Skip to content
MarkSchemeMarkScheme
Sign up — no card

9702 · 16.2

The first law of thermodynamics — practice questions

Practice and worked examples for 9702 The first law of thermodynamics. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A gas in a cylinder absorbs 200 J of heat from its surroundings. Simultaneously, the gas expands, performing 70 J of work on the piston. Calculate the change in the internal energy of the gas.

Show solution outline
  1. Identify the system and processes: The system is the gas. It absorbs heat and expands.
  2. Determine the sign of Q: Since the gas absorbs heat, QQ is positive. Q=+200 JQ = +200\text{ J}.
  3. Determine the sign of W: Since the gas performs work (expands and pushes the piston), work is done by the system, so the work done on the system WW is negative. W=−70 JW = -70\text{ J}.
  4. Apply the First Law of Thermodynamics: ΔU=Q+W\Delta U = Q + W ΔU=(+200 J)+(−70 J)\Delta U = (+200\text{ J}) + (-70\text{ J}) ΔU=200 J−70 J\Delta U = 200\text{ J} - 70\text{ J} ΔU=+130 J\Delta U = +130\text{ J}
  5. State the conclusion: The internal energy of the gas increases by 130 J.

Worked example 2

A fixed mass of an ideal gas is held in a cylinder by a piston at a constant pressure of 2.5 x 10^5 Pa. The gas is heated, and its volume increases from 1.2 x 10^-3 m^3 to 1.9 x 10^-3 m^3. During this process, 450 J of thermal energy is supplied to the gas. Calculate the change in the internal energy of the gas.

Show solution outline
  1. Identify given values and signs:
    • Heat supplied to the gas, so QQ is positive: Q=+450 JQ = +450\text{ J}.
    • Pressure is constant: p=2.5×105 Pap = 2.5 \times 10^5\text{ Pa}.
    • Initial volume: Vi=1.2×10−3 m3V_i = 1.2 \times 10^{-3}\text{ m}^3.
    • Final volume: Vf=1.9×10−3 m3V_f = 1.9 \times 10^{-3}\text{ m}^3.
  2. Calculate the change in volume (ΔV\Delta V): ΔV=Vf−Vi=(1.9×10−3)−(1.2×10−3)=0.7×10−3 m3\Delta V = V_f - V_i = (1.9 \times 10^{-3}) - (1.2 \times 10^{-3}) = 0.7 \times 10^{-3}\text{ m}^3.
  3. Calculate the work done by the gas: The gas expands, so it does work on the surroundings. The work done by the gas is pΔVp \Delta V. Work done by gas = (2.5×105 Pa)×(0.7×10−3 m3)=175 J(2.5 \times 10^5\text{ Pa}) \times (0.7 \times 10^{-3}\text{ m}^3) = 175\text{ J}.
  4. Determine WW for the First Law equation: The First Law uses WW, the work done on the system. Since the gas expanded and did 175 J of work, the work done on the gas is negative. W=−175 JW = -175\text{ J}.
  5. Apply the First Law of Thermodynamics (ΔU=Q+W\Delta U = Q + W): ΔU=(+450 J)+(−175 J)\Delta U = (+450\text{ J}) + (-175\text{ J}) ΔU=275 J\Delta U = 275\text{ J}.
  6. State the conclusion: The internal energy of the gas increases by 275 J.
Mark a 9702 question on this topicFull lessonOpen in course studio