Chemistry For irreversible process from state 1 to 2, why can the system not be isolated for reversible process from 1 to 2?

AI Thread Summary
The discussion centers on why a system cannot be isolated during a reversible process when transitioning from state 2 back to state 1 after an irreversible process from state 1 to state 2. It emphasizes that while the entropy change for the irreversible process is positive, the total entropy change for a reversible process must equal zero, necessitating heat exchange with the surroundings. This heat exchange implies that the system cannot be isolated, as it requires heat to leave the surroundings to maintain reversibility. The conversation also references the second law of thermodynamics, highlighting that certain states are inaccessible without heat transfer. Overall, the inability to isolate the system is crucial for maintaining the conditions required for a reversible process.
zenterix
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Homework Statement
There is a basic derivation about entropy that is shown below that I still have doubts about even though I am able to do various calculations involving entropy.
Relevant Equations
Consider an isolated system and an irreversible process A from state 1 to state 2.
Then

$$q_{irrev}=0\tag{1}$$

Take the system from state 2 back to state 1 using a reversible process B.

My first question is: why can the system not be isolated for this reversible process to be possible?

Assume we have a non-isolated system in process B.

Process A and process B together form an irreversible cycle.

$$\oint dS<0\tag{2}$$

$$\oint dS=\int_1^2 dS_{irrev}+\int_2^1 dS_{rev}\tag{3}$$

$$=\int_2^1 dS_{rev}\tag{4}$$

$$=S_1-S_2<0\tag{5}$$

$$\implies S_2-S_1>0\tag{6}$$

From (3) to (4) we used the fact that ##dS_{irrev}=\frac{\delta q_{irrev}}{T}=0##.

(6) tells us that the entropy always increases for an irreversible process in an isolated system.

This is the basis for saying that entropy never decreases in the universe.

The universe is an isolated system so any irreversible process in it generates entropy increase.

Any reversible process in the universe is done such that

$$\delta q_{sys,rev}=-\delta q_{surr,rev}$$

$$d S_{sys,rev}=\frac{\delta q_{sys,rev}}{T}$$

$$d S_{surr,rev}=\frac{\delta q_{surr,rev}}{T}=-\frac{\delta q_{sys,rev}}{T}$$

$$\implies dS_{univ}=dS_{sys,rev}+dS_{surr,rev}=0$$

Thus,

$$dS_{univ}\geq 0$$
 
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zenterix said:
Homework Statement: There is a basic derivation about entropy that is shown below that I still have doubts about even though I am able to do various calculations involving entropy.
Relevant Equations: Consider an isolated system and an irreversible process A from state 1 to state 2.

Then

$$q_{irrev}=0\tag{1}$$

Take the system from state 2 back to state 1 using a reversible process B.

My first question is: why can the system not be isolated for this reversible process to be possible?
The isolated system's entropy change is positive when it goes from state 1 to 2 irreversibly
and the surroundings' entropy change is zero since nothing happens in the surroundings.

For a reversible process between the same states (from 1 to 2) the system's entropy change is the same as for the irreversible process since entropy is state function.
For the process to be reversible the total entropy change has to be zero.
Therefore the surroundings' entropy change has to be negative.
This means heat has to leave the surroundings and enter the system and the system cannot be isolated.
 
"From (3) to (4) we used the fact that dSirrev=δqirrevT=0." That's not a fact, this is wrong. You always have to calculate the entropy change using the heat exchanged in a reversible process.

Btw, there is the formulation of the second law according to Caratheodory: "In the neighbourhood of any equilibrium state of a physical system with any number of thermodynamic coordinates, there exist states which are inaccessible by adiabatic processes."
The prototypic example is that of stirring a propeller in a viscous fluid: You are only doing work on the system, so dq=0. But this work is dissipated in the fluid, increasing its internal energy. The only way to bring back your system into the state where it started from, is by removing heat over the boundary.
 
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