In the process of constant external pressure, the volume (change) work W=-Pe(V2-V 1) of gas is strictly deduced according to the definition of mechanical work. Here, gas can only be regarded as a system. No matter whether the gas is compressed or expanded, it is necessary to calculate the work done by external force. When the gas expands, W is negative, so we call it that the system overcomes the external force to do work (essentially, the external force does work on the system, but does negative work).
If you must regard the outside world as a system, for example, we don't know anything about the outside world, which may be a mechanical device that produces constant pressure or the atmosphere. If we discuss the "system" (using quotation marks to indicate that we regard the outside world in general discussion as a system) under the action of "external force" from gas, when the "system" is an incompressible system, there is no concept of volume work at all, so naturally W=-∫pidV=0 (note that dV is the volume change of the "system"), if the "system" is We don't know exactly how the atmospheric volume changes (we can't simply think that the atmospheric volume is reduced by V2-V 1, and the atmosphere is an open system with no definite volume), so we can't discuss the volume work. Even if it can be discussed, it is meaningless to the gas itself. We are concerned about the gas in the container, not the atmosphere.
When the first law of thermodynamics is applied to the gas in a container, it can only be that the work done by the outside (atmosphere) on the gas+the heat absorbed by the gas from the outside (atmosphere) = the increment of the internal energy of the gas. If we want to take the atmosphere as the system, it can only be the work done by the outside (gas) to the atmosphere+the heat absorbed by the atmosphere from the outside (gas) = the internal energy increment generated by exchanging energy with the outside (gas). It cannot be the work done by gas to the atmosphere (atmosphere is the system)+the heat absorbed by gas from the atmosphere (gas is the system) = the increment of internal energy of gas (gas is the system). Therefore, when you regard gas as a system, it is meaningless to investigate the work done by gas to the outside world. What is meaningful is the amount of work done by the system to overcome external forces (that is, the negative value of work done by the outside world on the system). You regard the atmosphere as a system, and indirectly discuss the energy change of gas through the energy change of the atmosphere.
So the volume work we are talking about is always the work done by external force on gas. We don't need to care about what the external force is provided by. What we care about is the energy change of gas, not the external energy change. When we need to know the energy change of the outside world, we can infer it from the energy change and energy conservation of the system, and it is difficult to draw a conclusion by directly analyzing the work with the outside world as the system.
It's really not easy to get to the bottom of this problem. Here is a typical example to illustrate the difficulty of volume work in irreversible process. For example, there is a massless, frictionless adiabatic piston in the middle of an insulated, constant-volume cylinder (the piston is initially fixed by a pin). The left side of the cylinder is filled with the ideal gas O2 of 1mol, 2atm, 273K, and the right side is filled with 1mol, 1atm, 273KO2. After pulling out the needle, ask how much work the left gas did to the right (that is, how much work the external force did to the right gas) and how much work the left gas did to overcome the external pressure (that is, the negative value of how much work the external force added to the left gas). This problem cannot be discussed in equilibrium thermodynamics. The difficulty lies in that the processes of gas generation on both sides are irreversible, and the gas processes on both sides are in an unbalanced state, and there is no definite pressure (the internal pressure on both sides is different everywhere), that is to say, the external pressure exerted by one side on the other side is unclear and cannot be directly calculated by integral. In order to discuss this problem, the system must be divided into a large number of thin layers (each layer is very small in macro and very large in micro), and each layer has approximately certain pressure. The first law is applied to these layers to analyze the changes of work, heat, internal energy and kinetic energy (note that the kinetic energy of each layer will change in non-equilibrium state, and there are both energy exchange and material exchange between layers). Finally, the energy change of the whole system is obtained by adding up all the layers, so the problem is extremely complicated. However, according to the conservation of energy, we can conclude that when the left gas expands to the final equilibrium state (there is no macro kinetic energy injection at this time). It is impossible to achieve a strict equilibrium state. Before the equilibrium state, the gas on both sides makes reciprocating damping oscillation motion near the equilibrium position, damping the friction from the gas, that is, viscous resistance and viscous resistance gradually dissipate the macro kinetic energy into internal energy). The work done by the left gas to the right must be equal to the negative value of the work done by the right gas to the left = the work done by the left gas to overcome the pressure of the right gas.
When the macro kinetic energy changes are noticed, the expression of the first law of thermodynamics is no longer internal energy increment = work+heat, and a kinetic energy increment should be added to the left.
The irreversible process of constant external pressure and constant external pressure can easily calculate the situation that the system overcomes the external pressure to do work. At this time, you should define the hard work and directly calculate how much work the gas has done to the outside world. On the one hand, it is meaningless (we don't need it to analyze the change of gas energy), on the other hand, it is very difficult to calculate directly. The landlord thinks that "the force of gas acting on the baffle should be greater than Pe, and the gradual decrease of Pi will eventually equal Pe, that is to say, the pressure of gas acting on the baffle will gradually decrease and eventually equal the external atmospheric pressure, so I think the work done by gas should be greater than the formula listed in the above formula" is not necessarily true. You don't know exactly what pi is, how it changes, and how much the air volume has decreased. Because there is no adiabatic condition, we can't compare the relative magnitude of the work done by gas to the atmosphere and the work done by gas to overcome atmospheric pressure only according to the conservation of energy.
If you have any questions, please discuss them further.