Would entropy maximized at equilibrium if there were no irreversible processes? I have read this question, but it does not answer my question:
In a truly ideal isolated system (say an ideal gas), it is quite possible that there is no irreversible process such that the net entropy production is zero. In such a case, the entropy of the system would not change from its initial value. Does that mean the system will not evolve toward an equilibrium? or are there other reasons which force system to evolve toward an equilibrium and maximize entropy?
 A: After reading your post several times, and reading the various comments of others as well as your responses, I think the question as to whether or not the entropy of a "truly" isolated system is maximized depends on the role that constraints internal to the system play to prevent the initiation of irreversible processes within an isolated system. Here are my thoughts.
Let's say you have a system consisting of a single component ideal gas contained in a rigid, perfectly thermally insulated chamber. The rigid insulated walls of the chamber form the boundary between the system and surroundings. So the system (ideal gas) is considered isolated.
Now within our isolated system is a fixed partition that divides the gas into two equal volumes. Let's stipulate that the temperature and pressure of the gas on one side of the partition is the same as the other. We would then say that the gases on each side of the partition are in thermal and mechanical equilibrium with one another, and that our isolated system is internally in equilibrium. Moreover, if the partition was somehow carefully removed so that the act of removing it did not itself "disturb" the system, the system would still be in thermal and mechanical equilibrium. In other words, the partition is irrelevant and the entropy of the system is already maximized.
Now instead let's say the temperature of the gas on the left side of the fixed partition is greater than on the right side and that our partition is perfectly thermally insulated and fixed in place.  Since the volumes are the same, the pressure on the left side is also higher than the right side. So we ask once again, is our system internally in thermal and mechanical equilibrium? Is the entropy of our system "maximized"? 
For the gas on the left and right side to be in thermal equilibrium, in order to obey the zeroth law, there would have to be no net heat flow if the partition was permeable to heat. In our case, however, there is no heat flow because the partition is not permeable to heat. Similarly, for the gas on the left and right side to be mechanical equilibrium, if the partition was not fixed in place the higher pressure gas on the left would not do any work compressing the gas to the right. That would also not be the case. In short, if the partition were not thermally insulated nor fixed in place, there would be heat transfer and/or work done internal to the system. Moreover the processes would be irreversible, because heat will not spontaneously flow from the right side back to the left, and the gas on the right will not spontaneously do work compressing the gas on the left to return the system to its original state. Intervention by the surroundings would be needed, which in turn will leave the surroundings changed.
Although the system is not technically internally in thermal or mechanical equilibrium, as long as the constraint remains in place no irreversible processes can take place and no entropy produced. But we are relying on the internal constraint to prevent irreversible processes from taking place. We might also ask, if it is possible to alter the characteristics of the constraint, is the system then truly isolated? Is any system truly isolated?
As food for thought, I hope this helps.
