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I really don't understand why this needs to be. To be reversible, isn't all we need is that there is no net energy change in the system so, who cares how fast or slow it goes?

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marked as duplicate by Qmechanic Jan 28 '18 at 22:16

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/39589 A process that occurs and leaves the total entropy of the universe unaltered would be reversible, if you look at the definition of irreversible processes. In reality, a quasi-static change not not actually achievable , it's an idealisation. Energy is not really the thing here, it's entropy. No macroscopic system can actually undergo a reversible reaction, new entropy always results. $\endgroup$ – user181180 Jan 28 '18 at 18:46
  • $\begingroup$ Where did you get the idea that only the total energy change matters, and not how fast it occurs? $\endgroup$ – Chet Miller Jan 28 '18 at 19:44
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    $\begingroup$ Related: physics.stackexchange.com/q/168, physics.stackexchange.com/q/336289, physics.stackexchange.com/q/78405 and links therein $\endgroup$ – valerio Jan 28 '18 at 21:50
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Here's an extreme example… In order to double the volume of a gas in a cylinder fitted with a piston, I pull the piston out at a speed comparable with the rms speed of the molecules. In that case the pressure of the gas in the vicinity of the piston is less than the pressure, $p$, in the bulk of the gas and the work done by the gas for each increment $\Delta V$ of volume increase will be less than $p\Delta V$, the work done in the reversible case. And in a very obvious way the change is irreversible: however fast or slow I push the piston back in, I won't create the same pressure conditions in the gas, specifically I won't reduce the pressure next to the piston!

Even for less extreme speeds, there'll be turbulence in the gas, the work done by the gas will be different from the reversible case, and the same pressure conditions won't be gone through in reverse by pushing the piston back in.

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