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I know that: IF adiabatic and reversible THEN isentropic

First question: does the implication IF isentropic THEN reversible hold for adiabatic processes?

Second Question: if yes to the above, are there other processes other than adiabatic for which it is true?

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  • $\begingroup$ Related: link $\endgroup$
    – foxfield
    Jun 2, 2016 at 9:26
  • $\begingroup$ these are logic problem. Assuming the first statement is correct, the first question is the logic consequence of the assumption. Because the assumption is a sufficiency statement, it doesn't exclude other processes. However adiabatic is also a necessity. With heat exchange, entropy will change. $\endgroup$
    – user115350
    Jun 2, 2016 at 19:43

3 Answers 3

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Thank you to everyone who contributed to this thread. Now, at least the first question, has an answer.

As clearly described at this related question, the answer to the first question is: IF adiabatic and isentropic THEN reversible.

In other words, for adiabatic processes, "isentropic" and "reversible" are equivalent notions.

As an aside, I notice here that it is also possible to arrive at this conclusion using the approach to adiabatic processes of Lieb and Yngvasson, for which the entropy is the only functional that completely characterizes the existence of an adiabatic process between two thermodynamical states.

But the second question remains, namely:

Does "IF isentropic THEN reversible" only hold for adiabatic processes?

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Entropy is a state function. This means that in every cyclical transformation (i.e. a transformation in which the initial and final states are the same) we have

$$\Delta S=0$$

That is to say: every cyclical transformation is isentropic.

But clearly there are cyclical transformation that are neither adiabatic nor reversible. Let's take the following example: a cylindrical container containing a perfect gas, with a movable adiabatic piston and in thermal contact with a large heat reservoir at temperature T.

The state of the system is initially $(P,V,T)$ where $T$ is the temperature of the reservoir and $P$ is the external pressure (let's assume that the piston's weight is negligible). If you quickly compress the gas pushing the movable piston, heat will be produced, which will flow to the heat reservoir. Now if you remove the force pushing the piston, the system will return to its original $(P,V,T)$ state.

The process was clearly non-adiabatic and irreversible, but since initial and final state are the same, it was isentropic.

Another example: let's make the cylinder of the previous example adiabatic. From the initial state $(P,V,T)$, we irreversibly (i.e. quickly, suddenly) push the piston downwards, taking the system to the state $(P',V',T')$. Now we irreversibly pull the piston back to its original position so that the final volume is $V''=V$. Since $P''=P$ also, from the ideal gas law $T''=T$, so we have performed a cyclical, adiabatic, irreversible transformation.

So isentropic+adiabatic doesn't imply reversible.

Conversely, it is easy to see that a reversible adiabatic process must be isentropic.

That's because for a reversible process we have

$$\Delta S_{AB} = \int_A^B \frac{\delta Q}{T}$$

If the process is also adiabatic $\delta Q=0$, so that $\Delta S_{AB}=0$.


Update

I would like to add some more things and clarify what I tried to say. For an irreversible process, we have

$$\Delta S_{AB} > \int_A^B \frac{\delta Q}{T}$$

we have already seen that

$$\text{adiabatic}+\text{reversible} \Rightarrow \text{isentropic} \ (\Delta S =0)$$

From the previous equation we see that

$$\text{adiabatic}+\text{irreversible} \Rightarrow \Delta S >0$$

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. So we cannot, if my reasoning is correct, conclude that

$$\text{adiabatic}+\text{isentropic} \Rightarrow \text{reversible}$$

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  • $\begingroup$ Thanks, but that doesn't answer the question. Let me try to reformulate the first question: if a process is adiabatic and isentropic, is it necessarily reversible? $\endgroup$
    – foxfield
    Jun 2, 2016 at 12:42
  • $\begingroup$ You're right, I should address your question better. I will update the answer. $\endgroup$
    – valerio
    Jun 2, 2016 at 13:00
  • $\begingroup$ The Clausius inequality tells us that, for a system that has experienced a process change between two thermodynamic equilibrium states, $\Delta S\geq \int{\frac{dq}{T}}$, where the equal sign applies to a reversible process, the greater than sign applies to an irreversible process, and where dq and T are evaluated at the boundary of the system. If the process is adiabatic and irreversible, then $\Delta S\geq 0$ and, if it is adiabatic and reversible, $\Delta S=0$. So, it is not possible to have an adiabatic irreversible process for which $\Delta S=0$. $\endgroup$ Jun 2, 2016 at 17:52
  • $\begingroup$ @ChesterMiller I may be wrong, but to me it looks like your reasoning (which is correct) only shows that adiabatic+irreversible $\Rightarrow \Delta S >0$ and adiabatic+reversible $\Rightarrow$ isentropic. This doesn't mean that adiabatic+isentropic $\Rightarrow$ reversible. In fact, for an adiabatic, isentropic process you obtain from the Clausius inequality $0\geq 0$ which is useless. $\endgroup$
    – valerio
    Jun 2, 2016 at 18:48
  • $\begingroup$ @valerio92 What can I say. In my judgment, my conclusion was correct. But, if you can think of a single example of an adiabatic isentropic process that is not reversible, let's hear about it. $\endgroup$ Jun 2, 2016 at 19:53
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A process does not have to be adiabatic or reversible to be isentropic. If the difference in entropy between State A and State B is zero, since entropy is a state function, any oddball path (either reversible or irreversible) that can take you from State A to State B will be isentropic.

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