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My understanding is that in a reversible process there is no increase in entropy, it remains constant. So $\Delta S = 0$ no?

And since it's reversible we know from the second law that $\Delta S = \frac{\delta Q_{\mathrm{rev}}}{T}$.

So doesn't a process being reversible mean that $\delta Q_{\mathrm{rev}} = 0$ ?

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    $\begingroup$ One usually refers to the entire process. Thus the key factor for reversibility is that the system is passing through equilibrium states via infinitesimal changes. $\endgroup$ – Phoenix87 Sep 6 '16 at 20:32
  • $\begingroup$ An example of an adiabatic irreversible process: take a partitioned box with vacuum in one half, an ideal gas in the other, and a breakable partition between. Break the partition. Temperature is constant, volume doubles, pressure halves, and no heat is exchange. An example of a non-adiabatic reversible process: the isothermal phases of the Carnot engine. $\endgroup$ – Sean E. Lake Sep 6 '16 at 23:21
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The formula for the change in entropy $$ \mathrm{d}S = \frac{\delta Q_\mathrm{rev}}{T} $$ refers to the entropy change of the system. For a reversible process the total change in entropy of the system and its surroundings will be $0$. The reason for this is that for a reversible process temperature of the system and its surroundings must only differ infinitesimally (otherwise an infinitesimal change could not reverse the process) and clearly the heat transferred to the system is the negative of the heat transferred to the surroundings so \begin{align} \mathrm{d}S_\mathrm{sys} + \mathrm{d}S_\mathrm{sur} & = \frac{\delta Q_\mathrm{sys}}{T} + \frac{\delta Q_\mathrm{sur}}{T}\\ & = \frac{\delta Q_\mathrm{sys} - \delta Q_\mathrm{sys}}{T}\\ & = 0 \end{align}

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The change in entropy for a system undergoing an adiabatic reversible path is zero. For a system that undergoes an adiabatic irreversible process, you need to devise a reversible path between the same initial and final states of the system, and then calculate the integral of dq/T for that path. That's what we mean when we say that the change in entropy is the integral of $dq_{rev}/T$. If the irreversible path is adiabatic, the reversible path will not be adiabatic. You will find that it is impossible to devise an adiabatic reversible path between the same initial and final thermodynamic equilibrium states that were obtained with an irreversible adiabatic process.

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