Entropy Change in a Reversible Isothermal Expansion
1 · Predict
A gas expands isothermally and reversibly to a larger volume. Does its entropy increase, decrease, or stay the same?
- Entropy increases — the gas has more possible microscopic arrangements at larger volume.
- Entropy decreases as the gas spreads out.
- Entropy stays constant since the process is reversible.
2 · Set Up
- Open the entropy-isothermal preset and press Reset. 1 mol of argon starts at 0.02 m³, 300 K.
- Enable the heat-exchanged and entropy-change readouts.
- Set the target (final) volume for each trial, and record the heat exchanged and the entropy change.
3 · Collect Data
| Final volume V₂ (m³) | Heat exchanged Q (J) | Entropy change ΔS (J/K) |
|---|---|---|
| 0.03 | ||
| 0.04 | ||
| 0.06 |
Plot entropy change ΔS (y-axis) against ln(V₂/V₁) (x-axis) for your three trials, using V₁ = 0.02 m³.
4 · Analyze
- For one trial, compute Q = nRT·ln(V₂/V₁) using n = 1 mol, R = 8.314 J/(mol·K), T = 300 K, then ΔS = Q/T (equivalently ΔS = nR·ln(V₂/V₁) directly). Compare both to the table.
- Explain why ΔS is positive in every trial here, and why a bigger expansion (larger V₂) produces a bigger entropy increase.
5 · Extend
- This experiment assumes a reversible (infinitely slow) expansion. A free (uncontrolled) expansion into vacuum to the same final volume would produce the exact same ΔS for the gas, even though no heat is exchanged at all (Q = 0) and no work is done. Explain why entropy is a state function that doesn't care how the gas got from V₁ to V₂.
- The second law says total entropy (gas plus surroundings) can never decrease for a real process. For this reversible isothermal expansion, the surroundings (the heat reservoir supplying Q) lose exactly as much entropy as the gas gains. Explain why that makes the total entropy change zero — the signature of a reversible process.
The Physics Behind This Experiment
Entropy Change in a Reversible Isothermal Process
For a reversible isothermal expansion, entropy change equals heat exchanged divided by the constant temperature, which works out to ΔS = nR·ln(V₂/V₁) for an ideal gas — entropy grows with the logarithm of the volume ratio.