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...when you just drop the weighted piston, it will fall and oscillateIt helps me to examine the 'dropped' piston case with the slowly lowered piston case in a somewhat realistic way. If Bob holds the piston back from falling fast, he applies an up force to balance most of the weight, but he allows the piston to drop.
momentarily. But when it has finally comes to rest, where has all of the mgh energy gone? Under the stated constraints it can only have gone to the gas.
/snip/
In the reversible case, you slowly lower the weight by pulling up with the variable force F so that at every instant the net external downward pressure (mg-F)/A only infinitesimally exceeds the gas pressure P. The gas pressure P slowly rises in value and you continuously reduce the strength of F until finally F=0 and mg/A=P. The total work done on the gas is then the area under the (mg-F) curve. This will be less than mgh.
Bob Sciamanda