Read work from a pressure–volume graph
You will be able to: Calculate gas work from signed pressure–volume areas and compare paths.
Why does pushing farther change the work?
A gas pushes a piston outward through a small distance. Force times distance is work. Since force is pressure times piston area and area times distance is volume change, pressure times volume change gives the work for constant pressure.
A useful starting point: The first law: keep an energy account →
Words and symbols before equations
- P_ext
- External pressure resisting the piston, in Pa.
- ΔV
- V_final−V_initial; positive for expansion, in m³.
- Quasistatic process
- A slow sequence of near-equilibrium states; gas pressure approximately balances external pressure.
- PV graph
- Pressure on the vertical axis and volume on the horizontal; arrows show process direction.
What this picture assumes
Quasistatic constant-pressure gas process from V_i=2 L. Gas pressure approximates external pressure. Signed work on the gas is −PΔV; kPa·L=J. Shaded area shows magnitude; arrow establishes direction.
Connect the picture to the physics
For constant external pressure, W_on=−P_ext ΔV. Expansion has ΔV>0 and negative work on the gas; compression has ΔV<0 and positive work on the gas. With varying pressure, add contributions along the path. On a quasistatic gas-pressure graph, magnitude equals area under the path.
One kPa·L equals one joule because (1000 Pa)(0.001 m³)=1 J. A horizontal path encloses a rectangle down to the volume axis. A vertical segment has zero volume change and zero boundary work, even if pressure changes.
Two paths between the same endpoints can have different areas, so work is path-dependent. For a clockwise rectangular cycle with top pressure 200 kPa, bottom 100 kPa and volumes 1–3 L, net work by the gas is (200−100)(3−1)=200 J. Net W_on=−200 J and ΔU_cycle=0, so net Q=+200 J.
A worked example, step by step
A gas expands quasistatically at 100 kPa from 2 L to 5 L. Find work on the gas. If ΔU=450 J, find Q.
- ΔV=5−2=3 L=0.003 m³.
- W_on=−(100000)(0.003)=−300 J.
- Rearrange ΔU=Q+W_on to Q=ΔU−W_on.
- Q=450−(−300)=750 J: some incoming heat raises U and some leaves as work.
The PV area is work, not internal energy. A gas-pressure path represents boundary work only under the appropriate mechanical-equilibrium assumptions.
If a PV path is vertical, is work necessarily nonzero because pressure changes?
Compare with an explanation
No. Boundary work is zero when ΔV=0.
Predict. Change one thing. Explain.
Keep the initial volume at 2 L and change final volume through expansion, no change, and compression. Then change pressure. Read the arrow and signed work, not just the shaded area magnitude.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
ΔV=3 L; W_on=−PΔV=-300 J. Expansion: gas does positive work on surroundings.
Quasistatic constant-pressure gas process from V_i=2 L. Gas pressure approximates external pressure. Signed work on the gas is −PΔV; kPa·L=J. Shaded area shows magnitude; arrow establishes direction.
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 quasistatic rectangular cycle expands at 300 kPa from 1 L to 3 L, cools at fixed volume, compresses at 100 kPa to 1 L, then returns vertically. (a) Find W_by during expansion. (b) Find W_by during compression. (c) Find net W_on. (d) Find net Q over the cycle.
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Compare with the answer and four-point rubric
- 1 point: Expansion W_by=300(2)=600 J.
- 1 point: Compression W_by=100(−2)=−200 J.
- 1 point: Net W_on=−(600−200)=−400 J; vertical paths do no work.
- 1 point: ΔU_cycle=0, so net Q=+400 J.
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 1Area under a quasistatic PV path?
Magnitude of boundary work; direction sets its sign.
RECALL 2What does a vertical path contribute?
Zero boundary work because volume is constant.
RECALL 3Why is work path-dependent?
Different pressure histories can enclose different areas between the same volume endpoints.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Read work from a pressure–volume graph
- Constant P_ext: W_on=−P_ext(V_f−V_i).
- 1 kPa·L=1 J.
- Clockwise PV cycle: positive net W_by, negative net W_on.
Remember: The PV area is work, not internal energy. A gas-pressure path represents boundary work only under the appropriate mechanical-equilibrium assumptions.
Conditions: Quasistatic constant-pressure gas process from V_i=2 L. Gas pressure approximates external pressure. Signed work on the gas is −PΔV; kPa·L=J. Shaded area shows magnitude; arrow establishes direction.
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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