In article ,
says...

I think I understand both examples. I don't think either refutes my
assertion that the weight ends up at the same place either way so work
delivered is the same. The "area under the path" argument works out
the same in both cases. Delivered work is the total available energy
(area of the whole triangle ( 1/2 * 30 * 2 = 30 in-lb ) minus the
remaining energy which is the area of the rectangle (15 * 1 = 15
in-lb) or 15 in-lb delivered.


The weight on the piston at the end of the expansion is
zero. As the graph shows, the force that is exerted varies
over the two paths.

It declines linearly from 30 to 0 in the unregulated case
(red line).

For the regulated case (the green line) the force is
constant at 15 lbs until the pressure in the vessel falls
below 2 atm, then declines to zero as expansion continues.

The work in the unregulated case is the entire area under
the red line = 30 in-lb. With the regulator in the circuit
the work performed is the area under the green line;
(15 X 1) + (15 X 1 x 1/2) = 22.5 in-lb.

Ned Simmons