Home · A-Level 9702 · Thermodynamics · The first law of thermodynamics · Practice
A2 · Practice questions · The first law of thermodynamics

Get the signs right.

Six original Cambridge-style questions on work done at constant pressure, the first law ΔU = q + W with its sign convention, and the constant-volume, isothermal and adiabatic special cases.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Increase in internal energy = heating of the system + work done on the system.

01
Analysis
[3 marks]

A student writes the first law of thermodynamics as ΔU = q − W. State the first law in the form used in the syllabus, defining the sign convention for each term, and explain under what interpretation of W the student's version gives the same numerical results.

  • The increase in internal energy of a system equals the thermal energy transferred to the system plus the work done on the system: ΔU = q + W. ✓
  • q is positive when the system is heated and negative when it loses thermal energy; W is positive when work is done on the system (a compression) and negative when the system does work (an expansion). ✓
  • The student's version is only correct if W there means the work done by the system, which is the negative of the work done on it; the two statements then describe the same energy balance. ✓
02
Calculation
[3 marks]

A gas in a cylinder is compressed at a constant pressure of 2.4 × 10⁵ Pa from a volume of 3.5 × 10⁻³ m³ to 2.0 × 10⁻³ m³. During the compression 180 J of thermal energy leaves the gas through the cylinder walls. Calculate the work done on the gas and the change in its internal energy, and state whether the temperature of the gas rises or falls.

  • Work done on the gas W = pΔV = 2.4 × 10⁵ × (3.5 − 2.0) × 10⁻³ = +360 J (positive, because the volume decreases). ✓
  • ΔU = q + W = (−180) + 360 = +180 J. ✓
  • The internal energy increases, so the temperature of the gas rises even though thermal energy left it. ✓
03
Calculation
[3 marks]

At atmospheric pressure, 1.01 × 10⁵ Pa, 1.0 g of water at 100 °C is turned into 1.0 g of steam at 100 °C. The water occupies 1.0 × 10⁻⁶ m³ and the steam occupies 1.67 × 10⁻³ m³. The specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. Calculate the thermal energy supplied, the work done on the system, and the increase in its internal energy.

  • q = mL = 1.0 × 10⁻³ × 2.26 × 10⁶ = 2260 J (2.3 × 10³ J). ✓
  • The steam pushes the atmosphere back, so the work done on the system is W = −pΔV = −1.01 × 10⁵ × (1.67 × 10⁻³ − 1.0 × 10⁻⁶) = −169 J. ✓
  • ΔU = q + W = 2260 − 169 = 2.09 × 10³ J, so about 93% of the energy supplied goes into separating the molecules and only 7% into pushing back the air. ✓
04
Analysis
[2 marks]

When the valve of a cylinder of compressed carbon dioxide is opened, the gas escapes so rapidly that frost forms on the nozzle. Use the first law of thermodynamics to explain why the escaping gas becomes cold.

  • The expansion is so rapid that there is no time for thermal energy to enter the gas, so q ≈ 0, while the gas does work pushing back the surrounding air, so the work done on the gas W is negative. ✓
  • ΔU = q + W is therefore negative: the internal energy falls, and since the internal energy of a gas depends on its temperature, the temperature drops low enough to freeze water vapour on the nozzle. ✓
05
Calculation
[3 marks]

A fixed mass of gas is taken round a cycle of three stages and returns to its starting state. In stage 1 the gas receives 420 J of thermal energy and does 260 J of work. In stage 2 the gas is cooled at constant volume, losing 150 J of thermal energy. In stage 3, 190 J of work is done on the gas to return it to its starting state. Determine the thermal energy transferred in stage 3, stating its direction.

  • Stage 1: ΔU₁ = 420 + (−260) = +160 J. Stage 2: W = 0 at constant volume, so ΔU₂ = −150 J. ✓
  • Internal energy is a function of state, so over the complete cycle ΔU = 0: ΔU₃ = −(160 − 150) = −10 J. ✓
  • Stage 3: ΔU₃ = q₃ + W₃ gives −10 = q₃ + 190, so q₃ = −200 J: the gas gives out 200 J of thermal energy. (Check: net q = 420 − 150 − 200 = +70 J equals the net work done by the gas, 260 − 190 = 70 J.) ✓
06
Calculation
[4 marks]

A cylinder fitted with a frictionless piston contains 0.030 mol of an ideal monatomic gas at a constant pressure of 1.0 × 10⁵ Pa. The gas is heated slowly from 300 K to 400 K. The internal energy of the gas is U = ₃⁄₂ nRT.

(a) Use pV = nRT to calculate the increase in the volume of the gas. (b) Calculate the work done on the gas. (c) Calculate the thermal energy supplied to the gas.

  • (a) At constant pressure ΔV = nRΔT / p = 0.030 × 8.31 × 100 / (1.0 × 10⁵) = 2.5 × 10⁻⁴ m³. ✓
  • (b) The gas expands, so the work done on it is W = −pΔV = −1.0 × 10⁵ × 2.49 × 10⁻⁴ = −25 J. ✓
  • (c) ΔU = ₃⁄₂ nRΔT = 1.5 × 0.030 × 8.31 × 100 = 37.4 J. ✓
  • q = ΔU − W = 37.4 − (−24.9) = 62 J, more than at constant volume because 25 J of the energy supplied leaves again as work done by the gas. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.