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    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

    1. Define each non-flow process by the quantity held constant.
    2. Apply ΔU = Q − W correctly to isothermal, adiabatic, isobaric and isochoric changes.
    3. Interpret p–V diagrams showing each type of process.
    Show all 6 objectives
    1. Use pV = nRT and pV^γ = constant to solve gas-process problems.
    2. Calculate work done at constant pressure using W = pΔV.
    3. 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
    1. 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.
    2. 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.
    3. 3Day 5-7: Work through calculation questions on isothermal and adiabatic changes, including pV = constant and pV^gamma = constant, checking units and significant figures.
    4. 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.
    5. 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)
    State

    Give a brief, factual answer with no explanation or calculation; for example, 'State the value of Q for an adiabatic process' requires simply 'zero'.

    Explain

    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.

    Calculate

    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)
    Pitfall: Students confuse the sign convention for work done and heat transfer, or assume that Q = 0 for every non-flow process. A common error is to treat an isothermal expansion as adiabatic simply because the temperature is constant.
    ❌ Weak Answer (Loses Marks):In an isothermal expansion the gas does work, so the internal energy decreases and the temperature falls.
    Example improved answer:In an isothermal expansion of an ideal gas, the temperature remains constant, so the internal energy does not change (delta U = 0). By the first law, delta U = Q - W, therefore Q = W. The gas does work W on the surroundings and an equal amount of energy Q is transferred into the gas by heating, so the temperature stays constant.
    Examiner Tip: Always state the sign convention you are using (work done BY the gas is positive) and explicitly link each term in delta U = Q - W to the physical situation before substituting numbers.
    Pitfall: In adiabatic processes students forget that Q = 0 but W is not zero, and they incorrectly use pV = constant instead of pV^gamma = constant. They also fail to explain why the temperature changes when no heat is supplied.
    ❌ Weak Answer (Loses Marks):In an adiabatic compression no heat enters or leaves, so the temperature stays the same because Q = 0.
    Example improved answer:For an adiabatic process Q = 0, so the first law gives delta U = -W. When a gas is compressed adiabatically, work is done ON the gas, so W is negative and delta U is positive. The internal energy increases, and since internal energy of an ideal gas depends only on temperature, the temperature rises even though no heat is transferred.
    Examiner Tip: Use the phrase 'work is done on the gas' for compression and 'work is done by the gas' for expansion, and always justify the temperature change using delta U = Q - W rather than quoting pV^gamma without explanation.
    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. 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. 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. 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.
    Final Answer: Final pressure = 4.0 x 10^5 Pa. The gas is compressed isothermally, so work is done on the gas and energy is transferred out of the gas by heating to keep the temperature constant.

    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. 1.Step 1: For an adiabatic process Q = 0, so the first law becomes delta U = -W.
    2. 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. 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.
    Final Answer: The internal energy decreases by 300 J and the temperature of the gas falls, because the gas does 300 J of work with no heat input.
    Active Recall Memory Test
    State the first law of thermodynamics and define each symbol with its sign convention.
    Key Fact: delta U = Q - W, where delta U is the change in internal energy, Q is energy transferred to the system by heating, and W is work done by the system on the surroundings.
    For an adiabatic process, what are the values of Q and delta U in terms of W?
    Key Fact: Q = 0, so delta U = -W. If the gas does work (W positive), internal energy decreases; if work is done on the gas (W negative), internal energy increases.
    What is the shape of an isothermal curve on a p-V diagram and what equation describes it?
    Key Fact: A rectangular hyperbola described by pV = constant, because temperature is constant for a fixed mass of ideal gas.
    Why does the temperature of an ideal gas fall during an adiabatic expansion?
    Key Fact: Q = 0 and the gas does work on the surroundings, so delta U = -W is negative. Internal energy decreases, and since internal energy depends only on temperature for an ideal gas, the temperature falls.
    Frequently Asked Questions
    What is a non-flow process in A-Level Physics?
    A non-flow process is a thermodynamic change in which a fixed mass of fluid remains inside the system and no fluid crosses the boundary. This means the mass is constant throughout, so you can apply the ideal gas equation and the first law of thermodynamics directly to the system. Examples include the compression of gas in a sealed cylinder or the expansion of gas in a closed piston. AQA exam questions often present these processes on p-V diagrams and ask you to identify whether they are isothermal, adiabatic, isobaric or isochoric.
    What is the difference between an isothermal and an adiabatic process?
    In an isothermal process the temperature stays constant, so for an ideal gas delta U = 0 and Q = W; energy must be transferred by heating to balance the work done. In an adiabatic process no energy is transferred by heating, so Q = 0 and delta U = -W; any work done changes the internal energy and therefore the temperature. On a p-V diagram an isothermal curve follows pV = constant, while an adiabatic curve is steeper and follows pV^gamma = constant. A common exam trap is assuming that constant temperature means no heat transfer, which is incorrect.
    How do I calculate the work done in a non-flow process?
    For an isobaric process at constant pressure, work done by the gas is W = p delta V, where delta V is the change in volume in cubic metres. For other processes, the work done equals the area under the p-V curve, which you may need to estimate by counting squares or using the equation of the curve. In an isochoric process delta V = 0, so no work is done. Always state whether work is done by the gas (positive) or on the gas (negative) and give your answer in joules.
    Why does a gas cool when it expands adiabatically?
    In an adiabatic expansion no energy enters or leaves by heating, so Q = 0. The gas does work on the surroundings as it expands, so W is positive and the first law gives delta U = -W, meaning the internal energy decreases. For an ideal gas, internal energy is directly proportional to absolute temperature, so a decrease in internal energy means the temperature falls. This is why rising air in the atmosphere cools and can lead to cloud formation.
    What does pV^gamma = constant mean and when do I use it?
    The equation pV^gamma = constant describes an adiabatic change for an ideal gas, where gamma is the ratio of specific heat capacities Cp/Cv. You use it when a gas expands or is compressed with no heat transfer, such as in a well-insulated cylinder or a rapid process. For monatomic gases gamma is about 1.67, and for diatomic gases it is about 1.40. Do not use pV = constant for adiabatic changes; that equation only applies to isothermal processes.
    How is the first law of thermodynamics used in AQA exam questions on non-flow processes?
    AQA exam questions expect you to write delta U = Q - W and then link each term to the specific process. For an isothermal change you state delta U = 0 and conclude Q = W; for an adiabatic change you state Q = 0 and conclude delta U = -W; for an isochoric change you state W = 0 and conclude delta U = Q. Marks are awarded for correct sign conventions and for explaining the resulting temperature change in terms of internal energy. Always define whether work is done by or on the gas before substituting values.