Reversible vs. Irreversible Processes (same endpoints)
Two processes can start and end at the same thermodynamic states but differ in the
path. Path matters because quantities like work and heat depend on the path, while
state functions like internal energy \(U\) and entropy \(S\) depend only on endpoints.
Key ideas
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Reversible (ideal limit): quasi-static, no dissipation, can be reversed by an infinitesimal change.
It produces maximum work for a given expansion between fixed endpoints.
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Irreversible: finite gradients (pressure/temperature), friction, turbulence, free expansion, etc.
It generates entropy:
\( \Delta S_{\text{gen}} = \Delta S_{\text{univ}} > 0 \).
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Second law accounting:
\( \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \ge 0 \).
Equality corresponds to the reversible idealization.
Isothermal ideal-gas expansion example
Consider an ideal gas expanding isothermally from \(V_1\) to \(V_2\) at temperature \(T\).
The entropy change of the gas depends only on endpoints:
Reversible isothermal path (maximum work)
For a reversible isothermal expansion, the gas pressure satisfies \(P(V)=nRT/V\), so the work done by the gas is:
Free expansion (irreversible)
In a free expansion into vacuum (idealized insulated container), the gas does no boundary work:
\(W=0\), and receives no heat:
\(Q=0\).
Nevertheless, the gas entropy increases by the same endpoint formula:
Lost work (university note)
A common way to quantify irreversibility is “lost work”, comparing the reversible maximum work to what actually occurs:
What the PV plots mean
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The reversible isothermal PV curve is a true quasi-static path, so the area under \(P(V)\) matches the work.
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Free expansion is not quasi-static; it has no well-defined intermediate PV states. The calculator shows a dashed “jump”
to emphasize that the path is not a real equilibrium curve.