There is a crucial pedagogical point hidden in thermodynamic diagrams: the same gas can be taken from the same initial state AA to the same final state BB in completely different ways, one reversible and one irreversible. The natural question is: what changes and what stays the same between the two paths?

The answer separates two categories of quantities. The internal energy EintE_\text{int} and the entropy of the gas SgasS_\text{gas} are state functions: they depend only on the state of the gas, not on the path taken to reach it. So ΔEint\Delta E_\text{int} and ΔSgas\Delta S_\text{gas} are identical on the two paths, because AA and BB are the same. In contrast, the heat QQ and the work LL exchanged depend on the path: they differ from one process to another, even for the same endpoints.

Summary — From A to B: same gas, same change of state, but...

Reversible: Q, Lgas0Q,\ L_\text{gas} \neq 0, with ΔStot=0\Delta S_\text{tot} = 0. Irreversible (free expansion): Q, Lgas=0Q,\ L_\text{gas} = 0, with ΔStot>0\Delta S_\text{tot} > 0. In both cases ΔEint\Delta E_\text{int} and ΔSgas\Delta S_\text{gas} are equal, because they are state functions.

This distinction has a powerful practical consequence. In an irreversible process there is no well-defined path in the PP-VV diagram: there is no curve to integrate. But precisely because entropy is a state function, we can get round the obstacle.

The reversible case as a "useful path"

When ΔSgas\Delta S_\text{gas} must be calculated for an irreversible process, where there is no well-defined path in the PP-VV diagram, one can imagine a fictitious reversible path between the same initial and final states, because entropy is a state function. It is a practical technique: “treat the problem as if it were reversible” to calculate ΔS\Delta S, but remember that the real one is not.

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Topics: Termodinamica Concepts: Primo principio della termodinamica · Energia interna · Entropia Objects: Gas ideale

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