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A gas can expand, do no boundary work, receive no heat and still move to a state of higher entropy. That combination feels contradictory only when expansion is confused with useful work or when entropy is treated as energy. Free expansion is a clean test of thermodynamic bookkeeping: define the system boundary, separate path quantities from state functions, and use a reversible path only to calculate the entropy difference between the same endpoints.
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Chapter contents
Choose the route closest to the decision, or read from Chapter 1. The chapters remain open; these grouped routes are a navigable index.
Route 1 · Define the decision
Route 2 · Work the ledgers
Route 3 · Transfer the method
Route 4 · Check limits and evidence
Route 5 · Ask and continue
CHAPTER 1 OF 12 · DEFINE THE DECISION
1. Define the free-expansion system
Imagine an insulated rigid vessel split into two chambers. An ideal gas occupies one side and the other side is evacuated. When the partition breaks, the gas fills the entire vessel.
Take the system as all gas inside the outer rigid vessel. The external boundary does not move, no mass crosses it and insulation makes heat transfer negligible.
CHAPTER 2 OF 12 · DEFINE THE DECISION
2. Separate expansion from boundary work
Boundary work requires force acting through displacement at the system boundary. Internal gas motion after the partition fails is not, by itself, work delivered across the rigid outer wall.
For this system, W=0 and Q=0. The first law gives ΔU=Q−W=0 under the common sign convention that W is work done by the system.
CHAPTER 3 OF 12 · WORK THE LEDGERS
3. Work the ideal-gas endpoint calculation
One mole of ideal gas at 300 K expands freely from 10 L to 20 L inside an insulated rigid vessel.
| Quantity | Initial | Final | Reason |
|---|---|---|---|
| Volume | 10 L | 20 L | Partition removed |
| Boundary work | — | 0 | Outer vessel rigid |
| Heat transfer | — | 0 | Insulated |
| Internal energy change | — | 0 | First law |
| Temperature | 300 K | 300 K | Ideal-gas U depends only on T |
| Pressure | P1 | P1/2 | Ideal gas, doubled volume |
For a real gas, internal energy can depend on volume as well as temperature. The same Q=0 and W=0 balance still gives ΔU=0, but ΔT need not be zero.
CHAPTER 4 OF 12 · WORK THE LEDGERS
4. Calculate entropy with a reversible comparison path
Entropy is a state function, so its change depends on the endpoints even though the actual free expansion is irreversible. Connect the same ideal-gas endpoints with an imagined reversible isothermal expansion.
| Path | Heat to gas | Work by gas | ΔU | ΔS gas | Entropy generation |
|---|---|---|---|---|---|
| Actual insulated free expansion | 0 | 0 | 0 | R ln 2 | R ln 2 |
| Reversible isothermal comparison | RT ln 2 | RT ln 2 | 0 | R ln 2 | 0 |
For one mole, ΔS=R ln(20/10)=R ln 2≈5.76 J K−1. With no entropy transfer by heat, the entire increase is generated inside the system.
CHAPTER 5 OF 12 · WORK THE LEDGERS
5. Locate the lost work opportunity
A reversible isothermal expansion between these ideal-gas volumes could deliver RT ln 2 of work while receiving the same amount of heat. Free expansion produces none.
Relative to an environment at temperature T0, destroyed work potential is connected to T0Sgen. The exact availability balance still depends on the complete system and surroundings definition.
CHAPTER 6 OF 12 · TRANSFER THE METHOD
6. Use pressure carefully during the event
The initial and final equilibrium pressures are well defined. During rapid free expansion, a single uniform gas pressure may not describe the whole vessel, so writing W=∫Pgas dV along the actual path can be misleading.
The reliable work argument uses the rigid external system boundary. The reversible pressure integral belongs to the comparison path, not the uncontrolled transient.
CHAPTER 7 OF 12 · TRANSFER THE METHOD
7. Changed-context transfer: compressed gas into a receiver
A valve opens between a pressurised insulated tank and an evacuated receiver enclosed by one rigid control boundary. Gas redistributes without moving the combined outer wall.
The combined closed system again has zero external boundary work and negligible heat transfer. If the gas is not ideal, property data are needed to determine the final temperature from constant total internal energy.
CHAPTER 8 OF 12 · TRANSFER THE METHOD
8. Connect paths, endpoints and irreversibility
Read same endpoints, different pressure–volume paths to separate state changes from path-dependent work. Continue with finite bodies, entropy generation and the direction of heat flow for a second irreversibility example.
Use a rigid tank losing gas when mass crosses the boundary and the closed-system argument no longer applies.
CHAPTER 9 OF 12 · CHECK LIMITS AND EVIDENCE
9. Limits and common misconceptions
Zero work does not mean nothing happened, and constant internal energy does not always mean constant temperature. The latter step requires an ideal-gas model or equivalent property evidence.
Entropy is not conserved in an irreversible isolated process. Energy remains constant while entropy increases; the two quantities answer different questions.
CHAPTER 10 OF 12 · CHECK LIMITS AND EVIDENCE
10. Practical evidence and checking routine
Sketch the boundary, identify heat, work and mass crossings, state the sign convention, write the first-law balance and decide whether an ideal-gas relation is justified.
Calculate entropy from valid endpoint properties or a convenient reversible path. Check Sgen≥0 and keep actual-path work separate from comparison-path work.
CHAPTER 11 OF 12 · ASK AND CONTINUE
11. Frequently asked questions
Why is work zero if the gas volume increases?
The chosen system’s outer rigid boundary does not move. Expansion into an internal vacuum does not deliver work across that boundary.
Why can a reversible path calculate an irreversible entropy change?
Entropy is a state function. A reversible path supplies a valid integral between the same equilibrium endpoints.
Does the gas stay at one pressure during expansion?
Not necessarily. The transient can be nonuniform; only the initial and final equilibrium pressures are required here.
CHAPTER 12 OF 12 · ASK AND CONTINUE
12. Sources and next reading
MIT OpenCourseWare’s Unified Engineering thermodynamics chapter uses adiabatic free expansion to introduce entropy and irreversibility. MIT’s Second Law notes work through zero heat, zero work and positive ideal-gas entropy change. The vessel and receiver examples are original teaching cases.
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