Apply the first law of thermodynamics
Track heating and work done on a system with ΔU = q + W, using signs to interpret expansion, compression and cooling.
Section 16 · Thermodynamics
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 16. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use temperature, molecular kinetic energy, pressure, volume and energy-transfer conventions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In ΔU = q + W, what does W represent?
What you will learn
- Use ΔU = q + W with W defined as work done on the system
- Interpret q as energy transferred to the system by heating
- Combine heating and mechanical work to calculate internal-energy change
The first law accounts for two energy transfers
The first law is ΔU = q + W. Here ΔU is the increase in internal energy, q is energy transferred to the system by heating, and W is work done on the system.
With this sign convention, heating the system gives q > 0; cooling gives q < 0. Work done on the system is positive, such as during compression. Expansion work done by the gas means W for work done on the system is negative.
Add signed transfers to find the internal-energy change
For a gas expanding at constant pressure, W = −pΔV in the first-law equation. If heat flows into the gas, q is positive; compare both terms before calculating ΔU.
A positive ΔU means internal energy increases; a negative ΔU means it decreases. The first law is energy accounting: heat and work are transfers across the system boundary, not energy stores in the system.
Pause and explain
A gas receives 200 J by heating and has 50 J of work done on it. What is ΔU?
Worked example
A gas receives 500 J of energy by heating and expands by 2.0 × 10⁻³ m³ at 1.0 × 10⁵ Pa. Calculate ΔU.
Show the worked solution
- q = +500 J.
- W = −pΔV = −200 J because the gas expands.
- Use ΔU = q + W.
Answer ΔU = 500 − 200 = +300 J.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A gas receives 700 J by heating while 200 J of work is done on it. Find its internal-energy change.
Give me a hint
Use both transfers as positive inputs in ΔU = q + W.
Compare my reasoning
- q = +700 J.
- W = +200 J.
- ΔU = 700 + 200.
ΔU = +900 J.
Look for these in your work
- Heating sign correct
- Work-on sign correct
- Internal-energy change stated
A gas cools and loses 150 J by heating; 400 J of work is done on it. Calculate ΔU.
Give me a hint
Energy leaving by heating makes q negative; work on the system is positive.
Compare my reasoning
- q = −150 J.
- W = +400 J.
- ΔU = −150 + 400.
ΔU = +250 J.
Look for these in your work
- Heat-loss sign applied
- Work-on sign applied
- Signed sum correct
A gas expands at constant pressure 1.0 × 10⁵ Pa by 4.0 × 10⁻³ m³ while 100 J of energy is transferred out by heating. Calculate ΔU.
Give me a hint
For the first law, work done on the expanding gas is −pΔV and heat leaving is negative q.
Compare my reasoning
- W = −pΔV = −400 J.
- q = −100 J.
- ΔU = q + W = −100 − 400.
ΔU = −500 J; the gas's internal energy decreases.
Look for these in your work
- Expansion work sign correct
- Heat-loss sign correct
- Negative change interpreted
Common mistakes
- Using work done by the gas as W in ΔU = q + W
- Giving q a positive sign when energy leaves the system
- Treating q and W as energy stored rather than transferred
What can you explain now?
In ΔU = q + W, what does W represent?
Compare with the explanation
Work done on the system
That is the sign convention used in this first-law form.
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Make it stick
What do q and W mean in ΔU = q + W, and how do their signs show energy entering or leaving?
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