It is possible to deduce that in a thermodynamic process for an isolated system $\mathrm{d}S$ has to be greater than zero, from this it follows trivially that $ \Delta S > 0$. It is usually said then that in an isolated system, thermodynamic processes always increase entropy between the initial and final states. My question is: is the converse true? Meaning, is it true that if $\Delta S > 0 $ then the process is permitted by the second law?

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    $\begingroup$ define spontaneous please $\endgroup$
    – anna v
    Aug 5, 2015 at 16:41
  • $\begingroup$ My definition: a process is spontaneous if it occurs in an isolated system. $\endgroup$
    – hyportnex
    Aug 5, 2015 at 17:26
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    $\begingroup$ By "spontaneous" you may mean that a process occurs without external stimulus. In a thermodynamically isolated system, this would mean that the 2nd law of thermodynamics applies to the spontaneous process without need to adjust for energy leaving or entering the system, if I guess your meaning correctly. $\endgroup$
    – Ernie
    Aug 5, 2015 at 17:55
  • $\begingroup$ Yeah, I guess "spontaneous " is kind of redundant, I understand "spontaneous" as permitted by the laws of thermodynamics. $\endgroup$
    – Ignacio
    Aug 5, 2015 at 19:11
  • $\begingroup$ I will edit accordingly $\endgroup$
    – Ignacio
    Aug 5, 2015 at 19:12

1 Answer 1


Spontaneity of a thermodynamic process can be analysed from the perspective of free energy. Lets look at the Gibbs' free energy: $$dG=dH-TdS$$ For a process to be spontaneous, $dG<0$. Likewise, a non-spontaneous process is equivalent to $dG>0$. At equilibrium, we have $dG=0$.

Consider a couple of scenarios:

  1. $dH>0$ & $TdS>0$: Then $dG<0$ or $dG>0$ depending on the relative size of $dH$ vs $TdS$. If $dH>TdS$, then $dG>0$ and is non-spontaneous. If $dH<TdS$, then $dG<0$ and is spontaneous.
  2. $dH<0$ & $TdS>0$: Then $dG<0$ and is always spontaneous.
  3. $dH>0$ & $TdS<0$: Then $dG>0$ and is always non-spontaneous.
  4. $dH<0$ & $TdS<0$: Then $dG<0$ or $dG>0$ depending on the relative size of $dH$ vs $TdS$. If $dH<TdS$, then $dG>0$ and is non-spontaneous. If $dH>TdS$, then $dG<0$ and is spontaneous.

So to answer your question: No, it is not necessary to have $dS>0$ for your process to be 'permitted'. Instead you could have the same situation as case 4. However in any process the total entropy change has to be $dS_{tot}\ge0$ by the second law of thermodynamics.

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    $\begingroup$ $dG < 0 $ is only a criteria for spontaneous processes if $T$ and $P$ are held constant, so it does not apply here, the process in question concerns an isolated system but I never asked that $T$ and $P$ are held constant. $\endgroup$
    – Ignacio
    Aug 5, 2015 at 19:09
  • $\begingroup$ @Ignacio: Your converse statement is already not true if $T$ and $P$ are held constant. Why do you expect that to change when they are not held constant? $\endgroup$
    – nluigi
    Aug 5, 2015 at 19:38
  • $\begingroup$ I ment your answer applies to not isolated systems when $T$ and $P$ are held constant, my question is about an isolated system, lots of things can happen in a non isolated system. As you well said the total entropy has to be greater or equal to cero in any thermodynamic process, in an isolated system the total entropy is equal to the entropy of the system, so the entropy of the system always has to increase and you can never have your "case 4" scenario where the entropy decreases and the process happens anyway, that's a scenario that only applies to non isolated systems. $\endgroup$
    – Ignacio
    Aug 6, 2015 at 18:25

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