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.

  • $\begingroup$ Are we talking about reversible processes only? Entropy is conserved in all reversible processes. $\endgroup$
    – Greg
    Feb 28, 2017 at 7:04
  • $\begingroup$ @Greg Please read my question $\endgroup$
    – Nova
    Feb 28, 2017 at 7:59
  • $\begingroup$ 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. $\endgroup$ Feb 28, 2017 at 15:49

2 Answers 2


Is the total entropy change of all isothermal processes 0?

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

$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$


$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{w}{T}$$

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.


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.


Only reversible processes are iso-entropic.

  • $\begingroup$ $\Delta S_\mathrm{universe}$ is not always positive. $\endgroup$
    – ringo
    Feb 28, 2017 at 8:16
  • $\begingroup$ You might want to take a look at this discussion of how to properly format mathematical and chemical equations. It's really not hard to learn and it will make your answers much more readable. $\endgroup$
    – airhuff
    Feb 28, 2017 at 9:48

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