Why is the entropy production inside the system zero for a reversible process?

For a reversible process, why is the production of entropy within the system equal to zero?

It is not the production of “energy” that is zero in a reversible process but the generation of entropy that’s zero.

Entropy is generated in a process as a result of disequilibrium between a system and its surroundings. Examples are pressure disequilibrium and thermal (temperature) disequilibrium. A reversible process is carried out slowly so that the system and surroundings are always in equilibrium.

Hope this helps.

• Hahaha yes my bad, it was a typo. Thanks for your answer! – S. Tailor Jun 6 at 1:56
• Hi Bob. I think you'll like the analysis in the reference I mentioned in my Answer. It has filled many of the gaps that had existed in my understanding. – Chet Miller Jun 6 at 3:10
• @ChetMiller Hi Chet. Looked it up on line to see if any free down loads. Not sure I want to shell out a hundred bucks (being a cheap skate!) – Bob D Jun 7 at 19:42
• @Bob D During my long engineering career, I used this book more than all my others combined. It has a wealth of information. – Chet Miller Jun 7 at 20:11

See Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 11, Problem 11.D.1 Equation of change for entropy. They show that the local rate of entropy generation within a system is expressible as the sum of 3 terms, one proportional to the square of the temperature gradient, the second proportional to the square of the velocity gradient, and the third proportional to the square of the concentration gradient. In a reversible process, all three of these terms approach zero.

• Chet, do the first two terms roughly relate to temperature disequilibrium (temperature gradient) and pressure or force disequilibrium (velocity gradient)? And does the third term relate to chemical disequilibrium? In my head I have entropy generation as being due to disequilibrium which, in turn, is necessary to drive all real processes. I suppose it may be a little simplistic. – Bob D Jun 7 at 20:50
• @Bob D Yes. They relate to the rate of entropy generation per unit volume from temperature gradients, velocity gradients, and (I guess you found the other part of the development) concentration gradients. The presence of the various gradients provide the "driving forces" for finite rates of heat-, momentum-, and mass transfer. – Chet Miller Jun 7 at 23:01