# Entropy change during irreversible process

Entropy change is defined as $$\Delta S = \int \frac{\delta Q_{rev}}{T}$$, where $$Q_{rev}$$ is heat transferred through a reversible process. I’m a little confused how this works, so I have a couple of related questions which I hope will clarify what’s going on.

1. If a reversible process is defined as $$\Delta S = 0$$ then how can this integral be non-zero for a reversible process? Or is it that just $$\Delta S_{Univ} = 0 = \Delta S_{sys} + \Delta S_{surr}$$ and therefore $$\Delta S_{sys} = -\Delta S_{surr} \iff T_{sys} = T_{surr}$$, in which case

2. How would one calculate it for an irreversible process, in particular, some heat, Q, transferred across a finite temperature difference? I would try (considering isothermal transfer between infinite temperature reservoirs) $$\Delta S_{Univ} = Q \big( \frac{1}{T_{sys}} - \frac{1}{T_{surr}} \big)$$ But this is no longer reversible, so I need to do something else?

3. Since entropy is a state function, is the entropy change through any process that is irreversible, but between equilibrium states, equivalent to $$\int \frac{\delta Q}{T}$$ through a reversible process?

4. Is this equivalent to pretending I’m doing 2 reversible transfers to some fictitious bodies, the first taking Q reversibly from the system to the fictitious body, both at $$T_{sys}$$, and then magically dropping the fictitious body to $$T_{surr}$$ and transferring the heat Q to the surroundings? If so, why use $$\delta Q$$ rather than $$dQ$$? My understanding was that $$\delta Q$$ was for path-dependent processes, but there is only one fixed (adiabatic) path for reversible transfer?

Sorry for yet another entropy question, thanks in advance!

It simplifies things a little to think about how entropy of a system can change. There are only two ways: entropy transfer across the boundary of the system by heat flow in and out at the boundary temperature, and entropy generation (due to irreversibility) within the system. In a reversible process, only the first mechanism is present. In both cases, the transfer across the boundary is $$\int{dq/T_B}$$ where $$T_B$$ is the boundary temperature.
• Thank you for your reply. The linked tutorial is indeed helpful. Let me assemble my questions and thoughts... I think you are saying that 4. is correct, this is equivalent to doing 2 reversible transfers to fictitious bodies? This seems similar to what you describe in the Example 1: Thermal equilibration, where the hot object has $dS = \frac{MCdT}{T}$ going from $T_h$ to $T_f$, without caring what the temperature of object 2 is, and similarly for object 2 from $T_c$ to $T_f$. – Furrier Transform Jan 5 at 3:45
• In example 2, may I ask why you are using the constant-volume heat capacity during a non-constant-volume (expansion) process? Also is the denominator $C$ in eq. 6, $C_v$? I think so, but you labelled it in the numerator, so just making sure. Right before Step 3, has the $T_f$ in state 2 lost the “1-...” it had in eq. 5? Not trying to be finicky, just making sure I didn’t miss something. – Furrier Transform Jan 5 at 4:01
• Are Ex. 2+3 saying that the $\Delta S = \int\frac{\delta q_{rev}}{T}$ is the same for any reversible path between two equilibrium states, and therefore irreversible adiabatic processes may have $\Delta S > 0$, but a reversible adiabatic process ($q_{rev} = 0$) will always have $\Delta S = 0$, no matter the reversible path constructed? – Furrier Transform Jan 5 at 4:06