# 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.

• Are we talking about reversible processes only? Entropy is conserved in all reversible processes. – Greg Feb 28 '17 at 7:04
• @Greg Please read my question – Nova Feb 28 '17 at 7:59
• Please describe in more detail what you are referring to when you say isothermal expansion at constant pressure. Are you saying that you add heat to the system while you hold the pressure constant at the original value, or are you saying that you hold the system in contact with a constant temperature bath while, at time zero, you drop the external pressure to a lower value and then hold the external pressure at that constant value until the system re-equilibrates. – Chet Miller Feb 28 '17 at 15:49

Isothermal expansion at constant pressure is NOT a reversible process and hence ∆S(universe) is always greater than zero. I.e. $$\Delta S_\mathrm{univ}=\frac{Q_\mathrm{actual}}{T}-\frac{\int PdV}{T}$$
$$i.e. ~ \Delta S_\mathrm{univ} >= 0$$
• $\Delta S_\mathrm{universe}$ is not always positive. – ringo Feb 28 '17 at 8:16