# Why is reversible process quasi-static? [duplicate]

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?

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

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!