Confusion in the pressure-volume work in an irreversible process

Question

Let there be a certain amount of gaseous substance kept in a cylinder fitted with a weightless & frictionless piston. The initial volume and pressure of the gas are $V_1$ & $P_1$ respectively. The gas initially exists in thermodynamic equilibrium state with the surroundings(piston) . That means the gas is in mechanical equilibrium with the pressure from the surroundings which,therefore,initially exerts pressure $P_1$ .Now to take the gas at state where volume is $V_2$, suddenly the external pressure is reduced to $P_2$ . Since $P_2 < P_1$ , therefore the gas will do work on the piston to again equalize the pressure. Now,the question is how much work is done by the gas?

Here, discrepancy arose between my thinking and my book. What I did to find the work done was simple: since at equilibrium ,the pressure was $P_1$, no work was done. Now, the external pressure is reduced to $P_2$ . Therefore a net pressure arises on the piston which is equal to $$P_1 - P_2$$ ,which will be exerted by the gas( since it exerts the same previous pressure $P_1$ on piston and the external surroundings now exert pressure $P_2$ on the piston . Therefore,the gas is exerting greater pressure by an amount $P_1 - P_2$ and that will do work ,right?) . Therefore, work done $$w = \int_{V_1}^{V_2} (P_1 - P_2)\, dV $$ . This was my deduction. But my book did it like that: $$ w = \int_{V_1}^{V_2} P_{\text{ext}} \, dV \implies \int_{V_1}^{V_2} P_2\, dV $$ . So, what is the problem with my approach? I couldn't understand my book's approach. Is mine wrong? If so, why? And why did the book do so? Plz help.

Answer 1

For $PV$ work in which $P_{\text{ext}}$ is a constant, you could actually intepret it as $W=Fx$ where $F$ is the force on the piston (due to pressure) and $x$ is the distance travelled by the piston.

From here you can see that the force which you applied plays no part in determining the work done against opposing force.

Imagine lifting a mass off the ground, it does not matter how much force you exert in lifting the mass, all that matters is the distance the mass is lifted, as well as the gravitational force, mg. The work done is simply $W=mgh$. More explanation could be found here

Similarly, in this case, the internal pressure plays no part in determining the work done against an external pressure, as the work done against a constant opposing force(or opposing pressure) is only determined by the distance travelled by the piston (or change in volume).

Do note that this is true if and only if the piston is stationary before and after expansion. If expansion causes the piston to be moving in the final state, work done may include KE term.

Answer 2

Let us assume a imaginary world having a container filled with gas with piston over it and I am applying a force $(Pressure\times area)$ on piston externally and gas applying a foce on piston internally.

Now let us say that piston moves $dl$ distance outwards (opposite to force applied by me since assumed that force applied by gas on piston is greater than me). The work done by me will be $(-F_{me}.dl)$,which means I am gaining this amount of energy.

By conservation of energy, this amount of energy should lost by system, which is given by negative of work done by me. Hence, work done by the system is $P_{ex}.dl$.