As I understand it, a reversible process is required to be quasi-static because each infinitesimal step in a quasi-static process generates only infinitesimal amounts of entropy at a time which can be reversed with only an infinitesimal amount of work. But my question is: even if only infinitesimal amounts of entropy are generated at each step, when you integrate this over a finite path, doesn’t the work required to reverse the process integrate to a finite value, rendering the process irreversible? Given this, how can any process be irreversible?
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No real process is reversible, for precisely the reason you mention: a gradient (e.g., in temperature, pressure, or chemical potential) is required to drive a process, but energy moving down that gradient produces entropy. By skilled engineering (to reduce friction, for instance) and by slow operation, we can reduce entropy generation to an arbitrarily low level, but we cannot make it zero. The idealization of zero entropy generation and reversibility is nonetheless sometimes useful.
answered Mar 13, 2022 at 1:44
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$\begingroup$ Thanks for the insightful reply! But why would a slow process generate any less entropy than a fast process? $\endgroup$ Commented Mar 13, 2022 at 2:01
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2$\begingroup$ The rate of entropy generation per unit volume due to local temperature gradient goes as the square of the temperature gradient, not temperature gradient to the first power. The rate of entropy generation per unit volume due to velocity gradient goes as the square of the velocity gradient, not velocity gradient to the first power. The rate of entropy generation per unit volume due to local concentration gradient goes as the square of the concentration gradient, not concentration gradient to the first power. So when you integrate these with respect to time for a process, (continued) $\endgroup$ Commented Mar 13, 2022 at 2:50
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2$\begingroup$ it comes out higher for an irreversible process than for one that tis closer to reversible. For more details on this, see Transport Phenomena, Bird,, Stewart, and Lightfoot, Chapter 11, problem 11D1. $\endgroup$ Commented Mar 13, 2022 at 2:52
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$\begingroup$ Thanks for the comment. So the bottom line is: to prevent generating entropy, you want to minimize the gradient at all times, which is achieved by keeping the system close to equilibrium (i.e. quasi-static). Correct? $\endgroup$ Commented Mar 13, 2022 at 7:52
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1$\begingroup$ Yes. That is exactly correct. $\endgroup$ Commented Mar 13, 2022 at 9:28