I believe your answer is correct.
Because we are doing work in this case, we can imagine that this is simply a piston being expanded against the atmosphere.
As you have said, a reversible process corresponds to the maximum amount of work, which in this case also corresponds to the largest change in energy.
Because we are only interested in comparing the temperature $T_2$ and $T_2'$ because the pressures and volumes change by the same amount, we must ask ourselves if the temperature is going to increase or decrease?
Well, without any mathematical proof, I will say that while we are doing work, and thus giving out energy to the atmosphere, the temperature must decrease because our system must lose energy. Thus, while outputting work, the more work we do, the greater the change in temperature.
The above paragraph depends on the process being adiabatic though because if the process were not adiabatic, we would have to consider the relative amounts of heat flow and things could get weird.
But, we have said that $$W_{rev}>W_{irrev}$$thus, $$|\Delta T_{rev}|>|\Delta T_{irrev}|$$but, since temperature is decreasing as we give out energy, $$T_2<T_2'$$
So, you are correct unless we are both making a stupid mistake.
This answer makes sense though, because the for the reverse process, we would be compressing a piston and thus must get back as much energy as possible in the form of a temperature increase. With changes in pressure and volume being the same, we know immediately that the process which begins at the lower temperature and gets back to the original temperature $T_1$ has the largest change in internal energy which must be true for the reversible process.