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A2 · Practice questions · Internal energy

Add up every molecule.

Six original Cambridge-style questions on what internal energy is, why it is a function of state, and how it follows the temperature of an ideal gas.

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

Internal energy is a total, not a temperature.

01
Analysis
[2 marks]

A student writes that "a hot brick contains more heat than a cold brick". Explain why this statement is wrong, and state the quantity the student should have used instead, giving its meaning.

  • Heat (thermal energy) is energy in transfer because of a temperature difference; it is not something a body stores, so a brick does not contain heat. ✓
  • The stored quantity is the internal energy: the sum of the random distribution of kinetic and potential energies of all the molecules of the brick. ✓
02
Calculation
[3 marks]

A sealed flask contains 2.5 × 10²² atoms of helium, which may be treated as an ideal gas, at a temperature of 290 K. Calculate the mean kinetic energy of one atom and hence the internal energy of the helium. Explain why no other term is needed in the total.

  • Mean kinetic energy = ₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J. ✓
  • U = N × ₃⁄₂ kT = 2.5 × 10²² × 6.00 × 10⁻²¹ = 150 J (1.5 × 10² J). ✓
  • An ideal gas has no intermolecular forces, so its molecules have no potential energy and the internal energy is entirely kinetic. ✓
03
Calculation
[3 marks]

A block of ice of mass 0.025 kg at 0 °C melts completely to water at 0 °C. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹, and the change in volume is negligible. Calculate the increase in the internal energy of the water, and explain in molecular terms which part of the internal energy has changed.

  • With no change in volume no work is done, so the increase in internal energy equals the thermal energy supplied: ΔU = mL = 0.025 × 3.34 × 10⁵ = 8.4 × 10³ J. ✓
  • The temperature is unchanged at 0 °C, so the mean random kinetic energy of the molecules is unchanged. ✓
  • The whole 8.4 kJ goes into molecular potential energy: the molecules are pulled apart against the intermolecular forces as the rigid lattice breaks up. ✓
04
Calculation
[3 marks]

A fixed mass of gas is taken from state A to state B by two different routes. On route 1 the gas receives 850 J of thermal energy and does 300 J of work on its surroundings. On route 2 the gas receives 700 J of thermal energy. Determine the work done by the gas on route 2, explaining the principle you use.

  • Route 1: ΔU = q + W = 850 + (−300) = +550 J, taking W as the work done on the gas. ✓
  • Internal energy is a function of state, so ΔU from A to B is the same by every route: route 2 also has ΔU = +550 J. ✓
  • Route 2: 550 = 700 + W gives W = −150 J, so the gas does 150 J of work on its surroundings. ✓
05
Analysis
[3 marks]

Explain, in terms of the molecules, why the internal energy of a fixed mass of ideal gas rises when its temperature rises, and why the internal energy of a real gas can change when it expands at constant temperature although that of an ideal gas cannot.

  • For an ideal gas the internal energy is the total random kinetic energy of the molecules, and the mean kinetic energy ₃⁄₂ kT is proportional to the thermodynamic temperature, so U rises with T. ✓
  • At constant temperature the mean kinetic energy is fixed, and an ideal gas has no molecular potential energy, so its internal energy is unchanged whatever happens to p and V. ✓
  • A real gas has intermolecular forces; expansion increases the molecular separation against the attractive forces, raising the molecular potential energy, so U changes even at constant T. ✓
06
Calculation
[4 marks]

A rigid cylinder holds 0.040 mol of argon, treated as an ideal monatomic gas, at 280 K. It is heated at constant volume until its temperature is 350 K.

(a) Calculate the mean kinetic energy of one argon atom at 280 K. (b) Calculate the increase in the internal energy of the gas. (c) State the thermal energy that was supplied, giving a reason.

  • (a) ₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 280 = 5.8 × 10⁻²¹ J. ✓
  • (b) U = N × ₃⁄₂ kT = ₃⁄₂ nRT for the whole sample, so ΔU = ₃⁄₂ nRΔT = 1.5 × 0.040 × 8.31 × (350 − 280). ✓
  • ΔU = 1.5 × 0.040 × 8.31 × 70 = 35 J. ✓
  • (c) At constant volume no work is done on or by the gas, so the thermal energy supplied equals the rise in internal energy: q = 35 J. ✓

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.