The first law: keep an energy account
You will be able to: Apply ΔU=Q+W_on with a clearly defined system and sign convention.
Where does energy go when a gas pushes outward?
A gas receives 500 J from a heater while it pushes a piston outward, transferring 200 J mechanically to its surroundings. Only 300 J remains as increased internal energy. The first law keeps track of both paths across the boundary.
A useful starting point: Count the energy inside the system →
Words and symbols before equations
- Q
- Thermal energy transferred into the gas; positive inward, negative outward, in J.
- W_on
- Work done on the gas; positive when surroundings supply mechanical energy.
- W_by
- Work done by the gas on surroundings; W_by=−W_on.
- Closed versus isolated
- Closed: no matter crosses the boundary. Isolated: neither matter nor energy crosses.
What this picture assumes
Closed system; negligible change in bulk kinetic or gravitational energy. Positive transfers enter the gas; negative transfers leave. Bars are signed energy amounts, not rates or time histories.
Connect the picture to the physics
Choose the gas as a closed system and neglect changes of its bulk kinetic and gravitational energies. Then ΔU=Q+W_on. This expresses conservation of energy; a positive ΔU means an increase in internal energy.
Expansion against a resisting piston usually means the gas does work on the surroundings, so W_on is negative. Compression means positive W_on if the surroundings supply mechanical energy to the gas. Heat has its own independent sign.
Some books write ΔU=Q−W_by. It is the same law with a different work definition. Write the subscript or define the convention before substituting. A closed gas can lose or gain energy; closed is not the same as insulated or isolated.
| Transfer | Positive | Negative |
|---|---|---|
| Q | Energy enters thermally | Energy leaves thermally |
| W_on | Work supplies energy to gas | Gas transfers energy out as work |
A worked example, step by step
A gas receives 500 J of heat and does 200 J of work on the surroundings. Find ΔU. Then consider 100 J of cooling and 150 J of work done on the gas.
- First case: Q=+500 J; W_on=−200 J.
- ΔU=500−200=+300 J.
- Second case: Q=−100 J; W_on=+150 J.
- ΔU=−100+150=+50 J. Cooling can accompany a net internal-energy increase when work input is larger.
“Work” without a stated direction is ambiguous. Do not change conventions halfway through a problem.
Does cooling always make internal energy decrease?
Compare with an explanation
No. Positive work on the system may outweigh energy lost by cooling.
Predict. Change one thing. Explain.
Vary heat and work independently. The signed bars show transfers and their sum. Find two different processes with the same ΔU, and one process with Q<0 but ΔU>0.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
ΔU=Q+W_on=(500)+(-200)=300 J. Internal energy increases.
Closed system; negligible change in bulk kinetic or gravitational energy. Positive transfers enter the gas; negative transfers leave. Bars are signed energy amounts, not rates or time histories.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant particle, temperature or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA closed gas loses 80 J thermally while it is compressed by an external agent doing 200 J of work. (a) State Q. (b) State W_on. (c) Find ΔU. (d) For fixed monatomic ideal gas amount, state the temperature trend and justify.
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Compare with the answer and four-point rubric
- 1 point: Q=−80 J.
- 1 point: W_on=+200 J.
- 1 point: ΔU=+120 J.
- 1 point: Temperature increases because U is proportional to T at fixed n.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Our work convention?
W_on is positive for energy transferred mechanically into the gas.
RECALL 2Closed versus isolated?
Closed prevents matter flow; isolated prevents matter and energy flow.
RECALL 3Can different paths give the same ΔU?
Yes. ΔU is a state change; Q and W individually depend on path.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
The first law: keep an energy account
- ΔU=Q+W_on=Q−W_by.
- Heating: Q>0; cooling: Q<0.
- Compression work on gas is positive; expansion work on gas is negative.
Remember: “Work” without a stated direction is ambiguous. Do not change conventions halfway through a problem.
Conditions: Closed system; negligible change in bulk kinetic or gravitational energy. Positive transfers enter the gas; negative transfers leave. Bars are signed energy amounts, not rates or time histories.
Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.4, objectives 9.4.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the first AP Physics 2 unit; College Board numbers it Unit 9, continuing after AP Physics 1. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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