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    First law of thermodynamics (A-level only) — AQA A-Level Physics

    Teaching note checked against the official specification.

    The first law of thermodynamics states that the change in internal energy of a system equals the sum of the heat supplied to the system and the work done on the system: delta U = Q + W. It is a statement of conservation of energy applied to thermodynamic processes, where W is positive when work is done on the gas (compression) and negative when the gas does work (expansion).

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    First law of thermodynamics (A-level only) explained

    First law of thermodynamics (A-level only) covers 2 specification points, including Quantitative treatment of first law of… and Applications of first law of thermodynamics..

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    Open a point below for its explanation and revision guidance.

    Quantitative treatment of first law of thermodynamics, Q = ∆U + W where Q is energy transferred to the system by heating, ∆U is increase in internal energy and W is work done by the system.

    The first law is energy conservation for a thermodynamic system: Q = ∆U + W. Q is energy transferred to the system by heating; ∆U is the increase in internal energy; W is work done by the system. If a gas absorbs 500 J by heating and does 200 J of work expanding, then ∆U = Q − W = 500 J − 200 J = +300 J, so internal energy rises by 300 J and temperature rises for an ideal gas. If 300 J of work is done on the gas and it loses 100 J by heating, Q = −100 J and W = −300 J, giving ∆U = +200 J. Signs matter: heating to the system is positive Q; work done by the system is positive W. Use Q = ∆U + W to find any term, and state the sign convention explicitly.

    Applications of first law of thermodynamics.

    Apply Q = ∆U + W to real processes. In an isothermal expansion of an ideal gas, ∆U = 0, so Q = W: heating supplies the energy for the gas to do work. In adiabatic compression, Q = 0, so ∆U = −W; work done on the gas increases internal energy and temperature. In a rigid container, W = 0, so Q = ∆U. In free expansion into a vacuum, W = 0 and Q = 0, so ∆U = 0. For a cyclic process, ∆U = 0 over a full cycle, so net Q = net W. Use these cases to deduce energy transfers, temperature changes and work done, and to explain everyday examples such as a bicycle pump warming up or a gas cylinder cooling as it discharges.

    Your focus

    1. State the first law of thermodynamics and define each term with its sign convention.
    2. Apply Q = ∆U + W quantitatively to find an unknown energy transfer or change in internal energy.
    3. Interpret positive and negative values of Q and W in terms of heating, cooling, expansion and compression.
    Show all 6 objectives
    1. Apply the first law to isothermal, adiabatic and constant-volume processes.
    2. Deduce energy transfers and temperature changes in cyclic and free-expansion processes.
    3. Explain practical thermodynamic examples using the first law with correct sign conventions.

    Which command word is this?

    A concise, factual answer is required, often a definition or a simple statement of a law. For example, 'State the first law of thermodynamics.' Expected answer: 'The change in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.' No explanation needed.

    First law of thermodynamics (A-level only) exam tips

    Topic Overview

    The first law of thermodynamics is a fundamental principle in A-level Physics that extends the concept of energy conservation to thermal systems. It relates the change in internal energy of a system to the heat transferred to or from the system and the work done on or by the system. This law is essential for understanding how energy is transferred and transformed in processes ranging from car engines to biological systems.

    In the AQA A-level specification, this topic is part of the thermal physics section and is examined through calculations and explanations of thermodynamic processes. It builds on prior knowledge of energy transfer, work, and the kinetic theory of gases, and it provides a foundation for more advanced topics such as entropy and the second law of thermodynamics. Mastery of this law is crucial for analysing real-world applications and for achieving high marks in exam questions that require both quantitative and qualitative reasoning.

    Key Concepts
    • →Internal energy is the sum of the random kinetic and potential energies of the particles in a system, and for an ideal gas, it depends only on temperature.
    • →The first law of thermodynamics: delta U = Q + W, where delta U is the change in internal energy, Q is the heat supplied to the system, and W is the work done on the system.
    • →Sign convention: Q is positive when heat is supplied to the system, negative when lost; W is positive when work is done on the system (compression), negative when work is done by the system (expansion).
    • →Work done on or by a gas at constant pressure is given by W = -p delta V, where p is the external pressure and delta V is the change in volume.
    • →For an ideal gas, an isothermal process has delta U = 0, so Q = -W; an adiabatic process has Q = 0, so delta U = W.
    Marking Points
    • States the first law as Q = ∆U + W and identifies each symbol with its meaning and sign convention.
    • Applies the equation to a numerical problem, substituting Q, ∆U and W with correct signs and solving for the unknown.
    • Explains that ∆U is the increase in internal energy and that for an ideal gas it is directly proportional to the change in thermodynamic temperature.
    • Distinguishes between work done by the system (positive W) and work done on the system (negative W), and between heating to the system (positive Q) and cooling (negative Q).
    • Uses the law to deduce energy changes in a cyclic process, where ∆U = 0 over a complete cycle so Q = W.
    • Applies Q = ∆U + W to an isothermal process, stating ∆U = 0 and hence Q = W.
    • Applies Q = ∆U + W to an adiabatic process, stating Q = 0 and hence ∆U = −W, linking work done on a gas to a temperature rise.
    • Applies Q = ∆U + W to a constant-volume (isochoric) process, stating W = 0 and hence Q = ∆U.
    • Explains a cyclic process by stating that ∆U = 0 over a complete cycle, so the net heating equals the net work done by the system.
    • Uses the first law to explain a practical example, such as a bicycle pump warming when air is compressed quickly or a gas canister cooling as it discharges.
    Examiner Tips
    • 💡Always write the sign convention next to the equation before substituting numbers, so the examiner can see your reasoning.
    • 💡When a gas expands, W is positive; when it is compressed, W is negative. Check this before calculating.
    • 💡For a cyclic process, state that ∆U = 0 and therefore Q = W, which often simplifies the calculation.
    • 💡Identify the process type first (isothermal, adiabatic, isochoric, cyclic) and write the resulting simplification before calculating.
    • 💡For adiabatic processes, state clearly that Q = 0 and link the sign of W to whether the gas is compressed or expands.
    • 💡When explaining a practical example, name the process and justify the simplification using the first law.
    • 💡Always write down the first law equation and define your sign convention before substituting values. This shows the examiner you understand the physics and helps avoid sign errors.
    • 💡When explaining processes, refer to the first law explicitly. For example, for an adiabatic expansion, state that Q = 0, so delta U = W, and since W is negative, delta U is negative, leading to a temperature decrease.
    • 💡For calculation questions, ensure units are consistent (e.g., convert cm^3 to m^3 for volume) and always include the correct sign for Q and W. Marks are often awarded for correct substitution and final answer with units.
    Common Mistakes
    • Using the wrong sign for work done on the system: the error is treating work done on the gas as positive W; the correction is to use W = −(work done on the gas) because W is work done by the system.
    • Forgetting that ∆U is an increase, not a final value: the error is substituting the final internal energy instead of the change; the correction is to calculate ∆U = U_final − U_initial.
    • Mixing up Q and W in rearrangement: the error is writing ∆U = Q + W when work is done by the system; the correction is to use ∆U = Q − W for work done by the system.
    • Assuming Q = 0 for any rapid process: the error is treating a fast process as automatically adiabatic; the correction is to check whether the process is thermally insulated or whether heating has time to occur.
    • Forgetting that ∆U = 0 for an isothermal process only for an ideal gas: the error is applying ∆U = 0 to a real gas isothermal process; the correction is to state the ideal-gas assumption.
    • Confusing constant-volume with constant-pressure processes: the error is setting W = 0 for a process that expands at constant pressure; the correction is to use W = 0 only when volume is constant.
    • Students often think that internal energy is the same as temperature. Correction: Internal energy is the total kinetic and potential energy of particles, while temperature is a measure of the average kinetic energy per particle. For an ideal gas, internal energy is directly proportional to temperature, but for real substances, potential energy also contributes.
    • Students may believe that if a gas expands, it always cools down. Correction: A gas cools during an adiabatic expansion (no heat exchange), but if heat is supplied during expansion, the temperature may remain constant (isothermal) or even increase.
    • Students sometimes confuse the sign of work done. Correction: In the equation delta U = Q + W, W is positive when work is done on the gas (compression) and negative when the gas does work (expansion). Always check the direction of energy transfer.
    Revision Plan
    1. 1Day 1-2: Review the definition of internal energy and the first law equation. Learn the sign conventions for heat and work, and practice identifying them in different scenarios.
    2. 2Day 3-4: Study the application of the first law to isothermal, adiabatic, isobaric, and isochoric processes. Create a table summarising the values of Q, W, and delta U for each process for an ideal gas.
    3. 3Day 5-6: Work through calculation problems involving the first law, including those with changes in volume, pressure, and temperature. Pay attention to unit conversions and sign conventions.
    4. 4Day 7-8: Practice exam-style questions, especially those that require explanations of real-world applications, such as a bicycle pump heating up or a gas expanding in a cylinder. Focus on using precise terminology.
    5. 5Day 9-10: Complete a past paper section on thermal physics, marking your answers against the mark scheme. Review any mistakes and revisit the relevant concepts.
    Exam Question Types
    • 📋Calculation of change in internal energy: Given values of heat supplied and work done, or a scenario where you must determine work from pressure and volume change. Advice: Always write the equation, substitute with signs, and state the final answer with units.
    • 📋Explanation of a thermodynamic process: Describe what happens to the internal energy, heat, and work in a given process (e.g., adiabatic expansion). Advice: Use the first law explicitly and refer to the sign convention.
    • 📋Graph analysis: Interpret a pressure-volume graph to determine work done (area under the curve) and relate it to internal energy changes. Advice: Remember that work done on the gas is the negative of the area under the curve if the gas expands.
    • 📋Practical application: Explain the physics of a real-world device, such as a heat pump or a car engine, using the first law. Advice: Link the processes to the equation and discuss energy transfers.
    Command Word Expectations (AQA)
    State

    A concise, factual answer is required, often a definition or a simple statement of a law. For example, 'State the first law of thermodynamics.' Expected answer: 'The change in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.' No explanation needed.

    Calculate

    You must show your working, substitute correct values with signs, and give the final answer with appropriate units. Marks are awarded for correct equation, substitution, and answer. For example, 'Calculate the change in internal energy when 200 J of heat is supplied and 150 J of work is done by the gas.' Expected: delta U = Q + W = 200 + (-150) = 50 J.

    Explain

    You must provide a detailed reasoning that links the first law to the specific situation, using correct terminology. For example, 'Explain why the temperature of a gas decreases when it expands adiabatically.' Expected: 'In an adiabatic expansion, no heat is exchanged (Q = 0). The gas does work on the surroundings, so W is negative. By the first law, delta U = Q + W = W, so delta U is negative. Since internal energy is proportional to temperature for an ideal gas, the temperature decreases.'

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Students often confuse the sign convention for work done, leading to incorrect calculations of internal energy change. They may also incorrectly assume that if temperature is constant, internal energy must change, or vice versa.
    ❌ Weak Answer (Loses Marks):When a gas expands, it does work, so internal energy decreases. Therefore, delta U = Q - W, where W is work done by the gas.
    Example improved answer:Using the convention delta U = Q + W, where W is work done on the gas. For an expansion, the gas does work on the surroundings, so W is negative. If the expansion is isothermal for an ideal gas, delta U = 0, so Q = -W, meaning heat must be supplied to the gas to maintain constant temperature. For an adiabatic expansion, Q = 0, so delta U = W, and since W is negative, delta U is negative, causing a temperature drop.
    Examiner Tip: Always state the sign convention you are using. In AQA, the equation is given as delta U = Q + W, with W as work done on the system. Remember that for an ideal gas, internal energy depends only on temperature, so isothermal processes have delta U = 0.
    Pitfall: Students often fail to apply the first law to real-world contexts, such as a gas in a cylinder with a moving piston, or they ignore the work done against atmospheric pressure. They may also forget that heat and work are energy transfers, not properties of the system.
    ❌ Weak Answer (Loses Marks):The gas has internal energy because it is hot. When it expands, it loses some internal energy as work, so it cools down.
    Example improved answer:The internal energy of a system is the sum of the random kinetic and potential energies of its particles. When a gas expands and does work on the piston, energy is transferred from the gas to the surroundings, so the internal energy decreases if no heat is supplied. According to the first law, delta U = Q + W. If the expansion is adiabatic (Q = 0), then delta U = W, and since W is negative (work done by the gas), delta U is negative, leading to a decrease in temperature. If heat is supplied, it may offset the work done, and the internal energy could remain constant or increase.
    Examiner Tip: Use the equation delta U = Q + W explicitly. Define each term with its sign. For a gas expanding against a constant external pressure, W = -p delta V. Remember that internal energy is a state function, while heat and work are path-dependent.
    Step-by-Step Worked Solutions

    Question: A gas in a cylinder is compressed by a piston. The work done on the gas is 250 J, and during the compression, 120 J of heat is lost to the surroundings. Calculate the change in internal energy of the gas.

    1. 1.Step 1: Identify given values: work done on gas W = +250 J (since work is done on the system), heat supplied Q = -120 J (since heat is lost).
    2. 2.Step 2: Apply the first law: delta U = Q + W = (-120 J) + (+250 J).
    3. 3.Step 3: Calculate: delta U = +130 J. The internal energy increases by 130 J.
    Final Answer: The change in internal energy is +130 J, so the internal energy of the gas increases by 130 J.

    Question: An ideal gas expands isothermally at 300 K, doing 500 J of work on its surroundings. Calculate the heat supplied to the gas and explain why the internal energy does not change.

    1. 1.Step 1: For an isothermal process on an ideal gas, the temperature is constant, so the internal energy is constant: delta U = 0.
    2. 2.Step 2: Apply the first law: delta U = Q + W. Since delta U = 0, Q = -W.
    3. 3.Step 3: Work done on the gas W = -500 J (since the gas does work on the surroundings). Therefore, Q = -(-500 J) = +500 J.
    4. 4.Step 4: The heat supplied to the gas is 500 J. The internal energy remains constant because the heat supplied exactly equals the work done by the gas, so the energy transferred in as heat is used to do work, leaving the internal energy unchanged.
    Final Answer: Heat supplied = 500 J. Internal energy change = 0 because for an ideal gas, internal energy depends only on temperature, which is constant in an isothermal process.
    Active Recall Memory Test
    State the first law of thermodynamics and define each term with its sign convention.
    Key Fact: The first law states that the change in internal energy of a system equals the sum of the heat supplied to the system and the work done on the system: delta U = Q + W. Q is positive when heat is supplied to the system, negative when lost. W is positive when work is done on the system (compression), negative when work is done by the system (expansion).
    For an ideal gas, what is the change in internal energy during an isothermal process? Explain.
    Key Fact: For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature is constant, so the change in internal energy is zero (delta U = 0). Therefore, by the first law, Q = -W, meaning any heat supplied is used to do work by the gas.
    What is an adiabatic process, and how does the first law apply to an adiabatic expansion of a gas?
    Key Fact: An adiabatic process is one in which no heat is exchanged with the surroundings (Q = 0). For an adiabatic expansion, the gas does work on the surroundings, so W is negative. By the first law, delta U = Q + W = W, so delta U is negative, meaning the internal energy decreases, and for an ideal gas, the temperature decreases.
    How is work done on a gas calculated when it is compressed at constant pressure?
    Key Fact: Work done on a gas at constant pressure is given by W = -p delta V, where p is the external pressure and delta V is the change in volume (final minus initial). For a compression, delta V is negative, so W is positive, indicating work is done on the gas.
    Frequently Asked Questions
    What is the first law of thermodynamics in simple terms?
    The first law of thermodynamics is a version of the law of conservation of energy applied to thermal systems. It says that the change in a system's internal energy is equal to the heat added to the system plus the work done on the system. In equation form, delta U = Q + W. This means energy cannot be created or destroyed; it can only be transferred or transformed.
    How do I know when to use a positive or negative sign for work and heat in the first law?
    In the AQA convention, the first law is written as delta U = Q + W. Q is positive when heat is supplied to the system (the system gains energy by heating) and negative when heat is lost. W is positive when work is done on the system (e.g., compression) and negative when work is done by the system (e.g., expansion). Always check the direction of energy transfer to assign the correct sign.
    Why does a gas cool down when it expands adiabatically?
    In an adiabatic expansion, no heat is exchanged with the surroundings (Q = 0). The gas does work on the surroundings, so W is negative. According to the first law, delta U = Q + W = W, so the internal energy decreases. For an ideal gas, internal energy is directly proportional to temperature, so a decrease in internal energy means a decrease in temperature, causing the gas to cool.
    What is the difference between internal energy and temperature?
    Internal energy is the total energy stored in a system due to the random motion and positions of its particles, including both kinetic and potential energy. Temperature is a measure of the average kinetic energy per particle. For an ideal gas, internal energy is directly proportional to temperature, but for real substances, changes in potential energy can occur without a change in temperature, such as during a phase change.
    Can the first law of thermodynamics be applied to non-ideal gases?
    Yes, the first law is a general principle that applies to all systems, including non-ideal gases. However, for non-ideal gases, the internal energy depends on both temperature and volume (or pressure), because intermolecular forces contribute to potential energy. The equation delta U = Q + W still holds, but calculating delta U may require more complex equations of state rather than simply using the ideal gas law.
    How is the first law of thermodynamics used in real life?
    The first law is used to analyse many real-world devices and processes. For example, in a car engine, fuel combustion supplies heat, and the expanding gases do work on the pistons, converting internal energy into mechanical work. In a refrigerator, work is done on the refrigerant to transfer heat from the cold interior to the warmer surroundings. Understanding the first law helps engineers improve energy efficiency and design better thermal systems.