Entropy in Adiabatic Transformations
Theoretical Background
- Adiabatic process: $q = 0$
- First Law: $\Delta U = q + w \;\Rightarrow\; \Delta U = w$
- For an ideal gas: $U = U(T)$
- Reversible adiabatic: $\Delta S = 0$
- Irreversible adiabatic: $\Delta S > 0$
- Entropy change of the universe:
\(\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}\)
Important distinction.
$\Delta S_{\text{sys}}$ is always computed by comparing initial and final equilibrium states, even if the actual path is irreversible.
$\Delta S_{\text{univ}}$ tells us whether the transformation is reversible ($=0$) or produces entropy ($>0$).
Exercise
A sample of 1 mol of ideal gas expands adiabatically from $V_1 = 10.0\,\text{L}$ to $V_2 = 20.0\,\text{L}$.
- Evaluate the entropy change of the system in the case of a reversible adiabatic expansion.
- Repeat for an irreversible adiabatic free expansion against vacuum.
- Discuss the entropy change of the universe in both cases.
Step-by-Step Solution
Case 1 — Reversible adiabatic expansion
- By definition, $q_{rev} = 0$
- Entropy change of system: \(\Delta S_{\text{sys}} = \int \frac{dq_{rev}}{T} = 0\)
- So: $\Delta S_{\text{sys}} = 0$
Entropy of the surroundings:
\(\Delta S_{\text{surr}} = 0\)
Therefore:
\(\Delta S_{\text{univ}} = 0\)
Case 2 — Irreversible adiabatic free expansion
- Still $q = 0$, so $\Delta U = 0$ (no work done, no heat exchanged).
- For the system entropy we must compare initial and final equilibrium states.
For 1 mol ideal gas: \(\Delta S_{\text{sys}} = R \ln\!\left( \frac{V_2}{V_1} \right) = 8.314 \ln 2 = 5.76\,\text{J K}^{-1}\)
Surroundings are vacuum, so:
\(\Delta S_{\text{surr}} = 0\)
Therefore:
\(\Delta S_{\text{univ}} = +5.76\,\text{J K}^{-1}\)
Case 3 — Discussion
- Reversible adiabatic: entropy does not change (system and universe).
- Irreversible free expansion: entropy of the system increases, and so does that of the universe.
This illustrates the Second Law:
\(\Delta S_{\text{univ}} \ge 0\)
Notes
- In a reversible adiabatic expansion, the gas cools as it expands, but the process is perfectly balanced: entropy remains constant.
- In the free expansion, the gas ends up occupying a larger volume at the same temperature, increasing disorder → entropy rises.
- This is a textbook example that entropy is a state function, independent of path, but the entropy production distinguishes reversible vs irreversible paths.
- Pedagogical reminder: Free expansion is only adiabatic and irreversible in the idealized case of expansion into a vacuum. In real systems, some interactions with the surroundings are unavoidable.