- For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is
$$\Delta S = \int_a^b \frac{dQ}{T}$$
For reversible path between two states, entropy of the universe (Or any isolated system) is zero.
$$\Delta S + \Delta S_\text{surroundings} = 0$$
So You cannot just take any system and say that entropy change between two states for this system will be zero because it is zero for a reversible process. It is not. So when you say
Surely the total change of entropy is zero.
for reversible process of closed system, it is not true.
Answer to This question might help you here.
- As for the first part of your question, I don't understand what the question is. Could you edit it to be more specific?
Also, You said the following, which is false.
The entropy changes of the system are same for both cases, reversible
and irreversible processes because the first and final states are
unchanged. In this situation I think the surrounding also have the
same first and final states for both reversible and irreversible
processes.
We don't know whether surrounding has same first and final states or not. We only know about the system's first and final states. Think about it this way: In a reversible process, system is going from state A to B, and so is surrounding. Since it is reversible, $ \Delta S_{System} = - \Delta S_{Surrounding} $. So ultimately, $ \Delta S_{Universe} = 0$.
Now for an irreversible process, we know that through this irreversible path, the System goes from A to B. We don't know about surroundings. Now, since system's states are same, $ \Delta S_{System} $ will be same as above case. For the surrounding, you say that states are same as the reversible case. But then, here also $ \Delta S_{Surrounding} $ would be same as before and again $ \Delta S_{Universe} = 0$. But we know that that is not true for irreversible process. Hence, Surrounding's states must be different. So, in irreversible process, while the system goes A to B same as before, the surrounding must go from A to some C. There is no reason to believe that it would go from A to B again.
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Of course, thermodynamics says things about differences in entropy between reversible and irreversible processes. In fact, we can analyse how irreversible a process was and what to do in order to, for instance, extract more work from it. That's a very important part of thermodynamics.
But, no, there are no truly reversible processes — at least for daily macroscopic, bulk phenomena; I might be unaware of some crazy quantum-relativistic-boson-fermion experiment :). In practice, all real processes are irreversible, even though some can be good approximations of reversible ones (e.g. if you go really slowly).
Also, your statement "classical thermodynamics gives a lower bound on the increase of entropy" is a bit weird. That "lower bound" would be zero, so I'd prefer to say that it states that global entropy never decreases. Apart from that, classical thermodynamics also goes on to explore quantities like availability, irreversibility and efficiency, and it can do a good work of optimizing the hell out of your irreversible processes.
Best Answer
There is not just one reversible path between the initial and final thermodynamic equilibrium states of a system. There are an infinite number of reversible paths, and they all give the exact same value for the change in entropy (as well as for the changes in the other thermodynamic functions). The integral of $dq/T_{boundary}$ for all these paths is also greater than the corresponding integral for any irreversible path, where $T_{boundary}$ is the temperature at the boundary between the system and its surroundings. This is known as the Clausius inequality.