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Q & A: entropy and reversible processes

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Most recent answer: 09/01/2010
To determine the entropy change for a reversible process, a summation of the (Q/T)s are done, where the Qs are the heat transfers to and from the high and low temperature reservoirs, and the Ts are the temperatures at which those transfers are performed. This equation is valid only for reversible processes, so to find the entropy change for an irreversible process, we take advantage of the fact that entropy is a state variable and find a reversible process that connects the same two equilibrium end points. My confusion stems from the fact that the entropy change for a reversible process is zero, not so for an irreversible process, which is positive. So how can this work?
- Ozzie Vidal (age 53)
Homestead, Florida
This question almost always come up as one tries to first learn the fairly subtle art of thermodynamic reasoning. So it's great that you asked.

Reversible processes involve zero total entropy change, summing over all systems and reservoirs involved. Since the entropy changes of the fixed-temperature reservoirs are straightforward to calculate, as you describe, one can then infer the entropy change of the system.

Irreversible processes involve net positive change in total entropy. Without further detailed analysis to determine just how big that net change is, calculation of reservoir changes is not sufficient to figure out how much the system entropy changed.

Mike W.

(published on 09/01/2010)

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