A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Thermodynamic Processes and the First Law of Thermodynamics — mark scheme explained
The short answer
In this section, we will explore different types of thermodynamic processes: isothermal, adiabatic, constant pressure (isobaric), and constant volume (isochoric) changes. We will also apply the first law of thermodynamics to these processes and understand how work is done in each case. 1. Isothermal Process An isothermal process occurs at a constant temperature.
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
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculations, show all steps clearly and include units. For explanations, be concise but thorough, using key terms and equations where appropriate. Always check your final answer for reasonableness.
What the command words demand
- Calculate
- Perform a numerical calculation to find the answer.
- Explain
- Provide a clear and concise explanation of the concept or process.
- Derive
- Show the steps involved in deriving an equation or relationship.
- Compare
- Identify similarities and differences between two processes or concepts.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 5-7 minutes per question to ensure you have enough time to work through the problem methodically.
- Use the relationship for an isothermal process: pV = constant.1 mark
- Calculate the final pressure using p 1 V 1 = p 2 V 2 : 100 kPa × 2 m 3 = p 2 × 4 m 3 .1 mark
- Solve for p 2 : p 2 = (100 kPa × 2 m 3 ) / 4 m 3 = 50 kPa.1 mark
- 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.0 marks
- 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).1 mark
- Calculate the work: W = 200 kJ × ln(2) ≈ 138.6 kJ.1 mark
Final answer: 138.6 kJ
Work through every step correctly and you earn all 5 marks.
Another worked example
A gas undergoes an adiabatic compression from a volume of 0.5 m 3 to 0.25 m 3 . If the initial pressure is 100 kPa and γ = 1.4, calculate the final pressure.
- Use the relationship for an adiabatic process: pV γ = constant.1 mark
- Calculate the final pressure using p 1 V 1 γ = p 2 V 2 γ : 100 kPa × (0.5 m 3 ) 1.4 = p 2 × (0.25 m 3 ) 1.4 .1 mark
- Solve for p 2 : p 2 = [100 kPa × (0.5 m 3 ) 1.4 ] / (0.25 m 3 ) 1.4 .0 marks
- Calculate the final pressure: p 2 = 100 kPa × 2 1.4 = 100 × 2.639 ≈ 264 kPa.2 marks
Final answer: 264 kPa
Work through every step correctly and you earn all 4 marks.
Common mistakes
Forgetting that internal energy is a state function and depends only on the initial and final states, not the path taken.
Why it happens: Students often confuse the change in internal energy with the work done or heat added during a process. Internal energy is a state function, meaning it depends only on the initial and final states of the system, not the path taken between them.
Fix: 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.
Why it happens: Students sometimes overlook one of the terms in the first law equation (ΔU = Q - W) and make errors in their calculations.
Fix: 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.
Why it happens: Each process has a specific relationship between p, V, and T. Students can get confused and use the wrong equation for the given process.
Fix: 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.
Why it happens: Students may not fully understand that for an ideal gas at constant temperature, the internal energy does not change (ΔU = 0).
Fix: 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.
Why it happens: Students sometimes think that since there is no heat exchange (Q = 0) in an adiabatic process, the internal energy does not change. However, work can still be done, which affects ΔU.
Fix: 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.
Why it happens: Students may use a generic value for γ without considering the type of gas (monatomic, diatomic, etc.).
Fix: 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
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate work done | Find the work done by a gas during isothermal expansion using pressure and volume. | 5 |
| Adiabatic Process Calculation | Find the final pressure of a gas after an adiabatic compression using pV^γ. | 4 |
| Calculate Work Done | Use W = pΔV to find the work done during a constant pressure gas expansion. | 3 |
| First Law Thermodynamics | Apply the first law to find internal energy change during a constant volume process. | 2 |
| First Law Thermodynamics | Apply the first law of thermodynamics to find heat added during an isothermal process. | 2 |
| Total across these question types | 16 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.