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Since $ds=\frac{dq_{rev}}{T}$ for reversible processes it seems we can have reversible isentropic processes that are not adiabatic provided the temperature changes in such way that the sum of $\frac{dq}{T}$ is zero but the sum of $dq$ itself is not zero. Is this possible? What are some example of such a process used/found in real life?

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In conventional thermodynamic language "adiabatic" process means being adiabatic (no heat exchnage) at every instant of the process not just in the sense that the total heat exchange is zero. The process itself can be reversible or irreversible, only the heat exchanged must be zero. The so-called adiabatic legs of the Carnot cycle (isothermal-adiabatic-isothermal-adiabatic) are such processes during which there is no heat exchange at all; heat is exchanged during the two isothermal legs: absorb heat at the high temperature and release heat at the low temperature.

Using conventional language then adiabatic and reversible must mean that $\delta q_{rev}=0$, hence, you also have $dS=0$, see below the answer of @BobD; in other words the system entropy is kept constant at any instant, i.e, isentropic.

Except for a single book, namely Pippard: Elements of Classical Thermodynamics, every other book on thermodynamics under the sun uses the word adiabatic in the sense I described above. (In his book Pippard uses the word adiathermal in the sense that during any instant of the process that can be reversible or irreversible the heat exchanged is zero, and by adiabatic he means the special case of a process that is reversible and adiathermal that is isentropic.)

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  • $\begingroup$ But is it wrong to say a a cycle is not adiabatic overall because part of it is adiabatic and part of it is not? So because there is heat exchanged in the isothermal legs of the Carnot cycle, the (reversible) carnot cycle overall is isentropic but not adiabatic. Note I didn't ask if an adiabatic reversible process is isentropic, I asked if an isentropic process is adiabatic $\endgroup$
    – Skawang
    Jun 29, 2020 at 7:33
  • $\begingroup$ Also, does isentropic mean $dS=0$ at every instant or is it $\int dS =0$? Since entropy is a state variable the latter makes more sense to me $\endgroup$
    – Skawang
    Jun 29, 2020 at 7:47
  • $\begingroup$ Look, you are allowed to call anything whatever you want for "naming" does not change the physics. In conventional text-book language a Carnot cycle has two adiabatic and two isothermal legs. The isothermal legs are also diathermal for they involve heat exchange (diathermal). During an isentropic (adiabatic+reversible) stage you have $dS=0$ always and every instant meaning $S$ is kept constant; note that a Carnot cycle is always defined to be a reversible one at every stage. .... $\endgroup$
    – hyportnex
    Jun 29, 2020 at 11:57
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    $\begingroup$ In any cycle the initial and final states of the system are the same, meaning all its macroscopic parameters are returned to their initial values, that entropy, internal energy, pressure, temperature, magnetization, etc. The conventional language does not call such cycles iso; that attribution iso when applied is usually reserved for something that is completely unchanging not just at the beginning and at the end. $\endgroup$
    – hyportnex
    Jun 29, 2020 at 12:00
  • $\begingroup$ I appreciate the clarification, I wasn't aware this was the general notation. I'll accept @Bob D 's answer as it more directly addresses my question. thanks for your help $\endgroup$
    – Skawang
    Jun 29, 2020 at 16:54
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it seems we can have reversible isentropic processes that are not adiabatic..

No you cannot.

An isentropic process is by definition a process that is both adiabatic and reversible. So you can't have an isentropic process that is not adiabatic.

However, you can have an adiabatic process that is not isentropic, if it is not a reversible process. An example is an adiabatic process involving friction losses.

The equation defining a differential change in entropy

$$ds=\frac{dq_{rev}}{T}$$

can be used to determine the difference in entropy between any two states by assuming any convenient path between the two states and evaluating the integral, because entropy is a state function (independent of the path).

For example, suppose the actual process connecting the two states is an irreversible adiabatic process. Since it is irreversible, $ds$ is not zero and it is not isentropic even though there is no heat transfer.

We can determine the difference in entropy between the two states by evaluating the equation for any convenient reversible path between the two states since entropy is a state function. We might, for example, connect the two states with a combination of a reversible isothermal and reversible isochoric (constant volume) process, and evaluate the integral for both.

Hope this helps.

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  • $\begingroup$ What about cyclic processes like the other answer said? What's wrong with combining reversible isentropic and non isentropic processes to give an overall process thats reversible isetnropic but not adiabatic? Admittedly that's not what I had in mind when I asked the question but I don't see why that won't work $\endgroup$
    – Skawang
    Jun 27, 2020 at 15:27
  • $\begingroup$ I see reversible (carnot) cycles are by definition isentropic. But since they're not adiabatic(overall), wouldn't that fit my question of "processes that are isentropic but not adiabatic"? I think it comes down to terminology so I'd appreciate if you'd explain to me why that's not true as I'm not a physics student and I'm not entirely sure how those terms are really used $\endgroup$
    – Skawang
    Jun 29, 2020 at 7:42
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Any reversible cyclic process will be isentropic (since entropy is a state function), whereas it's not necessary that it's adiabatic as well. A famous example is the Carnot cycle.

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  • $\begingroup$ Could you clarify what you mean by a "cyclic" process? The Carnot cycle is a reversible cycle that includes two reversible adiabatic (isentropic) processes. $\endgroup$
    – Bob D
    Jun 27, 2020 at 14:45
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    $\begingroup$ @BobD A cyclic process is one where the initial and final states of the system are the same. Despite Carnot cycle being a composition of two adiabatic and two isotherma processes, still overall, Carnot cycle is not adiabatic, since $\Delta Q_{\rm cycle}\neq 0$. $\endgroup$
    – user258881
    Jun 27, 2020 at 16:34
  • $\begingroup$ Who said a Carnot cycle is adiabatic? $\endgroup$
    – Bob D
    Jun 27, 2020 at 20:31

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