) W = PΔV = 3.0 × 10⁵ × 2.0 × 10⁻³ [1] = 600 J [1] (b) P₁V₁ = 600 J, P₂V₂ = 1200 J [1]; ΔU = 3/2 × 600 = 900 J [1]; Q = 900 + 600 = 1500 J [1] (c) Volume is fixed, so W = 0 [1]; P₃V₃ = 600 J, so ΔU = 3/2 × (600 − 1200) = −900 J [1], and Q = −900 J [1] (d) P₃V₃ = P₁V₁ = 600 J, so ΔU = 900 − 900 = 0 and states 1 and 3 are at the same temperature [1] (e) ΔU would be the same as it depends only on the end states, while Q and W could differ because they depend on the path [1] [10] Examiner insight: Part (e) needs both halves: that ΔU is fixed by the end states and that Q and W are route dependent. Naming only one earns less than the full mark.

Where marks are usually lost

• Treating compression work as positive, which flips the sign of ΔU or Q.
• Forgetting that W = 0 on a vertical P-V line.
• Mixing kJ with J, or cm³ and litres with m³.
• Using Celsius temperatures in an equation that needs T in kelvin.
• Writing ΔU = 0 for an adiabatic change instead of Q = 0.
• Using PV^(5/3) = constant for an isothermal change, or PV = constant for an adiabatic change.
• Rounding intermediate values and then losing the third significant figure.
• Explaining a temperature change without naming the energy transfer (work or heat) that causes it.
• Saying Q and W are the same for any route between two states.

Next steps

• Review the revision notes and the study guide.
• Go to the IB DP Physics course page and the printable checklist.
• Take the free 10-minute diagnostics to find your weakest topics.
• Book a free trial class for help with thermodynamics.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. Topic B.4 (Thermodynamics) is additional higher level only.