Is it true that all processes where entropy increases are permitted by the second law, if the system is isolated?

Question

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?

Answer 1

This depends on what exactly you mean by permitted. If you actually have an isolated system initially out of equillibrium (say, a thermos with hot and cold water that hasn't mixed thoroughly), it evolves towards equilibrium in one way only. This can involve lots of complicated phenomena: at the very least heat convection and conduction, meaning that the possibility space is huge (you would have to describe the water as a vector field of velocities and a scalar field of temperatures), and only one path is realised. Which path that is depends on the exact initial conditions.

There exist observations that show some paths are consistently chosen. For example, the Leidenfrost effect: water droplets on a very hot surface create a layer of water vapor under them, which provides some insulation, makes the droplets dance around and survive for longer than on a colder surface.

That said, this is not the domain of classical thermodynamics. Classical thermodynamics only allows/forbids transitions between equilibrium states. The formal idea is that you have an equilibrium state with some inner constraint (such as an insulating wall between two volumes of gas), then you remove the constraint and wait to see what happens in the long run. Thermodynamics says that among the possible end states that are now accessible, the system will eventually reach the one with maximum entropy.

So in some sense, states in which entropy can be increased even further are forbidden by thermodynamics: you may pass through them, but not stop there. For example, if you throw an ice cube into a cup of hot tea, the cube will dissolve entirely, not just partially, even through that also increases entropy.

The question of what happens out of equilibrium belongs to non-equilibrium thermodynamics, though from my brief study of the subject, it seems that there is very little general and elegant to say; mostly it's models for specific systems.

Answer 2

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.