My question is very similar to this, but I decided to ask another question because I felt that the problem deserved to be addressed in a more specific and formal way and I also wanted to discuss a specific example.
Let's start from Clausius' inequality.
For a reversible transformation between the points $A$ and $B$ we have
$$\Delta S_{AB} = \int_A^B \frac{\delta Q}{T}$$
While for an irreversible transformation we have
$$\Delta S_{AB} > \int_A^B \frac{\delta Q}{T}$$
For an adiabatic process, $\delta Q =0$.
Is thus easy to see that
$$\text{adiabatic}+\text{reversible} \Rightarrow \text{isentropic} \ (\Delta S =0)$$
and
$$\text{adiabatic}+\text{irreversible} \Rightarrow \Delta S >0$$
My question is:
Is
$$\text{adiabatic}+\text{isentropic} \Rightarrow \text{reversible}$$
also true?
I think that this is not true in general because for a general process we have
$$\Delta S_{AB} \geq \int_A^B \frac{\delta Q}{T}$$
If the process is adiabatic and isentropic, we obtain
$$0 \geq 0$$
which is trivially verified and tells us nothing about reversibility or irreversibility.
I would also like to bring a possible example of an adiabatic, isentropic transformation which is also irreversible.
I'll start by remarking two facts upon which I hope we agree:
- A non quasi-static transformation cannot be reversible. Quasi-staticity is a necessary condition for reversibility.
- Since entropy is a state function, every cyclical transformation is isentropic.
Let's take an ideal gas enclosed in an adiabatic vessel with an adiabatic movable piston. From the initial state $(P,V,T)$, where $P$ is the ambient pressure (we suppose that the weight of the piston is negligible), we irreversibly (i.e. non quasi-statically) push the piston downwards, taking the system to the state $(P',V',T')$. Now we irreversibly (non quasi-statically) pull the piston back to its original position so that the final volume is $V''=V$. Since $P''=P$ also (because the pressure must be equal to ambient pressure in order to have mechanical equilibrium), from the ideal gas law $T''=T$, so we have performed a cyclical, adiabatic, irreversible transformation.
PS: In case of positive answer (adiabatic+isentropic implies reversible), it would be best if the answer pointed out the flaw in my example but provided also a formal proof of the above-mentioned implication.