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9702 · 16.1

Internal energy — practice questions

Practice and worked examples for 9702 Internal energy. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A gas in a cylinder absorbs 300 J of heat from its surroundings. At the same time, the gas expands, doing 120 J of work on the piston. Calculate the change in the internal energy of the gas.

Show solution outline
  1. Identify given values with correct signs:
    • Heat absorbed by the gas (enters system), so Q is positive: Q = +300 J.
    • Work done by the gas (expansion), so W is negative: W = -120 J.
  2. Apply the First Law of Thermodynamics:
    • ΔU=Q+W\Delta U = Q + W
  3. Substitute the values:
    • ΔU=(+300 J)+(−120 J)\Delta U = (+300 \text{ J}) + (-120 \text{ J})
    • ΔU=300 J−120 J\Delta U = 300 \text{ J} - 120 \text{ J}
    • ΔU=+180 J\Delta U = +180 \text{ J}
  4. State the conclusion:
    • The internal energy of the gas increases by 180 J.

Worked example 2

A piston compresses a gas in an insulated cylinder, doing 500 J of work on the gas. During the compression, the gas loses 200 J of heat to the surroundings. Calculate the change in the internal energy of the gas.

Show solution outline
  1. Identify given values with correct signs:
    • Work is done on the gas (compression), so W is positive: W = +500 J.
    • Heat is lost by the gas (leaves system), so Q is negative: Q = -200 J.
  2. Apply the First Law of Thermodynamics:
    • ΔU=Q+W\Delta U = Q + W
  3. Substitute the values:
    • ΔU=(−200 J)+(+500 J)\Delta U = (-200 \text{ J}) + (+500 \text{ J})
    • ΔU=300 J\Delta U = 300 \text{ J}
  4. State the conclusion:
    • The internal energy of the gas increases by 300 J.
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