For free expansion into a vacuum, the process is irreversible, but the entropy change of the gas depends only on its initial and final equilibrium states. We compute ΔS using a reversible isothermal path between the same states.
- Ideal gas: ΔS = n R ln(V2 / V1)
- Van der Waals (covolume b): ΔS ≈ n R ln((V2 − n b)/(V1 − n b)) (requires V − n b > 0)
- Select a model. Use the ideal option for most textbook problems.
- Choose a volume unit, then enter V1 and V2 (with V2 greater).
- Enter the amount of gas n. Temperature is optional context.
- If using van der Waals, enter b and its unit.
- Press Calculate to see results above the form.
- Use the export buttons to download CSV or PDF results.
| # | Model | n (mol) | V1 (L) | V2 (L) | b (L/mol) | ΔS (J/K) |
|---|---|---|---|---|---|---|
| 1 | Ideal | 1.0 | 2 | 10 | — | 13.38 |
| 2 | Ideal | 0.5 | 1 | 4 | — | 5.76 |
| 3 | Van der Waals | 1.0 | 2 | 10 | 0.04267 | 13.65 |
1) What free expansion means
Free expansion is a rapid increase in volume without useful boundary work. A rigid vessel split by a valve is common; one side has gas, the other is evacuated. After opening, the gas redistributes until uniform equilibrium is reached.
2) Why entropy rises in vacuum expansion
Even when no heat enters and no work leaves, the gas gains more accessible microstates after it fills the larger volume. That larger phase space produces a higher thermodynamic entropy. This is why free expansion is a standard example of an irreversible process with positive entropy generation.
3) State variables behind the calculation
Entropy is a state function, so only the initial and final equilibrium states matter. This calculator uses amount of substance n and volumes V1 and V2. Temperature is included for context and reporting consistency.
4) Ideal‑gas path equivalence
The real free expansion path is irreversible, so you cannot integrate δQ/T along it directly. We instead choose a reversible isothermal path between the same endpoints. For an ideal gas, this gives ΔS = nR ln(V2/V1). Because the endpoints are equilibrium, the computed ΔS is unique for the stated inputs.
5) Real‑gas option using covolume
At higher densities, finite molecular size reduces the available free volume. The van der Waals covolume parameter b approximates this by replacing V with (V − nb). The estimate becomes ΔS ≈ nR ln((V2 − nb)/(V1 − nb)). It is most relevant when V is not much larger than nb.
6) Interpreting entropy generation
For vacuum expansion, the surroundings exchange negligible heat, so the entropy change of the universe is dominated by the gas. Under this common idealization, entropy generation is reported as approximately equal to the system entropy increase. This helps link computed results to the second‑law inequality.
7) Typical classroom and lab ranges
Many exercises use ratios like V2/V1 = 2 to 10. With n = 1 mol, ΔS is about 5.8 J/K for doubling volume and about 19.1 J/K for a tenfold increase. Results scale linearly with n.
8) Best practices for reliable inputs
Use consistent units and verify that V2 > V1. If you enable the real‑gas option, confirm that V − nb stays positive to avoid unphysical states. When comparing models, keep the same n and volume ratio to isolate non‑ideality effects. Export CSV for spreadsheets and PDF for lab notes.
1) Is free expansion adiabatic?
In the classic vacuum case, there is no heat transfer and no boundary work, so it is often treated as adiabatic. However, “adiabatic” describes heat flow, not reversibility; the process is still highly irreversible.
2) Why can we use a reversible formula?
Entropy is a state function. You may compute ΔS using any convenient reversible path that connects the same initial and final equilibrium states, even when the real process is irreversible.
3) What if V2 equals V1?
If there is no change in volume, the entropy change from volume expansion is zero. This calculator enforces V2 > V1 to focus on expansion scenarios.
4) Does temperature matter for ΔS here?
For the volume‑only expressions used here, temperature cancels for the isothermal reference path. Temperature is shown as context and for better reporting, not because it changes the ideal‑gas ΔS formula.
5) When should I use the van der Waals option?
Use it when densities are higher and ideal behavior is questionable, or when you want a simple non‑ideality sensitivity check. It adds covolume correction through b and requires positive effective volumes.
6) What is the entropy generation value?
Entropy generation measures irreversibility. For free expansion into vacuum with negligible surroundings change, it is commonly approximated as equal to the gas entropy increase, since there is no compensating entropy decrease elsewhere.
7) Why is ΔS always positive for expansion?
Because ln(V2/V1) is positive when V2 > V1. Physically, more accessible volume means more microstates, which corresponds to higher entropy for the same amount of gas.