Why Use the Derivative When Calculating Work? [ENDORSED]

[Gabriel Wolf 3I]

I was a bit confused on why a reversible process is more specific than an irreversible. I should clarify, just in case there is a fundamental issue with my understanding, that to my knowledge a reversible process is one that is in constant equilibrium with its surroundings throughout the process and an irreversible process is one that is not until the end. If this is the case, with the reactions in both methods of measurement reacting and pushing until equilibrium is reached, why does it matter if you take the integral? From what I understand, the reversible process converts all work done by the gas into work pushing the piston, as the pressure is the same inside and out, meaning the only thing the gas is pushing against is the piston's weight. This makes sense as a very accurate way to measure work. However, in an irreversible process, where the outside pressure is constant, can't we calculate the exact amount of work necessary to push the piston any amount of distance? Assuming we use the same piston as before, meaning the amount of energy lost to friction is the same between the two, how is it any less accurate? The only extra thing the gas is pushing against is a constant outside pressure, which to my understanding essentially equates to additional downward push, which would functionally increase the piston's weight by a constant amount. Or are we assuming that an irreversible reaction is SO sudden that the piston gains momentum, and thus will lose some amount of energy to the surroundings in an overshoot that cannot be accounted for? Or am I missing the point entirely?

[Chem_Mod]

We use the derivate when looking at small changes.
e.g., Small change in volume: dV

Small reversible changes occur when a system is in equilibrium with its surroundings.
e.g., System P = Surroundings P = 2atm
Need to sum the many small changes (take the integral) to obtain the total change.
System does more work as it is pushing against 2atm.

Large irreversible changes occur when a system is not in equilibrium with its surroundings.
e.g., System P = 2atm
Surroundings P = 1atm
Piston must be held in place by something (pin).
Remove pin and system undergoes a large, sudden expansion (use Delta V not dV).
System is pushing against 1atm and does less work.

As I commented in class, introductory thermodynamics always ignores friction and mass of piston, etc.