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I am confused whether the cyclic integral of dQ/T is zero for "Internally" reversible process or for a "totally" reversible process and why ?

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    $\begingroup$ please define what you mean by "internally vs. externally" reversible processes. $\endgroup$ – hyportnex Jan 29 '17 at 12:57
  • $\begingroup$ Internally Reversible process is one wherein the system comes back to its original state via the same path but the surrounding doesn't. $\endgroup$ – Stealth Jan 29 '17 at 14:03
  • $\begingroup$ External irreversibility refers to the surrounding coming back to its original state via the same path... $\endgroup$ – Stealth Jan 29 '17 at 14:04
  • $\begingroup$ In an irreversible process $\oint \frac{\delta Q}{T} > 0$, so somewhere the excess generated entropy must go, and the place where it goes the state does not return to its starting (original) state because there the entropy will have locally increased. $\endgroup$ – hyportnex Jan 29 '17 at 14:50
  • $\begingroup$ What exactly do you mean by "return to its original state via the same path"? I assume you don't mean that you a path that simply retraced its steps, as this would exclude even the Carnot cycle. $\endgroup$ – By Symmetry Aug 19 '18 at 9:45
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If the process is cyclic and internally reversible (i.e., reversible for the system), the integral evaluated over the system boundary is zero. However, it may not be zero if evaluated over the boundary of the surroundings, or over the combined boundary of the system and surroundings. The latter will happen if there are irreversibilities in the surroundings during the cycle.

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