- 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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What you are saying is correct: a reversible and adiabatic process between two states, $A$ and $B$, does not change the entropy either of the system or of its environment (surroundings). An irreversible and adiabatic process between two states, $A'$ and $B'$, increases the system's entropy. The two statements are reconciled by noting that if $A=A'$ then $B\ne B'$ and vice versa.
Another way of stating the same, is that an irreversible and adiabatic cycle is impossible. In fact, this statement is almost equivalent to Caratheodory's axiom (a standard formulation of the 2nd law), namely, that in any neighborhood of any state there are states inaccessible via a purely adiabatic process.
The apparent one-sidedness of this, is a verbalization of the increase of entropy function, whose existence is a mathematical consequence of the same. You can sense the physical intuition of an entropy increase as a manifestation of the excess work expended to compensate for the irreversibility of the process to reach a certain state.
Best Answer
Your professor is using a framework in which the entropy change of a closed system is equal to the sum of the entropy created within the system (by irreversibilities, such as viscous dissipation) plus the entropy entering and leaving the system through its boundaries. The entropy entering through each boundary of the system is given by $Q/T_\textrm{boundary}$ where $Q$ is the heat passing through that part of the boundary and $T_\textrm{boundary}$ is equal to temperature at the boundary through which the heat is flowing. So, in this framework, $$\Delta S=S_\textrm{created}+\sum{\frac{Q}{T_\textrm{boundary}}}$$ If the process is reversible, then $S_\textrm{created}=0$ and $T_\textrm{boundary}=T$, where T is the (uniform) temperature of the system.
For more on this powerful approach, see Fundamentals of Engineering Thermodynamics by Moran et al.