A Level Physics 16.2 First Law of Thermodynamics MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 16.2 of the Cambridge International AS and A Level Physics (9702) syllabus: the first law of thermodynamics. It is an A Level (A2) topic, and every question uses the Cambridge sign convention, in which ΔU = q + W with W the work done on the system.
Formula recap – the first law of thermodynamics
Increase in internal energy = thermal energy supplied to the system + work done on the system, that is:
ΔU = q + W
Here ΔU is the increase in internal energy of the system, q is the thermal energy transferred into the system (q is positive when thermal energy enters the system and negative when it leaves), and W is the work done on the system (W is positive for a compression, when work is done on the gas). When a gas expands it does work on its surroundings, so W is negative: for an expansion at constant pressure p in which the volume changes by ΔV, the work done by the gas is pΔV, so the work done on the gas is W = −pΔV. The internal energy of an ideal gas depends only on its temperature. See the full A Level Physics Formula Sheet for all the equations in this topic.
The First Law & Sign Conventions
Q1. Which statement correctly expresses the first law of thermodynamics, using the Cambridge sign convention?
- (A) Increase in internal energy = thermal energy supplied to the system − work done on the system
- (B) Increase in internal energy = thermal energy supplied to the system + work done on the system
- (C) Increase in internal energy = work done by the system − thermal energy removed from the system
- (D) Increase in internal energy = thermal energy supplied to the system × work done on the system
Show Answer
Answer: B – The first law is a statement of the conservation of energy: ΔU = q + W, where q is the thermal energy transferred into the system and W is the work done on the system. Both energy inputs increase the internal energy, so both terms are added.
Q2. A gas in a cylinder is heated by a flame and nothing else happens to it. In the equation ΔU = q + W, the term q for the gas is
- (A) positive, because thermal energy is transferred into the gas
- (B) negative, because thermal energy is transferred into the gas
- (C) positive, because thermal energy leaves the gas
- (D) zero, because heating alone cannot change the internal energy
Show Answer
Answer: A – q is defined as the thermal energy transferred INTO the system. Heating the gas transfers thermal energy into it, so q is positive.
Q3. A piston is pushed inwards, compressing a gas. No thermal energy is transferred during the compression. The term W in ΔU = q + W is
- (A) negative, because work is done on the gas
- (B) zero, because the volume of the gas decreases
- (C) positive, because work is done on the gas
- (D) positive, because work is done by the gas
Show Answer
Answer: C – W is the work done ON the system. The piston does work on the gas as it compresses it, so W is positive (and here ΔU = W, since q = 0).
Q4. A gas expands and pushes a piston outwards, doing work on its surroundings. No thermal energy is transferred. The sign of W for the gas is
- (A) positive, because the gas does work
- (B) positive, because the volume of the gas increases
- (C) zero, because no thermal energy is transferred
- (D) negative, because the work is done by the gas, not on it
Show Answer
Answer: D – W counts only work done ON the system. In an expansion the gas does work on the surroundings, so W is negative and, with q = 0, the internal energy of the gas falls.
Q5. A hot block is left in a cold room and loses thermal energy to its surroundings. For the block, the term q is
- (A) positive
- (B) negative
- (C) zero
- (D) equal to the work done on the block
Show Answer
Answer: B – Thermal energy is transferred OUT of the system, which is the opposite direction to the one defined as positive, so q is negative.
Q6. A system absorbs 50 J of thermal energy and 30 J of work is done on it. The increase in its internal energy is
- (A) +80 J
- (B) +20 J
- (C) −20 J
- (D) +30 J
Show Answer
Answer: A – ΔU = q + W = (+50) + (+30) = +80 J. Both transfers are into the system, so the internal energy rises by the total energy supplied.
Q7. A gas absorbs 100 J of thermal energy and expands, doing 60 J of work on its surroundings. The change in internal energy of the gas is
- (A) +160 J
- (B) −40 J
- (C) +40 J
- (D) 0 J
Show Answer
Answer: C – The gas does work, so W = −60 J. Then ΔU = q + W = (+100) + (−60) = +40 J.
Q8. A gas gives out 40 J of thermal energy while 25 J of work is done on it during a compression. The change in internal energy of the gas is
- (A) +65 J
- (B) −65 J
- (C) +15 J
- (D) −15 J
Show Answer
Answer: D – Thermal energy leaves the gas, so q = −40 J; work is done on the gas, so W = +25 J. ΔU = (−40) + (+25) = −15 J.
Q9. Which combination of q and W leaves the internal energy of a system unchanged?
- (A) q = +25 J, W = +25 J
- (B) q = +25 J, W = −25 J
- (C) q = −25 J, W = −25 J
- (D) q = 0 J, W = +25 J
Show Answer
Answer: B – ΔU = q + W = (+25) + (−25) = 0 J. The energy brought in as thermal energy is exactly balanced by the work the system does on its surroundings.
Q10. During a change, a gas does 80 J of work on its surroundings and loses 30 J of thermal energy. The change in internal energy of the gas is
- (A) −110 J
- (B) −50 J
- (C) +50 J
- (D) +110 J
Show Answer
Answer: A – Work is done BY the gas, so W = −80 J, and thermal energy leaves, so q = −30 J. ΔU = (−30) + (−80) = −110 J.
Work Done by and on a Gas
Q11. A gas expands at a constant pressure of 1.0 × 105 Pa and its volume increases by 2.0 × 10−3 m3. The work done by the gas is
- (A) 20 J
- (B) 200 J
- (C) 50 J
- (D) 2000 J
Show Answer
Answer: B – Work done by the gas at constant pressure = pΔV = (1.0 × 105) × (2.0 × 10−3) = 200 J.
Q12. A gas expands at constant pressure and does 200 J of work on its surroundings. The value of W (the work done on the gas) to use in ΔU = q + W is
- (A) +200 J
- (B) 0 J
- (C) +400 J
- (D) −200 J
Show Answer
Answer: D – The 200 J of work is done BY the gas, so the work done ON the gas is the negative of this: W = −200 J.
Q13. A gas is compressed at a constant pressure of 2.0 × 105 Pa and its volume decreases by 1.5 × 10−3 m3. The work done on the gas is
- (A) +300 J
- (B) −300 J
- (C) +150 J
- (D) +30 J
Show Answer
Answer: A – Here ΔV = −1.5 × 10−3 m3, so the work done by the gas is pΔV = (2.0 × 105) × (−1.5 × 10−3) = −300 J. The work done ON the gas is therefore W = +300 J.
Q14. A p-V graph shows a gas expanding at a constant pressure of 3.0 × 105 Pa from a volume of 1.0 × 10−3 m3 to a volume of 4.0 × 10−3 m3. The area under the graph, and so the work done by the gas, is
- (A) 300 J
- (B) 1200 J
- (C) 900 J
- (D) 90 J
Show Answer
Answer: C – ΔV = (4.0 − 1.0) × 10−3 = 3.0 × 10−3 m3. Work done by the gas = area under the graph = pΔV = (3.0 × 105) × (3.0 × 10−3) = 900 J.
Q15. A gas is allowed to expand freely into an evacuated chamber, so its volume doubles but it pushes against zero external pressure. The work done by the gas is
- (A) pΔV, using the pressure of the gas
- (B) positive and large
- (C) zero
- (D) negative
Show Answer
Answer: C – In a free expansion there is no external pressure resisting the expansion and nothing is pushed back, so no work is done even though the volume increases: W = 0. (For an ideal gas this means ΔU = 0 and the temperature stays constant.)
Q16. A gas expands at a constant pressure of 1.5 × 105 Pa, its volume increasing by 4.0 × 10−3 m3. During the expansion the gas absorbs 900 J of thermal energy. The increase in its internal energy is
- (A) +1500 J
- (B) +600 J
- (C) +900 J
- (D) +300 J
Show Answer
Answer: D – Work done by the gas = pΔV = (1.5 × 105) × (4.0 × 10−3) = 600 J, so W = −600 J. Then ΔU = q + W = 900 + (−600) = +300 J.
Q17. A gas is compressed at a constant pressure of 1.0 × 105 Pa and its volume decreases by 2.0 × 10−3 m3. The internal energy of the gas increases by 50 J. The thermal energy transfer q is
- (A) −150 J, so 150 J of thermal energy is transferred out of the gas
- (B) +250 J, so 250 J of thermal energy is transferred into the gas
- (C) +150 J, so 150 J of thermal energy is transferred into the gas
- (D) −250 J, so 250 J of thermal energy is transferred out of the gas
Show Answer
Answer: A – Work done on the gas: W = −pΔV = −(1.0 × 105) × (−2.0 × 10−3) = +200 J. From ΔU = q + W, q = ΔU − W = 50 − 200 = −150 J, meaning 150 J of thermal energy leaves the gas during the compression.
Q18. On a p-V diagram, a gas expands while its pressure falls steadily from 4.0 × 105 Pa to 2.0 × 105 Pa and its volume increases by 2.0 × 10−3 m3. The work done by the gas (the area under the graph) is approximately
- (A) 400 J
- (B) 600 J
- (C) 800 J
- (D) 1200 J
Show Answer
Answer: B – The area under a straight-line p-V graph is the average pressure multiplied by ΔV. Average pressure = (4.0 + 2.0) × 105 / 2 = 3.0 × 105 Pa, so the work done by the gas = (3.0 × 105) × (2.0 × 10−3) = 600 J.
Q19. A gas is heated inside a sealed, rigid container. Its pressure rises but its volume cannot change. The work done on or by the gas is
- (A) equal to pΔV and positive
- (B) positive, because the pressure increases
- (C) negative, because the gas pushes on the container walls
- (D) zero, because ΔV = 0
Show Answer
Answer: D – Work is only done when the volume changes. With ΔV = 0, W = 0 however large the pressure change, so all the thermal energy supplied goes into internal energy.
Q20. During an expansion, a gas absorbs 500 J of thermal energy and its internal energy increases by 120 J. The work done by the gas is
- (A) 620 J
- (B) 120 J
- (C) 380 J
- (D) 500 J
Show Answer
Answer: C – From ΔU = q + W, W = ΔU − q = 120 − 500 = −380 J. The work done ON the gas is −380 J, so the work done BY the gas is 380 J.
Special Changes
Q21. An ideal gas expands isothermally (at constant temperature). The change in its internal energy is
- (A) zero, because the internal energy of an ideal gas depends only on its temperature
- (B) positive, because its volume increases
- (C) negative, because the gas does work
- (D) equal to the thermal energy supplied to the gas
Show Answer
Answer: A – For an ideal gas there are no intermolecular forces, so the internal energy is the total kinetic energy of the molecules, which depends only on temperature. At constant temperature, ΔU = 0.
Q22. An ideal gas expands slowly and isothermally, doing 250 J of work on its surroundings. To keep its temperature constant, the gas must
- (A) release 250 J of thermal energy
- (B) absorb 250 J of thermal energy
- (C) absorb 500 J of thermal energy
- (D) have no thermal energy transfer at all
Show Answer
Answer: B – For the isothermal change ΔU = 0, and W = −250 J. So 0 = q + (−250), giving q = +250 J: the gas absorbs thermal energy exactly equal to the work it does.
Q23. An ideal gas is compressed isothermally and 180 J of work is done on it. The thermal energy transfer for the gas is
- (A) +180 J (180 J absorbed)
- (B) 0 J
- (C) +360 J (360 J absorbed)
- (D) −180 J (180 J released)
Show Answer
Answer: D – ΔU = 0 for the isothermal change, so q = −W = −180 J. The work done on the gas must escape as thermal energy to keep the temperature constant.
Q24. The barrel of a bicycle pump becomes warm when air is compressed quickly inside it. The compression is almost adiabatic and 90 J of work is done on the air. The change in internal energy of the air is
- (A) 0 J
- (B) −90 J
- (C) +90 J
- (D) +180 J
Show Answer
Answer: C – The compression is fast, so there is little time for thermal energy to escape: q ≈ 0 (adiabatic). Then ΔU = q + W = 0 + 90 = +90 J, and the air heats up.
Q25. An insulated gas expands rapidly and does 120 J of work, with no thermal energy supplied (the change is adiabatic). The change in internal energy and the temperature change are
- (A) ΔU = −120 J; the temperature falls
- (B) ΔU = −120 J; the temperature rises
- (C) ΔU = +120 J; the temperature falls
- (D) ΔU = 0 J; the temperature is unchanged
Show Answer
Answer: A – With q = 0, ΔU = W = −120 J: the work is done at the expense of the internal energy, so the gas cools. This is why gas escaping from a spray can or a tyre valve feels cold.
Q26. A gas is taken through a complete cycle of changes and returns to its initial pressure, volume and temperature. The total change in its internal energy over the whole cycle is
- (A) equal to the net work done by the gas
- (B) zero, because internal energy depends only on the state of the gas, which is the same at the start and end
- (C) equal to the total thermal energy absorbed during the cycle
- (D) impossible to find without more information
Show Answer
Answer: B – Internal energy is a function of state. The gas ends the cycle in exactly the state it started in, so ΔU = 0 over the cycle, whatever happens in between.
Q27. Over one complete cycle, a gas does a net 300 J of work on its surroundings. The net thermal energy absorbed by the gas over the cycle is
- (A) 0 J
- (B) −300 J
- (C) +300 J
- (D) +600 J
Show Answer
Answer: C – Over a cycle ΔU = 0. The net work done on the gas is W = −300 J (the gas does the work), so 0 = q + (−300) and q = +300 J: the net thermal energy absorbed equals the net work done by the gas.
Q28. A gas in a rigid container is supplied with 400 J of thermal energy. Its volume cannot change. The increase in its internal energy is
- (A) 0 J
- (B) less than 400 J
- (C) more than 400 J
- (D) +400 J, because no work is done
Show Answer
Answer: D – At constant volume W = 0 (since ΔV = 0), so ΔU = q + W = 400 + 0 = +400 J. All the thermal energy supplied raises the internal energy and the temperature.
Q29. Ice at 0 °C absorbs thermal energy and melts to form water at 0 °C. The temperature does not change, yet the internal energy of the H2O increases. The best explanation is that
- (A) the kinetic energy of the molecules increases
- (B) the thermal energy increases the potential energy of the molecules, as the ordered structure of the solid breaks down and the molecules separate against intermolecular forces
- (C) the ice does work on its surroundings as it melts
- (D) internal energy always changes when thermal energy is supplied, even at constant temperature, because of increased molecular kinetic energy
Show Answer
Answer: B – At constant temperature the average kinetic energy of the molecules is unchanged. This is a real substance, not an ideal gas, so its internal energy also includes molecular potential energy, and that increases during melting: the supplied latent heat (with W ≈ 0) goes into separating the molecules.
Q30. In a diesel engine, air is compressed very rapidly just before the fuel is injected, and it becomes hot enough to ignite the fuel. The rapid compression is best modelled as
- (A) adiabatic: q ≈ 0, so the work done on the air increases its internal energy and its temperature
- (B) isothermal: ΔU = 0, so the temperature of the air stays constant
- (C) a change with W = 0, because the piston moves too quickly for work to be done
- (D) a free expansion, so no work is done on the air
Show Answer
Answer: A – The compression is so fast that thermal energy has no time to escape, making it approximately adiabatic. Then ΔU = W > 0: the large work done on the air raises its internal energy and temperature sharply – the same effect used in a fire piston to ignite tinder.
Keep practising
Thermodynamics marks are mostly won or lost on signs, so keep using the convention until it feels automatic: q is positive into the system, W is positive for work done on the system. Revise every equation for this topic on the A Level Physics Formula Sheet, and find more topic-by-topic practice on the Physics MCQs hub.
These are original practice questions written for Quaidianpoint.com for revision purposes. They are not Cambridge past paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.