Work done in Irreversible Isothermal Expansion

As a followup to the answer I gave previously and to my discussions with @Theorist, I have analyzed in detail the alternate irreversible process paths (1) specified by Theorist in his answer and (2) specified in the problem statement (and referred to in my answer as well as that of @Buck Thorn). To provide a common basis for comparison, I have derived the integral of $dQ/T_R$ for each of the two alternate irreversible process paths, where $T_R$ is the reservoir temperature exchanging heat with the gas (this is the temperature that should be used in conjunction with the Clausius inequality, as pointed out in Thermodynamics by Fermi).

For the process path specified by Theorist, it is found that $$int{frac{dQ}{T_R}}=nRleft(1-frac{V_1}{V_2}right)-nC_vleft(frac{V_2}{V_1}+frac{V_1}{V_2}-2right)$$ For the process path specified in the problem statement (and by Buck Thorn and myself), it is found that $$int{frac{dQ}{T_R}}=nRleft(1-frac{V_1}{V_2}right)$$ For any reversible path, the integral is equal to the entropy change, which, of course, is the same for all three paths:$$Delta S=nRln{(V_2/V_1)}$$For an irreversible path, the amount of entropy generated in the process is given by $sigma$: $$sigma=Delta S-int{frac{dQ}{T_R}}$$So, for the path specified by Theorist, $$sigma_{Theorist}=nRln{(V_2/V_1)}-nRleft(1-frac{V_1}{V_2}right)+nC_vleft(frac{V_2}{V_1}+frac{V_1}{V_2}-2right)$$and, for the path specified in the problem statement, $$sigma_{problem statement}=nRln{(V_2/V_1)}-nRleft(1-frac{V_1}{V_2}right)$$Note that the first two terms on the RHS of these entropy generated expressions are identical; however, the third term in expression for Theorist's path is positive for all values of the volume ratio. This indicates that, in this sense, Theorist's path is more irreversible since it involves more entropy generation. This is primarily the result of the first step in his process which involves irreversible cooling of the gas at constant volume by use of a lower temperature reservoir. This is not a criticism of the irreversible path he chose to specify (although it is not consistent with the problem statement). It is just a feature of that particular irreversible path.

In the parlance of thermodynamics, an irreversible isothermal process is defined as one in which the system and constant temperature reservoir start out at the same temperature and the system is held in contact with the constant temperature reservoir throughout the process, so that, in the end, they re-equilibrate at the same temperature.

In an irreversible process, one cannot use the ideal gas law to describe the gas behavior, because the ideal gas law is valid only at thermodynamic equilibrium. A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states, so the ideal gas law can be used for such a case.

In an irreversible process, in addition to the pressure being a function of the gas volume, it is also a function of the rate of change of volume. Fluid Mechanics tells us that this is caused by viscous stresses which contribute to the force per unit area exerted by the gas on the piston. In an irreversible expansion, these stresses lower the pressure that the gas exerts on the piston, so that, for a massless frictionless piston, the lower pressure exactly matches the externally applied reduced constant pressure. To provide the required match, the gas has to be expanding rapidly and at the right rate so that the viscous stresses are just right.

During the irreversible expansion, the gas temperature is not spatially uniform. The part of the gas in contact with the constant temperature reservoir is, of course, at the reservoir temperature, but the interior of the gas becomes cooler. However, by the end of the expansion, when thermodynamic equilibrium has been re-established, the gas temperature is again uniform and equal to the reservoir temperature.

This is all a one-stage process, and, by controlling the pressure on the piston so that it suddenly drops initially and then stays constant throughout the deformation, this can be considered a constant-pressure expansion.

In short, the answer given by @theorist isn't even close to being correct. @Buck Thorn's answer is much closer to the target.

There is some confusion about terminology. The figure shows two processes. The curved path represents a reversible expansion in which there is pressure balance between system and surroundings throughout, so that the applied pressure can be described by the ideal gas law (or Boyle's law $p=frac{p_1V_1}{V}$). This process is clearly isothermal, since $pV=nRT$.

The second process shown as two straight paths consists of a pressure drop and and an expansion against constant external pressure and is also isothermal (in thermodynamic lingo). The fact that the initial and final points are at equal T makes it isothermal. The temperature (and other properties except presumably composition) of the system during the process is on the other hand not well-defined. This process is irreversible.