Non-flow processes (A-level only) — AQA A-Level Physics
Teaching note checked against the official specification.
A non-flow process is a thermodynamic change where no working fluid enters or leaves the system during the process, so the mass of the system stays constant. In AQA A-Level Physics you analyse these using the first law of thermodynamics (delta U = Q - W) and the p-V diagram, distinguishing isothermal, adiabatic, isobaric and isochoric changes.
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Non-flow processes (A-level only) explained
Non-flow processes (A-level only) covers 2 specification points, including Isothermal, adiabatic, constant pressure and… and PV = constant isothermal change : pV =….
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Open a point below for its explanation and revision guidance.
Isothermal, adiabatic, constant pressure and constant volume changes.
A non-flow process changes the state of a fixed mass of gas in a cylinder. Isothermal means constant temperature, so internal energy of an ideal gas stays fixed and any heat added equals work done by the gas. Adiabatic means no heat enters or leaves, so work done changes internal energy and temperature. Constant pressure (isobaric) heating lets the gas expand, doing work pΔV while temperature rises. Constant volume (isochoric) heating does no work because ΔV is zero, so all heat raises internal energy and temperature. Link each name to the quantity held fixed, then to the first law.
pV = nRT γ adiabatic change : pV = constant isothermal change : pV = constant at constant pressure W = pΔV Application of first law of thermodynamics to the above processes.
The ideal gas equation pV = nRT links pressure, volume, amount and temperature. For an isothermal change, T is fixed so pV is constant, giving a smooth p–V curve. For an adiabatic change, no heat flows, so pV^γ is constant, where γ = C_p/C_V; the curve is steeper than an isotherm. At constant pressure, work done by the gas is W = pΔV. The first law ΔU = Q − W then explains each path: isothermal has ΔU = 0 so Q = W; adiabatic has Q = 0 so ΔU = −W; constant pressure has both heat and work; constant volume has W = 0 so Q = ΔU.
Your focus
- Define each non-flow process by the quantity held constant.
- Apply ΔU = Q − W correctly to isothermal, adiabatic, isobaric and isochoric changes.
- Interpret p–V diagrams showing each type of process.
Show all 6 objectives
- Use pV = nRT and pV^γ = constant to solve gas-process problems.
- Calculate work done at constant pressure using W = pΔV.
- Apply the first law to isothermal, adiabatic, constant pressure and constant volume changes.
Which command word is this?
Give a brief, factual answer with no explanation or calculation; for example, 'State the value of Q for an adiabatic process' requires simply 'zero'.
Non-flow processes (A-level only) exam tips
Topic Overview
Non-flow processes describe thermodynamic changes in a fixed mass of fluid where no fluid crosses the system boundary. In AQA A-Level Physics you study four idealised processes on p-V diagrams: isothermal (constant T), adiabatic (Q = 0), isobaric (constant p) and isochoric (constant V). Each is analysed using the first law of thermodynamics, delta U = Q - W, which is the energy conservation statement at the heart of this topic.
This topic matters because it connects macroscopic gas behaviour to energy transfers in engines, refrigerators and atmospheric processes. It builds directly on the ideal gas equation and kinetic theory, and it prepares you for more advanced work on heat engines, efficiency and entropy in later study. Exam questions frequently require you to interpret p-V graphs, calculate work from the area under the curve, and justify temperature changes using the first law.
Key Concepts
- →First law of thermodynamics: delta U = Q - W, where Q is energy transferred to the system by heating and W is work done by the system on the surroundings.
- →Isothermal process: temperature constant, so delta U = 0 for an ideal gas and Q = W; the p-V curve is a rectangular hyperbola with pV = constant.
- →Adiabatic process: no heat transfer (Q = 0), so delta U = -W; compression raises temperature and expansion lowers it.
- →Isobaric process: pressure constant, so work done W = p delta V and the p-V graph is a horizontal line.
- →Isochoric process: volume constant, so W = 0 and delta U = Q; the p-V graph is a vertical line and all energy goes into changing internal energy.
Marking Points
- Identifies the quantity held constant in each process: temperature for isothermal, heat flow for adiabatic, pressure for isobaric, volume for isochoric.
- Applies the first law ΔU = Q − W to each process, noting W = pΔV only when pressure is constant.
- Explains that for an ideal gas isothermal implies ΔU = 0, so Q = W, while adiabatic implies Q = 0, so ΔU = −W.
- Distinguishes constant volume, where W = 0 and Q = ΔU, from constant pressure, where both Q and W are non-zero.
- Uses a p–V diagram to represent each path: isothermal and adiabatic curves, horizontal isobar, vertical isochore.
- Relates the microscopic picture, such as molecular speed distribution, to temperature changes in each process.
- Uses pV = nRT to relate state variables and to show that pV is constant when T is constant.
- States and applies pV^γ = constant for an adiabatic change, identifying γ as the ratio of principal molar heat capacities.
- Calculates work done at constant pressure using W = pΔV, with consistent units.
- Applies ΔU = Q − W to each process, setting the appropriate term to zero.
- Compares isothermal and adiabatic curves on a p–V diagram, noting the adiabatic curve is steeper through the same point.
- Uses the first law to deduce temperature change from work done when Q = 0.
Examiner Tips
- 💡State the fixed quantity first, then write the first law with the correct term set to zero.
- 💡Sketch the p–V path and shade the work area to justify the sign of W.
- 💡Use the words system and surroundings to make clear where heat and work cross the boundary.
- 💡Quote the equation you use and state the condition that makes it valid.
- 💡Check units of p and V before substituting into W = pΔV.
- 💡For adiabatic questions, start from Q = 0 and link work done to internal energy change.
- 💡Always write the first law as delta U = Q - W and define your sign convention clearly before substituting values; many marks are lost through inconsistent signs.
- 💡When interpreting a p-V diagram, state the shape of the curve and what it tells you about the process (for example, a vertical line means W = 0 because delta V = 0).
- 💡For calculation questions, show the substitution with units and give the final answer to an appropriate number of significant figures with the correct unit, such as J or Pa.
Common Mistakes
- Saying an adiabatic process must be at constant temperature; correction: adiabatic means Q = 0, so temperature generally changes unless the process is also isothermal.
- Using W = pΔV for every process; correction: this expression applies only when pressure is constant, so use the area under the p–V curve otherwise.
- Assuming constant volume means constant pressure; correction: at constant volume the gas is sealed, pressure can rise as temperature rises.
- Writing pV = constant for an adiabatic change; correction: use pV^γ = constant, where γ is greater than 1.
- Treating W = pΔV as valid for all processes; correction: it applies only at constant pressure, otherwise use the area under the p–V curve.
- Forgetting that γ is dimensionless and greater than 1 for gases; correction: γ = C_p/C_V and is about 1.4 for diatomic gases.
- Students think that if temperature is constant then no energy is transferred. Correction: in an isothermal process Q is not zero; energy enters or leaves by heating to balance the work done, keeping delta U = 0.
- Students believe adiabatic means constant temperature. Correction: adiabatic means Q = 0, so any work done changes the internal energy and therefore the temperature of an ideal gas.
- Students use pV = constant for adiabatic changes. Correction: pV = constant applies only to isothermal changes; for adiabatic changes of an ideal gas, pV^gamma = constant where gamma = Cp/Cv.
Revision Plan
- 1Day 1-2: Revise the first law of thermodynamics and the ideal gas equation, then write definitions of isothermal, adiabatic, isobaric and isochoric processes with their p-V graph shapes.
- 2Day 3-4: Practise sketching and interpreting p-V diagrams for each process, labelling the area under the curve as the work done and identifying where Q, W and delta U are zero.
- 3Day 5-7: Work through calculation questions on isothermal and adiabatic changes, including pV = constant and pV^gamma = constant, checking units and significant figures.
- 4Day 8-10: Attempt past-paper questions on non-flow processes, then mark them using the official mark scheme and write down every term or justification you missed.
- 5Day 11-14: Create a one-page summary sheet linking the four processes to the first law, and test yourself using active recall prompts and flashcards without notes.
Exam Question Types
- 📋Calculation questions using pV = constant for isothermal changes or pV^gamma = constant for adiabatic changes, often asking for a final pressure or volume; always convert units to SI and show working.
- 📋Explanation questions requiring you to use delta U = Q - W to justify a temperature change, such as why a gas cools during adiabatic expansion; quote the law and link each term to the situation.
- 📋Graph interpretation questions where you identify the process from a p-V curve and calculate work done from the area under the curve; state whether work is done on or by the gas.
- 📋Six-mark structured questions combining a calculation with an explanation, for example comparing isothermal and adiabatic expansions; plan your answer to cover definition, equation, substitution and conclusion.
Command Word Expectations (AQA)
Give a brief, factual answer with no explanation or calculation; for example, 'State the value of Q for an adiabatic process' requires simply 'zero'.
Give a reasoned justification linking cause and effect, using correct terminology such as 'internal energy', 'work done by the gas' and 'energy transferred by heating'; marks are awarded for each valid point in a logical chain.
Show the equation, substitute the correct values with units, and give the final answer to an appropriate number of significant figures with the correct unit; method marks are available even if the final answer is wrong.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A fixed mass of ideal gas is compressed slowly at constant temperature from 2.0 x 10^-3 m^3 to 0.5 x 10^-3 m^3. The initial pressure is 1.0 x 10^5 Pa. Calculate the final pressure and state the energy transfer that occurs.
- 1.Step 1: Identify that the process is isothermal (constant temperature) and the mass is fixed, so Boyle's law applies: p1 V1 = p2 V2.
- 2.Step 2: Rearrange for p2: p2 = p1 V1 / V2 = (1.0 x 10^5 x 2.0 x 10^-3) / (0.5 x 10^-3).
- 3.Step 3: Calculate p2 = 4.0 x 10^5 Pa. Since delta U = 0 for an isothermal change, Q = W: work is done on the gas and an equal amount of energy is transferred from the gas by heating.
Question: A gas in a cylinder expands adiabatically against a constant external pressure of 1.5 x 10^5 Pa, increasing its volume from 1.0 x 10^-3 m^3 to 3.0 x 10^-3 m^3. Explain what happens to the internal energy and temperature of the gas.
- 1.Step 1: For an adiabatic process Q = 0, so the first law becomes delta U = -W.
- 2.Step 2: Work done by the gas W = p delta V = 1.5 x 10^5 x (3.0 x 10^-3 - 1.0 x 10^-3) = 300 J.
- 3.Step 3: Since W = +300 J (work done by the gas), delta U = -300 J. Internal energy decreases, so for an ideal gas the temperature decreases.