First Law of Thermodynamics

Key idea: Apply the first law of thermodynamics ΔU = Q + W (work done on the system), including constant-volume and constant-pressure (pΔV) cases (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply work conventions and the zeroth and first laws of thermodynamics.

1. Definitions (Must Know)

A. First law of thermodynamics (syllabus sign convention)

The first law of thermodynamics is conservation of energy for a thermodynamic system:

Δ U = Q + W

where (for the system):

  • Δ U is the change in internal energy (J).
  • Q is the energy transferred to the system by heating (J).
  • W is the work done on the system (J).

B. Work done by vs work done on

  • Work done by the gas, W_by: energy transferred from the gas to the surroundings.
  • Work done on the gas, Wₒₙ: energy transferred to the gas by the surroundings.

They are related by: Wₒₙ = -W_by

C. Internal energy is a state function

Internal energy U is a state function: it depends only on the system’s state (e.g. p, V, T), not on the path taken.

So for the same initial and final states, Δ U is the same even if Q and W are different.

2. Key Ideas (What Earns Marks)

  • In this syllabus, use:
    • Δ U = Q + Wₒₙ
  • Many textbooks use the alternative form:
    • Δ U = Q - W_by
    • which is the same equation, since Wₒₙ = -W_by.
  • At constant volume, Δ V = 0 ⇒ W_by = 0 so:
    • Δ U = Q
  • For expansion/compression against constant external pressure p:
    • W_by = pΔ V
    • Wₒₙ = -pΔ V
  • Unit check: Pa·m³ = J.
Exam pitfall: inconsistent sign convention for work

This lesson uses Δ U = Q + Wₒₙ. If a question instead defines W as work done by the gas, use Δ U = Q-W_by. Never change the meaning of W midway through a solution.

3. Detailed Explanations

A. A safe workflow for first-law questions

  1. Define the system (usually “the gas”).
  2. Decide the sign convention you will use (this site uses Wₒₙ).
  3. Identify what the process tells you:
    • isochoric (Δ V = 0) → W_by = 0,
    • no heating (no net energy transfer by heating) → Q = 0,
    • constant temperature (Δ T = 0) → for an ideal gas, Δ U = 0.
  4. Compute W (if needed) using the external pressure given.
  5. Substitute into Δ U = Q + Wₒₙ.

B. Work term for constant external pressure: W_by = pΔ V

If a gas expands against a constant external pressure p: W_by = pΔ V

So in the first law (syllabus convention): Wₒₙ = -pΔ V

Sign check:

  • expansion: Δ V > 0 ⇒ W_by > 0 and Wₒₙ < 0
  • compression: Δ V < 0 ⇒ W_by < 0 and Wₒₙ > 0

C. What Δ U means in words (for explanation marks)

  • Δ U > 0: internal energy increases.
  • Δ U < 0: internal energy decreases.

For an ideal gas, internal energy is tied mainly to particle kinetic energy, so Δ U tracks temperature change.

4. Common Mistakes

  • Mixing sign conventions (using Δ U = Q + W but treating W as work done by the system).
  • Using gas pressure instead of the constant external pressure specified for pΔ V.
  • Forgetting V must be in m³ (convert from L or cm³).
  • Writing “heat” as a stored quantity (it is a transfer; internal energy U is stored).

5. Exam Tips

  • State the sign convention explicitly if the question looks ambiguous (“W is work done on the gas”).
  • Always write a final energy-balance sentence, e.g. “Q adds energy, expansion does work out, net internal energy increases.”
  • Quick special cases:
    • constant volume → Δ U = Q
    • no heating (Q = 0) → Δ U = Wₒₙ

6. Worked Examples

Modelled example 1

Constant volume heating

Core

Problem

A fixed mass of gas is heated at constant volume. Q = +450 J is supplied. Find Δ U.
Study the worked solution
  1. Use the process constraint

    Method

    At constant volume, Δ V = 0 and no pressure–volume work is done.

    Reason

    A stationary boundary gives W_by = pΔ V = 0 and therefore Wₒₙ = 0.

    Working

    Δ V = 0 ⇒ Wₒₙ = 0
  2. Apply the first law

    Method

    The internal energy increases by 450 J.

    Reason

    With no work transfer, all energy supplied by heating increases internal energy.

    Working

    Δ U = Q + Wₒₙ = 450 + 0 = +450 J

Guided practice 2

Expansion with heating (use pΔ V)

About 6 min

Problem

A gas expands against constant external pressure 1.5 × 10⁵ Pa from 2.0 × 10⁻³ m³ to 3.5 × 10⁻³ m³ while Q = +600 J is supplied. Find Δ U.

Try this before viewing the solution

Unit: J

Hints

Hint 1: find the signed work term
Calculate pΔ V as work by the gas, then reverse its sign for Wₒₙ.
Hint 2: complete the balance
Use Δ U = Q + Wₒₙ without changing the meaning of W.
View solution step by step
  1. Find the volume change

    Method

    Δ V = +1.5 × 10⁻³ m³.

    Reason

    The final volume exceeds the initial volume.

    Working

    Δ V = (3.5-2.0) × 10⁻³ m³
  2. Convert work viewpoint

    Method

    The gas does 225 J of work, so Wₒₙ = -225 J.

    Reason

    Expansion transfers energy to the surroundings by work.

    Working

    W_by = pΔ V = 225 J, Wₒₙ = -225 J
  3. Balance the transfers

    Method

    The internal energy increases by 375 J.

    Reason

    Of the 600 J supplied by heating, 225 J leaves as expansion work.

    Working

    Δ U = 600-225 = +375 J

Energy balance for Example BA bar chart showing Q positive, W_on negative for expansion, and the resulting positive change in internal energy.Energy balance for Example BEnergy (J)
Using the syllabus convention ΔU = Q + W_on: the negative W_on during expansion reduces the internal-energy gain from the supplied heat.
Data table
CategoryExample B
Q (heating in)600
W_on (work on gas)-225
ΔU (change)375

Common misconception 3

Adiabatic compression (no heating)

Find and correct the mistake

Learner claim

A gas is compressed with 300 J of work done on it and no energy transferred by heating. A learner says “adiabatic means no energy change”, so Δ U = 0. Diagnose the claim.

Try this before viewing the solution

Unit: J

View solution step by step
  1. Interpret adiabatic correctly

    Method

    Adiabatic means Q = 0.

    Reason

    It excludes transfer by heating, not transfer by work.

    Working

    Q = 0
  2. Assign the work sign

    Method

    Wₒₙ = +300 J.

    Reason

    Compression transfers energy from the surroundings into the gas.

    Working

    Wₒₙ > 0
  3. Apply the first law

    Method

    The internal energy increases by 300 J.

    Reason

    Work is the only energy transfer in this process.

    Working

    Δ U = 0 + 300 = +300 J

Examiner practice 4

Compression with heat lost

4 marks

Examination question

A gas is compressed against constant external pressure 2.0 × 10⁵ Pa from 4.0 × 10⁻³ m³ to 2.5 × 10⁻³ m³ while losing 100 J by heating. Find Δ U using Δ U = Q + Wₒₙ. [4 marks]

Try this before viewing the solution

View solution step by step
  1. Assign heating sign

    1 mark

    Method

    Q = -100 J.

    Reason

    Energy leaves the gas by heating.

    Working

    Q = -100 J
  2. Find volume change

    1 mark

    Method

    Δ V = -1.5 × 10⁻³ m³.

    Reason

    The process is a compression.

    Working

    Δ V = (2.5-4.0) × 10⁻³ m³
  3. Calculate work on

    1 mark

    Method

    Wₒₙ = +300 J.

    Reason

    Wₒₙ = -pΔ V, so the negative volume change makes the work-on term positive.

    Working

    Wₒₙ = -(2.0 × 10⁵)(-1.5 × 10⁻³) = +300 J
  4. Complete the balance

    1 mark

    Method

    The internal energy increases by 200 J.

    Reason

    Compression supplies more energy than is lost by heating.

    Working

    Δ U = -100 + 300 = +200 J

Challenge 5

Isothermal ideal-gas process (internal energy unchanged)

Minimal support

Independent transfer

An ideal gas absorbs Q = +250 J while its temperature remains constant. Find Wₒₙ and interpret the result as work done by the gas.

Try this before viewing the solution

Hints

Hint 1: use the ideal-gas state result
For a fixed amount of ideal gas, constant temperature means Δ U = 0.
View solution step by step
  1. Set the state-function change

    Method

    Δ U = 0.

    Reason

    For a fixed amount of ideal gas, internal energy depends only on thermodynamic temperature.

    Working

    Δ T = 0 ⇒ Δ U = 0
  2. Infer work on the gas

    Method

    Wₒₙ = -250 J.

    Reason

    The first law requires the positive heating transfer to be balanced by an equal energy transfer out through work.

    Working

    0 = 250 + Wₒₙ ⇒ Wₒₙ = -250 J
  3. Change viewpoint

    Method

    The gas does + 250 J of work on the surroundings.

    Reason

    Work by and work on have equal magnitudes and opposite signs.

    Working

    W_by = -Wₒₙ = +250 J

7. Mind Stretchers

Mind stretcher 1: Expansion while losing heatExtension

A gas expands and does 200 J of work on the surroundings. During the process, it loses 120 J of heat to the surroundings. Find Δ U and interpret the sign.

Show Answer

Heat lost means Q = -120 J.

Work done by the gas is W_by = +200 J, so work done on the gas is Wₒₙ = -200 J.

First law: Δ U = Q + Wₒₙ = -120 - 200 = -320 J

Internal energy decreases: the gas loses energy overall.

Mind stretcher 2: Same Δ U, different pathsExtension

Two different processes take a gas from state 1 to state 2. Explain why Δ U is the same even if Q and W are different.

Show Answer

Internal energy is a state function: it depends only on the initial and final states.

So Δ U depends only on state 1 and state 2, not on the path. Different paths can have different Q and W, but the first law forces Q + Wₒₙ to give the same Δ U.

Mind stretcher 3: Optional (Enrichment)Extension

A. Heat capacity notation for gases

Some resources use molar heat capacities:

  • constant volume: Q = nC_VΔ T
  • constant pressure: Q = nC_PΔ T

This is not required for the core first-law learning outcome, but you may see it in extension problems.

8. Practice (Quiz)

Practice (Quiz)

Practice sign conventions and Δ U = Q + W questions:

A Level Thermodynamic Systems Quiz

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027