Is delta S of universe always 0 when an ideal gas undergoes an isothermal irreversible process?

Since for an irreversible process,
$$dS\_\mathrm{Surrounding} =-\frac{\text{dq}\_\mathrm{irr,sys} }{T\_{\text {surr }}}\tag{1}$$
where $\text{dq}\_\mathrm{irr,sys}$ is heat exchange of system
and $-dq\_\mathrm{irr}$ is heat absorbed by the surrounding.
Also $d u=d q+d w$.
Now in an irreversible isothermal process involving only ideal gas.
$$
\begin{aligned}
& du=d w+d q \quad . \quad(d v=C v \cdot d T) \\
& -d w=d q \\
& P\_{\text {ext }} \cdot d v=d q 
\end{aligned}
$$
Now Substituting in (i) Now $T\_{\text {syst }}=T\_{\text {surr }}$ As the process is Isothermal . Clearly $dS\_{\text {Surr}}=-dS\_{\text {Sys}}$. Does that mean change in entropy of universe in irreversible isothermal process is always $0$?




For the irreversible case, we are talking about a situation in which the ideal surroundings reservoir is maintained at the same temperature as the initial temperature T of the system, such that the final equilibrium temperature of the system is also T. This is what we mean by an isothermal irreversible process, even though, within the bulk of the system during the process, the temperature may not be spatially uniform at T.


If, in determining the change in entropy of the system, you devise a reversible path between the same initial and final thermodynamic equilibrium states of the system as for the irreversible process, you will find that the final thermodynamic equilibrium state of the surroundings will not be th[e same as for the irreversible process. So this approach gives you the correct change in entropy for the system in the irreversible process, but it will not give you the correct value for the surroundings. Since the reservoir is ideal, the change in entropy of the surroundings for the irreversible process will be
$$\Delta S\_{surroundings, irreversible}=\frac{q\_{surroundings, irreversible}}{T}=-\frac{q\_{system, irreversible}}{T}$$
The change in entropy of the system for both the irreversible and reversible path will be $$\Delta S\_{system,irreversible}=\frac{q\_{system,reversible}}{T}$$
So, $$\Delta S\_{total, irreversible}=\frac{(a\_{system,reversible}-q\_{system,irreversible})}{T}$$


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