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).
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The core idea
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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.
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
- Define the system (usually “the gas”).
- Decide the sign convention you will use (this site uses Wₒₙ).
- 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.
- Compute W (if needed) using the external pressure given.
- 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
Problem
Study the worked solution
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 = 0Apply 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)
Problem
Try this before viewing the solution
Hints
Hint 1: find the signed work term
Hint 2: complete the balance
View solution step by step
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 m³Convert 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 JBalance 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
Data table
| Category | Example B |
|---|---|
| Q (heating in) | 600 |
| W_on (work on gas) | -225 |
| ΔU (change) | 375 |
Common misconception 3
Adiabatic compression (no heating)
Learner claim
Try this before viewing the solution
View solution step by step
Interpret adiabatic correctly
Method
Adiabatic means Q = 0.Reason
It excludes transfer by heating, not transfer by work.Working
Q = 0Assign the work sign
Method
Wₒₙ = +300 J.Reason
Compression transfers energy from the surroundings into the gas.Working
Wₒₙ > 0Apply 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
Examination question
Try this before viewing the solution
View solution step by step
Assign heating sign
1 markMethod
Q = -100 J.Reason
Energy leaves the gas by heating.Working
Q = -100 JFind volume change
1 markMethod
Δ V = -1.5 × 10⁻³ m³.Reason
The process is a compression.Working
Δ V = (2.5-4.0) × 10⁻³ m³Calculate work on
1 markMethod
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 JComplete the balance
1 markMethod
The internal energy increases by 200 J.Reason
Compression supplies more energy than is lost by heating.Working
Δ U = -100 + 300 = +200 J
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark both transfer signs, the work calculation and the final balance.
Challenge 5
Isothermal ideal-gas process (internal energy unchanged)
Independent transfer
Try this before viewing the solution
Hints
Hint 1: use the ideal-gas state result
View solution step by step
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 = 0Infer 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 JChange 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 sign conventions and Δ U = Q + W questions:
A Level Thermodynamic Systems QuizContinue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027