Free B.4 Thermodynamics questions for IB Physics (SL & HL), HL only, part of B The particulate nature of matter. Each question has the answer and the marking points an examiner would look for.
Generate a full B.4 Thermodynamics worksheet →All B The particulate nature of matter questionsAn ideal gas expands at a constant pressure of 1.0 × 105 Pa, increasing its volume by 2 × 10−3 m3. 1000 J of thermal energy is supplied to the gas.
What is the change in internal energy of the gas?
Answer: B
A monatomic ideal gas is taken round a cycle A → B → C → D → A. At A the pressure is 1 × 105 Pa, the volume is 2 × 10−3 m3 and the temperature is 300 K. AB: pressure rises to 3× at constant volume. BC: volume rises to 3× at constant pressure. CD: pressure falls back to the value at A at constant volume. DA: volume falls back to that at A at constant pressure.
Sketch the cycle on a p–V diagram and calculate the temperatures at B and C.
Calculate the net work done by the gas in one cycle.
Calculate the thermal energy supplied to the gas in stages AB and BC.
Determine the efficiency of the cycle and compare it with the Carnot efficiency between the highest and lowest temperatures.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 900 K; 2700 K rectangle drawn, clockwise ✓ TB = 900 K ✓ TC = 2700 K ✓ | 3 |
| (b) | 800 J area enclosed = Δp × ΔV ✓ 800 J ✓ | 2 |
| (c) | 600 J; 3000 J n = 0.0802 mol ✓ AB: Q = ΔU = 32nRΔT = 600 J ✓ BC: Q = ΔU + pΔV = 52nRΔT ✓ = 3000 J ✓ | 4 |
| (d) | 22.2% vs 88.9% efficiency = WQin = 22.2% ✓ Carnot: 1 − 3002700 = 88.9% ✓ the real cycle is less efficient, as expected ✓ | 3 |
An ideal gas undergoes an isothermal compression from 2.0 × 10−3 m3 to 1.0 × 10−3 m3.
State the change in internal energy of the gas and explain your answer.
Explain why thermal energy must leave the gas.
State what happens to the entropy of the gas.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | zero; temperature constant ΔU = 0 ✓ internal energy of an ideal gas depends only on temperature ✓ | 2 |
| (b) | work done on gas, Q = W out work is done on the gas ✓ since ΔU = 0 the same energy must leave as heat ✓ | 2 |
| (c) | decreases decreases (energy transferred out at constant T) ✓ | 1 |
A heat engine operates between reservoirs at 600 K and 300 K.
Calculate the maximum possible efficiency.
The engine absorbs 2000 J per cycle and has an actual efficiency of 30 %. Calculate the work done and the energy rejected per cycle.
Calculate the entropy change of the cold reservoir per cycle.
State the second law of thermodynamics in terms of entropy.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 50 % 1 − 300/600 = 0.50 ✓ | 1 |
| (b) | 600 J; 1400 J W = 600 J ✓ Qc = 1400 J ✓ | 2 |
| (c) | 4.7 J K−1 ΔS = Q/T ✓ = 1400/300 = 4.7 J K−1 ✓ | 2 |
| (d) | total entropy never decreases the entropy of an isolated system never decreases ✓ | 1 |
0.50 mol of a monatomic ideal gas at 300 K expands at a constant pressure of 1.0 × 105 Pa from 1.25 × 10−2 m3 to 1.75 × 10−2 m3.
Calculate the work done by the gas.
Calculate the final temperature.
Calculate the change in internal energy of the gas.
Calculate the thermal energy supplied to the gas.
State and explain the sign of the change in entropy of the gas.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 500 J W = pΔV = 500 J ✓ | 1 |
| (b) | 420 K T = 300 × 1.75/1.25 = 420 K ✓ | 1 |
| (c) | 750 J ΔU = 3/2 nRΔT ✓ = 1.5 × 0.50 × 8.31 × 120 = 748 J ✓ | 2 |
| (d) | 1250 J Q = ΔU + W ✓ = 1248 J ✓ | 2 |
| (e) | increases entropy increases ✓ thermal energy is transferred into the gas (ΔS = ΔQ/T > 0) ✓ | 2 |
A monatomic ideal gas is taken round the following cycle.
A → B: adiabatic compression from volume VA to VA/8
B → C: 150 J of thermal energy is supplied at constant volume
C → D: adiabatic expansion back to volume VA
D → A: cooling at constant volume
At A the pressure is 1.0 × 105 Pa, the volume is 1.20 × 10−3 m3 and the temperature is 300 K. For an adiabatic change of a monatomic ideal gas, pV5/3 is constant. R = 8.31 J K−1 mol−1.
Determine the temperature and the pressure of the gas at B.
Determine the temperatures at C and at D.
The engine is redesigned. The gas at A is unchanged and the highest temperature reached in the cycle is kept at 1450 K, but the volume ratio VA/VB may be chosen. Determine the volume ratio that gives the greatest work done per cycle, and that work.
Calculate the efficiency of the redesigned cycle and explain why it is lower than that of the original cycle even though it does more work.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1200 K; 3.2 × 10⁶ Pa pB = 1.0 × 105 × 85/3 = 3.2 × 106 Pa ✓ combining with pV = nRT: TV2/3 is constant ✓ TB = 300 × 82/3 = 1200 K ✓ | 3 |
| (b) | 1450 K; 363 K n = pV/RT = 0.0481 mol ✓ ΔT = Q/(1.5nR) = 250 K, TC = 1450 K ✓ TD = 1450/82/3 = 363 K ✓ | 3 |
| (c) | 3.26; 259 J W = 1.5nR[(TC − TB) − (TD − TA)] with TD = TATC/TB ✓ W = 1.5nR(TC + TA − TB − TATC/TB): maximum when TB = √(TATC) (= TD) ✓ TB = 660 K; VA/VB = (660/300)3/2 = 3.26 ✓ W = 1.5nR(√TC − √TA)2 ✓ W = 259 J ✓ | 5 |
| (d) | 0.545 efficiency = 1 − 300/660 = 0.545 (heat supplied 474 J) ✓ lower compression ratio gives lower efficiency; more work comes from far more heat supplied per cycle ✓ | 2 |
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in