irreversible process

[Henry_2A]

Hi!

Why is the greatest work done when the process is irreversible? In particular, I understand the idea that external pressure is matched with the pressure of the gas at every stage (infinitely small changes), but I can't seem to grasp the connection between that and maximum work.

[Jose Angelo Grajeda 2D]

Hello! So actually I believe the greatest work occurs in a reversible process. When ∆V_i changes infinitesimally like that (in very small changes), then Pressure likewise changes in very small fluctuations, to the point where P_internal ≈ P_external like you said! Since pressure internally is matched to pressure externally, the pressures along the pathway are approximately the same at each step, so we consider this to be a "reversible" expansion. The work done is calculated by P*∆V at every infinitely small step, so this process is really slow, but it does more work (because there are more P*∆V calculations, especially at a higher value for both).

Now, in an irreversible reaction, P_ex < P_internal, so there is one fast change, but it does less work (because there is simply one constant P and P*∆V calculated). Therefore, reversible expansions do more work than irreversible ones, I think. Hope this helps!