AQA A-Level Physics: Thermodynamic Processes and the First Law of Thermodynamics — mark scheme explained

The question

A gas undergoes an isothermal expansion from a volume of 2 m 3 to 4 m 3 . If the initial pressure is 100 kPa, calculate the work done by the gas. [Paraphrased for study — not reproduced from any exam paper.]

Mark scheme, decoded

What each mark is really for — in plain English — and the wording trap that loses it.

• S1

  Use the relationship for an isothermal process: pV = constant.

• S2

  Calculate the final pressure using p 1 V 1 = p 2 V 2 : 100 kPa × 2 m 3 = p 2 × 4 m 3 .

• S3

  Solve for p 2 : p 2 = (100 kPa × 2 m 3 ) / 4 m 3 = 50 kPa.

• S4

  Note: the isothermal work formula W = nRT ln(V 2 /V 1 ) is beyond the AQA A-level specification (only W = pΔV for constant-pressure work is examinable). It is shown here for context only.

• S5

  Since pV = nRT, we can use the initial conditions: W = p 1 V 1 ln(V 2 /V 1 ) = 100 kPa × 2 m 3 × ln(4/2).

• S6

  Calculate the work: W = 200 kJ × ln(2) ≈ 138.6 kJ.

Model answer

Worked through, with each step tagged to the mark it earns.

1. S1

   Use the relationship for an isothermal process: pV = constant.

2. S2

   Calculate the final pressure using p 1 V 1 = p 2 V 2 : 100 kPa × 2 m 3 = p 2 × 4 m 3 .

3. S3

   Solve for p 2 : p 2 = (100 kPa × 2 m 3 ) / 4 m 3 = 50 kPa.

4. S4

   Note: the isothermal work formula W = nRT ln(V 2 /V 1 ) is beyond the AQA A-level specification (only W = pΔV for constant-pressure work is examinable). It is shown here for context only.

5. S5

   Since pV = nRT, we can use the initial conditions: W = p 1 V 1 ln(V 2 /V 1 ) = 100 kPa × 2 m 3 × ln(4/2).

6. S6

   Calculate the work: W = 200 kJ × ln(2) ≈ 138.6 kJ.

7. Final answer: 138.6 kJ

Common mistakes

• Forgetting that internal energy is a state function and depends only on the initial and final states, not the path taken. — Always remember that ΔU = Q - W and focus on the initial and final states when calculating internal energy changes.
• Misapplying the first law of thermodynamics by forgetting to account for work done or heat added. — Double-check that you have included both the heat added (Q) and the work done (W) when applying the first law of thermodynamics.
• Confusing the relationships between pressure, volume, and temperature in different processes. — Memorize the key equations for each process: isothermal (pV = constant), adiabatic (pV γ = constant), isobaric (W = pΔV), and isochoric (Q = ΔU).
• Forgetting that in an isothermal process, the internal energy change is zero. — Always remember that ΔU = 0 for an isothermal process and use this to simplify your calculations.
• Misinterpreting the adiabatic process as having no change in internal energy. — In an adiabatic process, Q = 0, so ΔU = -W. The internal energy changes due to the work done by or on the system.
• Using the wrong value for γ in adiabatic processes. — Always check the specific heat ratio (γ) for the type of gas you are dealing with. For monatomic gases, γ = 5/3; for diatomic gases, γ ≈ 7/5.

Where the marks go

• Full worked solution (all marking points)5 marks

Related questions

• AQA A-Level Physics: Fundamental and Derived Units in Physics — mark scheme explained →
• AQA A-Level Physics: Errors and Uncertainties in Measurements — mark scheme explained →