Work, zeroth law and first law
Key idea: H2 Physics lessons on temperature, ideal gases, internal energy and thermodynamic systems.
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The core idea
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Big question: How do heating and work change a system's internal energy?
The zeroth law makes temperature meaningful: systems separately in equilibrium with a third system share a temperature. The first law accounts for energy, ΔU = Q + W when W is work done on the system. Against constant external pressure, work done by a gas is pΔV; translate this carefully into work done on the gas.
Use equilibrium to define a shared temperature
The zeroth law says that if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. This transitive relation lets a thermometer C compare temperatures without placing every pair of systems together.
Equilibrium means equal temperature and no net heating. It does not mean equal internal energy or an absence of microscopic energy exchanges.
Check your understanding: How does a thermometer use the zeroth law?
When its reading becomes steady in contact with a system, they share a temperature; identical readings imply the systems would be mutually in equilibrium.
Choose and keep one work convention
The syllabus form ΔU = Q + W uses Q positive when energy enters by heating and W positive when work is done on the system. Expansion against constant external pressure has work by the gas pΔV, so work on the gas is −pΔV.
For constant external pressure, the rectangle pΔV on a p–V graph represents work done by the gas. Translate the wording first: expansion normally gives negative W in ΔU = Q + W, while compression gives positive W.
Check your understanding: A gas is compressed while 40 J of work is done on it and no energy is transferred by heating. What is ΔU?
W = +40 J and Q = 0, so ΔU = +40 J.
Key ideas to keep
- State the sign convention before using the first law.
- Expansion gives positive work by the gas but negative work on the gas.
- At constant external pressure, the rectangular pΔV area gives work done by the gas.
See the reasoning
Worked example
Keep the first-law sign convention through a compression
Question: A gas is compressed by 2.0 × 10⁻⁴ m³ at constant external pressure 3.0 × 10⁵ Pa while 25 J leaves by heating. Find ΔU with ΔU = Q + W.
Step 1: Assign the volume-change sign
Why: Compression has negative ΔV in the work-by-gas expression.
Working: ΔV = −2.0×10⁻⁴ m³, so work by the gas is pΔV = −60 J.
Step 2: Convert to work on the gas
Why: The syllabus equation uses W as work done on the system.
Working: W = +60 J for the compression.
Step 3: Include heating
Why: Energy leaving by heating makes Q negative.
Working: Q = −25 J, hence ΔU = Q + W = −25 + 60 = +35 J.
Answer: ΔV = −2.0 × 10⁻⁴ m³. Work by gas is pΔV = −60 J, so work on gas W = +60 J. Q = −25 J, hence ΔU = −25 + 60 = +35 J.
Check: Compression work raises U more than the 25 J heating loss lowers it, so a positive result is sensible.
Use a hint if needed
Practise with support
Try this
State the zeroth law and explain how it justifies a thermometer.
Hint: Use three systems and the transitive equilibrium relation.
Check your answer
If A and B are each in thermal equilibrium with C, A and B are in thermal equilibrium with each other. A thermometer is C: equal readings correspond to a shared temperature and therefore mutual thermal equilibrium.
Now work without the hint
Practise independently
Your turn
For each of expansion and compression, relate the signs of pΔV, work by the gas, work on the gas and the W in ΔU = Q + W.
Check your answer
Expansion has ΔV > 0 and work by gas +pΔV, so work on gas and W are negative. Compression has ΔV < 0 and work by gas negative, so work on gas and W are positive.
Avoid these traps
Common mistakes
Common mistake
Work by and work on the gas have the same sign.
What is wrong with this reasoning?
Show better thinking
They have opposite signs. In ΔU = Q + W, W is work done on the system.
Write for the examiner
Exam guidance
Annotate whether energy enters or leaves as heating or work before assigning algebraic signs.
Exam-style practice [7 marks]
A gas receives 260 J by heating and does 90 J of work. Use the syllabus convention to find ΔU and state the zeroth law.
Plan before you answer
- State the chosen convention.
- Translate 'does work' into work on the gas.
- Apply ΔU = Q + W, then state the zeroth law.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Work on the gas is W = −90 J, so ΔU = Q + W = 170 J. If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.
Come back in three days
Check what stayed with you
Recall question
A gas loses 40 J by heating while 75 J of work is done on it. Find ΔU.
Check the answer
Q = −40 J and W = +75 J in ΔU = Q + W, so ΔU = +35 J.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Temperature and ideal-gas ideas lead into the first law of thermodynamics. Use ΔU = Q + W, where W is work done on the system; for expansion against constant external pressure, work done by the gas is pΔV and work done on the gas is −pΔV.
- GCE A-Level H2 PhysicsTopic 13(d) / Topic 13(e) / Topic 13(f) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027