# Can an adiabatic, isentropic transformation be irreversible?

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:

1. A non quasi-static transformation cannot be reversible. Quasi-staticity is a necessary condition for reversibility.
2. 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.

• In your example, what makes you think that, when V"=V, P"=P. After the process you have described, T" will be > T, so P" will be > P. So the process you have described cannot be cyclical. The only way you can make it cyclical is if you allow for exchange of heat with the surroundings during one of the steps (probably the second step). – Chet Miller Jun 3 '16 at 14:03
• The fact that the pressure of the gas must be equal to ambient pressure to have mechanical equilibrium makes me think that $P''=P$. – valerio Jun 3 '16 at 14:08
• I will edit the question to clarify. – valerio Jun 3 '16 at 14:12
• Then if you allow the gas to expand irreversibly such that P" = P, the final volume can not be the same as the initial volume. You can't have both. If you want, I will solve the exact problem that you are referring to that we can see exactly how this plays out. – Chet Miller Jun 3 '16 at 14:29
• @ChesterMiller Mmh, ok. You are probably right, that is a valid objection. In order to have $P''=P$ I have to free the piston and the volume will change. But this could mean that my example is badly formulated. Can we formally show that adiabatic isentropic implies reversible? – valerio Jun 3 '16 at 14:38

We know that:

irreversible+adiabatic = $\Delta S>0$,

thus, if $\Delta S=0$ the process is either:

Then you can conclude that:
$\Delta S=0$+adiabatic=reversible.

Regarding you example of an irreversible adiabatic cycle: It is impossible and the example is flawed. The reason is that if the process is irreversible you will not come back to the same state: if the volume is the same, pressure and temperature will differ, as the work made by the piston when coming back and forth will not net zero, and the internal energy will change. Example: let it expand with an irreversible free expansion (no change in internal energy nor work by the system) and compress by any means you like (positive work on the system, increase in internal energy)

• I realized that it is wrong to try to do what I was doing (formally demonstrate the irreversibility starting from $\Delta S=0$ and $\delta Q=0$ and using Clausisus' inequality). As you said, if I managed to do an irreversible, cyclical, adiabatic transformation it would be a logical absurdum because we know that adiabatic+irreversible would imply $\Delta S>0$. I was mislead. – valerio Jun 5 '16 at 20:29

Perhaps a slightly more mathematical answer would also add something. We can write the entropy of a simple system as:

$$\mathrm{d} S = \frac{\mathrm{d} U}{T} + \frac{P}{T} \mathrm{d} V$$

According to the first law of thermodynamics, the internal energy can be written as:

$$\mathrm{d} U = \delta Q + \delta W$$

Where $\delta Q$ and $\delta W$ are not necessarily "reversible work and heat" (work and heat done in a reversible path). Joining both expressions together one finds that:

$$\mathrm{d} S = \frac{\delta Q + \delta W}{T} + \frac{P}{T} \mathrm{d} V$$

For an adiabatic, isoentropic path we have $\delta Q = 0$ and $\mathrm{d} S = 0$, this leaves us with $\delta W = -P \mathrm{d} V$, but this is nothing more than the expression for reversible work. This means that an adiabatic, isoentropic path must be a reversible path.