Is the total entropy change of all isothermal processes 0?

We define total entropy as: $$\Delta S\_\mathrm{total}=S\_\mathrm{system}+S\_\mathrm{surroundings}$$


I know that for an isothermal AND reversible expansion: $S\_\mathrm{total}= 0$


But what about an isothermal expansion **at a constant pressure**? Does $S\_\mathrm{total}$ also equal 0? If not, then would the equation look something like this: 
$$\Delta S\_\mathrm{total}=\frac{q\_\mathrm{rev}}{T}+\frac{q\_\mathrm{surroundings}}{T}$$


where $q\_\mathrm{surroundings}$ is the heat lost or gained by the surroundings by a process and $q\_\mathrm{rev}$ is the heat lost or gained by the system **IF** that process was to become reversible.





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> Is the total entropy change of all isothermal processes 0?
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No. Only for reversible processes.


However your equation is correct.


For an irreversible process in which an ideal gas expands isothermally against a constant external pressure


$$\Delta S\_\mathrm{total}=\frac{q\_\mathrm{rev}}{T}+\frac{q\_\mathrm{surroundings}}{T}$$


becomes


$$\Delta S\_\mathrm{total}=\frac{q\_\mathrm{rev}}{T}+\frac{w}{T}$$


where $w=-p \Delta V$, since $q\_\mathrm{surroundings}=-q\_\mathrm{system}$ and for the isothermal process the first law stipulates that $w=-q\_\mathrm{system}$.


For a free expansion $p\_\textrm{ext}=0$ and no work is done. No heat is exchanged and the entropy of the surroundings does not change. The change in entropy of the system then drives the process.


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Isothermal expansion at constant pressure is **NOT** a reversible process and hence ∆S(universe) is always greater than zero.
I.e. $$\Delta S\_\mathrm{univ}=\frac{Q\_\mathrm{actual}}{T}-\frac{\int PdV}{T}$$


$$i.e. ~ \Delta S\_\mathrm{univ} >= 0 $$


Equality holds only for *reversible processes*.


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Only reversible processes are iso-entropic.


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