2
$\begingroup$

Many people say that a reversible process must be quasi-static and infinitely slow. I (think I) understand the examples involving gases inside pistons to demonstrate the point, but I don't understand why people say this has to be true for all possible processes.

Consider a simple situation: a lossless spring launches a tennis ball vertically upward, the ball reaches the peak, and falls back on the spring. Are the classifications below by me correct?

  • It is quasi-static because all relevant variables are always well-defined (spring extension, spring extension velocity, ball position, ball velocity all take definite values).
  • It is reversible, because the motion of the spring and the ball adhere to Newtonian mechanics when run forward and backward.
  • The process does not take an infinite time to carry out.

If the above is true, then doesn't this serve as a counterexample to the idea that reversible processes must be infinitely slow? What am I misunderstanding?

$\endgroup$
2
  • $\begingroup$ Yes. Processes like the one you described that do not involve viscous dissipation,, internal rapid heat conduction, internal rapid molecular diffusion, and chemical reaction at finite rate do not need to be carried out slowly/quasistatically. $\endgroup$ Commented Apr 7 at 11:25
  • $\begingroup$ "Many people say that a reversible process must be quasi-static and infinitely slow". Infinitely slow and quasi-static are not separate conditions. A process must be infinitely slow to be quasi-static. $\endgroup$
    – Bob D
    Commented Apr 7 at 14:42

1 Answer 1

1
$\begingroup$

The quasistatic/indefinitely slow condition is relevant for thermal contact exchange only and is used to define equilibrium thermodynamics.

In order exchange heat by a heat flow between to systems in thermal contact, there must be a temperature difference.

But if the two temperatures of two reservoirs in contact are different by a small amount $dT$, by the basics, an exchange of $dQ$ of heat energy increases the entropy by $$-\frac{dQ}{T+dT}+\frac{dQ}{T}=\frac{dQ}{T} \frac{dT}{T}.$$

So all equations concerning entropy bilances are valid only in the limit $dT\to 0$. Time simply must not play a role in the ideal Carnot processes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.