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For both reversible and irreversible adiabatic processes, $dQ =0$,and by the definition of entropy $dS=dQ/T$, it should imply that entropy is constant for both. Why it is not so?

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    $\begingroup$ Your definition of entropy change is incorrect. It should read $$dS=\frac{dQ_{rev}}{T}$$ where the subscript rev refers exclusively to a reversible path. $\endgroup$ Dec 22, 2020 at 18:17

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In general, you can write that for all processes $\frac{\delta q}{T} \le dS$ (Clausius) but in the case of a reversible process you have equality $\frac{\delta q}{T}|_{rev} = dS$. If the process is adiabatic then by definition $\delta q =0$ hence for the reversible case you have the equality $dS=0$, i.e., an isentropic process, but for an irreversible process you can only say that $0<dS$ which is usually verbalized by saying that the entropy can only increase (never decreases) in an adiabatic process that is not reversible.

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Although entropy is defined for a reversible transfer of heat as $dS=dQ_{rev}/T$, entropy can be generated without heat transfer. An example of an irreversible adiabatic process that generates entropy is the rapid expansion or compression of a gas.

The classic example is the case of an insulated rigid chamber that is divided into two parts, one containing a gas and the other a vacuum. An opening is created in the barrier between the two parts allowing the gas in one part to spontaneously and rapidly expand into the evacuated part. This is an irreversible process (You would not expect the gas to spontaneously return to its original part leaving a vacuum behind) that involves no work, no heat transfer and no change in internal energy. If the gas is an ideal gas, then there would also be no temperature change.

For this example you can calculate the entropy change by assuming a reversible process between the initial and final states and applying the definition for entropy change. The obvious choice is reversible isothermal expansion process. You can do this because entropy is a state function. The difference in entropy between two equilibrium states does not depend on the path (process) connecting the states.

Hope this helps.

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