The entropy of the surroundings does change infinitesimally. But the surroundings are large and such a change does not change the total entropy of the surroundings in any sensible way.
Indeed, one already uses that fact in putting the system through a series of reversible steps. As you point out, if the temperature of system and surrounding were in fact identical, no heat would flow. But they are infinitesimally different and so an infinitesimal amount of heat does flow.
The same applies to the surroundings. It too is undergoing a reversible change.
- 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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Best Answer
The OP in the link is making erroneous statements, such as "Since the the surroundings remain at constant pressure". For an isothermal expansion it is the temperature of the surroundings that is constant. Now, getting to your questions.
The OP is referring to a reversible isothermal expansion where the surroundings is a thermal reservoir. For a reversible process the total change in entropy (system + surroundings) is zero. Therefore the change in entropy of the surroundings has to be the negative of the change in entropy of the system.
For a reversible isothermal expansion, the temperature of the surroundings is considered the same as the temperature of the system, since the difference is infinitesimal.
Hope this helps.