Change of entropy of surroundings in reversible and irreversible processes

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

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.

• 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? Feb 2 at 10:44
• @JeffJefferson "Assuming an irreversible process" where? Feb 2 at 12:28
• the answer to your question is "yes." Feb 2 at 13:00
• @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 Feb 2 at 14:10
• @JeffJefferson the answer is yes as Chet said but only because the system has not undergone a complete cycle Feb 2 at 15:18