Compare four thermodynamic processes
You will be able to: Connect isochoric, isobaric, isothermal and adiabatic conditions to the first law.
Can a gas expand without getting hotter?
The same gas can respond differently depending on its container and surroundings. A rigid container prevents expansion. A movable piston can allow it. Insulation suppresses heat transfer. A thermal reservoir can maintain temperature during a slow process.
A useful starting point: Read work from a pressure–volume graph →
Words and symbols before equations
- Isochoric or isovolumetric
- Constant volume; no PV boundary work.
- Isobaric
- Constant pressure.
- Isothermal
- Constant temperature; for fixed ideal gas amount, ΔU=0.
- Adiabatic
- No thermal energy transfer, Q=0; temperature may change.
What this picture assumes
Fixed monatomic ideal gas, initial P=100 kPa, V=2 L; n=200/(8.31×300) mol so T_i=300 K. Quasistatic ideal paths. Fixed-V path raises P; other paths expand V up to twice its initial value. Reversible adiabatic exponent 5/3 and isothermal logarithmic work are model calculations, not required formulas to memorize. Slider is not time.
Connect the picture to the physics
Always start with the physical constraint, then simplify the first law. Isochoric: W_on=0, so ΔU=Q. Isothermal for fixed ideal gas: ΔU=0, so Q=−W_on. Adiabatic: Q=0, so ΔU=W_on. Constant pressure alone does not set Q or ΔU to zero.
During an isothermal ideal-gas expansion, heat enters to replace energy transferred out as work. During an adiabatic expansion that does work, internal energy falls and the ideal-gas temperature falls. These are different processes despite both involving expansion.
On a PV graph an isochoric path is vertical, an isobaric path horizontal, and an ideal-gas isotherm follows P=nRT/V. A reversible monatomic adiabatic expansion drops in pressure faster than the isotherm from the same start. The explorer uses this additional monatomic model to calculate curves; memorizing its exponent or a logarithmic work formula is not required for these lessons.
| Process | Constraint | First-law consequence |
|---|---|---|
| Isochoric | Volume fixed | W_on=0; ΔU=Q |
| Isobaric | Pressure fixed | W_on=−PΔV |
| Isothermal ideal gas | Temperature fixed | ΔU=0; Q=−W_on |
| Adiabatic | No thermal transfer | Q=0; ΔU=W_on |
A worked example, step by step
A monatomic ideal gas receives 300 J at fixed volume. In a different process the same amount of gas expands adiabatically and does 120 J of work. Find ΔU in each.
- Fixed volume gives W_on=0. ΔU=300+0=+300 J.
- Adiabatic means Q=0. Work by the gas is +120 J, hence W_on=−120 J.
- ΔU=0−120=−120 J in the expansion.
- At fixed n, the first process raises T and the second lowers T.
Adiabatic means Q=0, not ΔT=0. Isothermal means ΔT=0, not Q=0.
Why must an isothermal ideal-gas expansion absorb heat when it does work?
Compare with an explanation
Its U remains constant, so positive heat input balances the negative work done on the gas.
Predict. Change one thing. Explain.
Select one of four constraints and change the progress parameter. Compare final P, V, T, Q and W_on from the same initial state. The progress slider is not time; each curve describes a different idealized process.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Isothermal: final P=50 kPa, V=4 L, T=300 K. Q=138.6 J; W_on=-138.6 J; ΔU=0 J. Check ΔU=Q+W_on.
Fixed monatomic ideal gas, initial P=100 kPa, V=2 L; n=200/(8.31×300) mol so T_i=300 K. Quasistatic ideal paths. Fixed-V path raises P; other paths expand V up to twice its initial value. Reversible adiabatic exponent 5/3 and isothermal logarithmic work are model calculations, not required formulas to memorize. Slider is not time.
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 fixed amount of ideal gas expands isothermally and does 250 J of work. (a) State ΔU. (b) State W_on. (c) Find Q. (d) Contrast the internal-energy change if the gas instead did the same work adiabatically.
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Compare with the answer and four-point rubric
- 1 point: ΔU=0 because ideal-gas T and amount stay fixed.
- 1 point: W_on=−250 J.
- 1 point: Q=+250 J.
- 1 point: Adiabatic gives Q=0 and ΔU=−250 J; its temperature falls.
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 1Isochoric constraint?
V constant, so PV work is zero.
RECALL 2Isothermal versus adiabatic?
Constant temperature versus no heat transfer; they are not synonyms.
RECALL 3Can isobaric gas change temperature?
Yes. For fixed n, V is proportional to T at constant P.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Compare four thermodynamic processes
- Isochoric: W_on=0; ΔU=Q.
- Isothermal ideal gas: ΔU=0; Q=−W_on.
- Adiabatic: Q=0; ΔU=W_on.
- Isobaric: W_on=−PΔV.
Remember: Adiabatic means Q=0, not ΔT=0. Isothermal means ΔT=0, not Q=0.
Conditions: Fixed monatomic ideal gas, initial P=100 kPa, V=2 L; n=200/(8.31×300) mol so T_i=300 K. Quasistatic ideal paths. Fixed-V path raises P; other paths expand V up to twice its initial value. Reversible adiabatic exponent 5/3 and isothermal logarithmic work are model calculations, not required formulas to memorize. Slider is not time.
Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.4.B · Review edition
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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.
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