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Why the entropy change is not zero in the irreversible adiabatic process? while it is defined as the integral of the heat added to the system over its temperature

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  • $\begingroup$ The reversible path between the two end states of an irreversible adiabatic process is not adiabatic. $\endgroup$ – Chet Miller Mar 2 '19 at 19:59
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Although entropy change is defined in terms of a reversible differential transfer of heat divided by the temperature at which the heat is transferred, you can have entropy change without heat transfer. Since entropy is a state function independent of the path between states, you can calculate it by assuming any reversible path connecting the states.

A classic example of an irreversible process causing entropy change not involving heat transfer is the free adiabatic expansion of an ideal gas.

A rigid insulated chamber is partitioned into two equal parts. Half the chamber contains an ideal gas. The other half is a vacuum. An opening is created in the partition allowing the gas to freely expand into the evacuated half. Since the chamber is insulated, there is no heat transfer ($Q=0$). Since the expansion of the gas does not expand the boundaries of the chamber, there is no boundary work ($W=0$). Consequently, per the first law, the change in internal energy is zero ($\Delta U=0$). Being an ideal gas, where a change in internal energy depends only on a change in temperature, there is therefore no change in temperature.

The end result is the volume has doubled the pressure has halved and the temperature is unchanged.

Although no heat transfer has occurred, the process is obviously irreversible (you would not expect the gas to spontaneously return to its original half of the chamber). But we can determine the entropy generated by taking any convenient reversible process to return the gas to its initial conditions so that the total entropy change of the system is zero. The obvious choice here is to remove the insulation and perform a reversible isothermal compression. To do that requires heat transfer to the surroundings. That amount of heat represents “lost work”, that is, the work that could have been done if the free expansion of the gas was replaced by a reversible adiabatic expansion.

Hope this helps.

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If you have an irreversible adiabatic process between two thermodynamic equilibrium end states of a system, there exists no possible reversible adiabatic process between these same two end states. So to get the entropy change for the irreversible adiabatic process, you need to devise an alternative reversible path between the same two end states, and this reversible path will not be adiabatic. On the reversible path, you will have to add heat to the system in order to transition between the same two end states. So the reversible path will give a positive change in entropy.

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The statement $$ dS = \frac{\delta Q}{T} $$ is only true of reversible processes. The generally true statement is $$ dS \ge \frac{\delta Q}{T}. $$ For an adiabatic process, $\delta Q = 0$, but that still allows for $dS > 0$.

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  • $\begingroup$ but if delta Q= 0, ds will be zero as well which is only true for a reversible process $\endgroup$ – AHMED KRS Mar 5 '19 at 11:54
  • $\begingroup$ @AHMEDKRS Yes, that's what I said. Equality only holds for reversible processes. For irreversible processes, the inequality is strict. $\endgroup$ – eyeballfrog Mar 5 '19 at 17:28

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