Difficulties with understanding total entropy change and unavailabillty

Question

Of course, I know the fact that the entropy of an isolated system never decreases. Neverthless what makes me confused about the entropy(or change of entropy) of an isolated system is the explanation used to tell the unavailable heat/work in a irreversible process.

With a closed system and its surrounding imagine that the system expanding isothermally(at T) then for

1) reversible process

Surely the total change of entropy is zero. Here I say the process is from a state A to a state B and the heat system absorbs is q.

2) irreversible process (from state A to B)

With very large surrounding compared to the system, the temperature of the surrounding can be assumed to be constant and same as that of system. Because the entropy is a state function the entropy change of the system is still q/T. It is obvious that q and q'(heat absorbed by system in irreversible process) are different. The problem is that a lot of books and lecture notes explain that the entropy change of the surrounding is -q'/T.

By the definition of entropy I cannot handle this statement because -q' is the heat extracted from the surrounding "irreversibly". If q' can be treated as a heat transferred reversibly why not use this quantity for the system? In the view of 2nd law of thermodynamics the difference between the magnitudes of entropy change of the system and the surrounding should exist in irreversible process but if we can say that -q'/T is the change of entropy of the surrounding we can also say that the change of entropy of the system is q'/T. As a result the total entropy change will be zero, which doesn't obey the law of thermodynamics.

Books/lecture notes examples for this problem:

Introduction to the thermodynamics of materials, D. R. Gaskell, 5th ed., p.44. & A note for lost work

In addition if the entropy of the system is a state function can't we regard the entropy of the surrounding as a state function? 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. In this way the total entropy change will be zero since q/T - q/T = 0. If reversible this result is not strange, but state function doesn't depend on paths but states...

In short, my troubles are

1. Many books/notes argue that q-q' is the heat unavailable or consumption of free energy, and I think this statement is plausible but the reason that we can take q' is not permissible thing. q' is not the reversibly transferred heat. It has irreversible property and thus we can't use q' for calculating entropy.

2. Entropy is a state function and this is true for both entropies of systems and surroudings. If we take an irreversible path between states(A to B) the total entropy change would be zero for we can always draw a reversible path between the states(also A to B). However this violates the law of thermodynamics.

Answer 1

2. 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.

1. 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.

.

Answer 2

The entropy change in going from A to B in reversible case is Zero because system and surroundings are in equilibrium all along like a controlled very slow expansion of gas inside a cylinder by piston but in the irreversible case system and surrounding are not in equilibrium at all which meant even if the system goes from A to B the surrounding might go from A to C but not from A to B it is the total change in entropy of system and surrounding which is greater than zero because the behavior is now more random as compared to reversible case in which it is fully predictable.