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I am currently in high school and trying to gain some conceptual clarity on entropy and the second law of thermodynamics which led me to read various answers on this forum and i have referred some other texts as my textbook does not explain much about the differences in irreversible and reversible process. I have written my understanding so far and would appreciate if anyone could help point out any mistakes and clarify some doubts.

I have learnt that change entropy in a reversible process is a state function and therefore $\Delta$ S over a cyclic process should be zero and entropy change of a system in an irreversible process can be calculated by imagining a suitable reversible process that takes you between the initial and final states of the system. Further in a reversible process $\Delta$ $S_{sys}$ + $\Delta$ $S_{surr}$ = 0. In an irreversible process this does not hold true due to $S_{gen}$ created due to internal irreversible changes.

I have read that due to the clausius inequality $\Delta$ $S$ = $\int \frac{dQ}{T}$ + $S_{gen}$ but here i am confused about whether the dq they are talking about is due to an irreversible path or a reversible one and if the change in entropy is of the system or surroundings. I thought it would be of the surroundings but i am confused as my textbook just says $\Delta$ $S_{surr}$ = $\int \frac{dQ_{irrev}}{T}$.

Also since entropy change of the system is a state function regardless of whether the path is reversible or not is it correct to say that $\Delta$ $S_{surr}$ in an irreversible process is a path function as $\Delta$ $S_{uni}$ >0 in a cyclic process

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I have read that due to the clausius inequality $\Delta$ $S$ = $\int \frac{dQ}{T}$ + $S_{gen}$ but here i am confused about whether the dq they are talking about is due to an irreversible path or a reversible one

The $dQ$ is for any path. The temperature $T$ in the equation is the temperature at the boundary between the system and the surroundings, which is in this case a thermal reservoir. It is helpful to write it as $T_{B}$ to account for the possibility that the system temperature $T$ may not be the same as the temperature at the boundary $T_B$.

$dQ$ is reversible if the system is in thermal equilibrium with the reservoir, that is, when $T_{B}=T$ where $T$ is the equilibrium temperature of the system. In that case no entropy is generated and $S_{gen}=0$. If $T \ne T_{B}$ then the heat is irreversible and $S_{gen}\gt 0$.

Also since entropy change of the system is a state function regardless of whether the path is reversible or not is it correct to say that $\Delta$ $S_{surr}$ in an irreversible process is a path function as $\Delta$ $S_{uni}$ >0 in a cyclic process

Entropy is a state function for both the system and the surroundings. (One could just as well reverse the definitions of the system and surroundings).

For an irreversible system cycle, $\Delta S_{surr}\gt0$ is due to the fact that the surroundings has not completed a cycle as it has acquired the entropy that was generated in the system in order for the system to complete its cycle so that $\Delta S_{sys}=0$.

Hope this helps.

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  • $\begingroup$ Oh ok this helps, i was under the misconception that system and surrounding cycles would be the same. So is it possible to have $\Delta$ $S_{surr}$ = 0 assuming an irreversible process but here the system would not have a net zero change? $\endgroup$ Feb 2 at 10:44
  • $\begingroup$ @JeffJefferson "Assuming an irreversible process" where? $\endgroup$
    – Bob D
    Feb 2 at 12:28
  • $\begingroup$ the answer to your question is "yes." $\endgroup$ Feb 2 at 13:00
  • $\begingroup$ @BobD It was an assumption on my part, i apologize if it wasnt framed well. I wanted to make sure what i had asked was valid for an irreversible cycle of the surrounding $\endgroup$ Feb 2 at 14:10
  • $\begingroup$ @JeffJefferson the answer is yes as Chet said but only because the system has not undergone a complete cycle $\endgroup$
    – Bob D
    Feb 2 at 15:18

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