Limits on Adiabatic Process

Bob D pointed out that stirring work is irreversible. With that insight, perhaps the following point is why my claim in the question is false:

Referring to the rightmost graph of the figure, suppose point $2$ may be reached adiabatically from point $1$ and suppose point $2_{text{rev}}$ can reach point $1$ adiabatically. Consider the cycle $1rightarrow 2 rightarrow 2_{text{rev}}rightarrow 1$. If stirring work is used to move from point $2$ to $2_{text{rev}}$ in the rightmost graph, there is no entropy loss in the cycle. Thus, the counterclockwise cycle is reversible.

However, because stirring work is irreversible, the counterclockwise cycle is not reversible. So we have a contradiction.

I am providing a new answer in view of the additional details you have provided and our related discussions. I have also retracted my vote to close the post.

First of all, I believe we both agree that process $2$ to $2_{rev}$ in the third diagram and process $1$ to $2$ in the fourth diagram are impossible if adiabatic, based on the first law. The issue has to do the author's claim that process $1$ to $2$ in the fourth diagram would necessarily violate the Kelvin-Planck statement of the second law and why that is necessarily so, given the possibility of adiabatic stirrer work on the fluid going from state $2$ to $2_{rev}$ as you have suggested.

First, the author's argument-

To recall, the Kelvin-Planclk statement (from Wikipedia) is:

“It is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work”

If one assumes that process $1$ to $2$ in the fourth diagram were in fact possible, then one can device a heat engine cycle in which energy is absorbed from a single thermal reservoir and the cycle delivers an equivalent amount of net work, as follows:

After the process $1$ to $2$ expansion is completed the adiabatic boundary is removed. Then heat is transferred to (absorbed by) the fluid reversibly at constant volume from state $2$ to state $2_{rev}$. The adiabatic boundary is then replaced and the fluid is then compressed adiabatically and reversibly back to state $1$. We now have a heat engine cycle that produces net work by absorbing heat from a single reservoir during process $2$ to $2_{rev}$ in violation of the Kelvin-Planck statement.

Now, your argument-

To the above, you have raised the following question in regard to the fourth diagram:

However, couldn't we just let the "work device" that the piston rod is connected to rotate the piston rod to do rotational work on the fluid in order to bring the fluid from $(V_2, U^{'}_2)$ to $(V_2, > U_{2,text{rev}})$? Then, by making the path a "nonzero-work process", $(V_2, U^{'}_2)$ may still be reached adiabatically from $(V_1, U_1)$ and we would not be violating Kelvin-Planck's statement. In summary, what forbids the same thing from happening in the rightmost diagram from point 2 to point $2_{rev}$?

You are correct this would not violate the Kelvin-Planck statement since no heat transfer occurs. However, the difference between this and the author's argument is that there is also no net work done in the cycle you propose. If no net work is done, the Kelvin-Planck statement does not apply.

Applying the first law to the complete cycle, we have

$$Delta U_{cycle}=sum Q_{i}-sum W_{i}=0$$

Since the cycle is adiabatic, $sum Q_{i}=0$ and

$$sum W_{i}=0$$

Or

$$W_{Net}=0$$

$$W_{Net}=W_{1-2}+W_{2-2r}+W_{2r-1}=0$$

The first term is positive since the fluid does work in expanding against resistance of the apparatus. The second term is negative since the stirrer work is done on the fluid. The third term is also negative since the reversible adiabatic compression is work done on the fluid. Since the overall net work has to be zero, the work done by the fluid has to equal the work done on the fluid for a net work of zero.

In spite of no violation of the second law, process $1$ to $2$ in the last diagram still violates the first law.

Hope this helps.