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Is there an irreversible process, where the entropy of the isolated system does not change?

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  • $\begingroup$ I am assuming that you actually mean isolated (no exchange of matter and energy) instead of closed (no exchange of matter). $\endgroup$
    – valerio
    Feb 19, 2018 at 21:22
  • $\begingroup$ Yes, I do! Changed it. $\endgroup$
    – minum
    Feb 19, 2018 at 21:30

2 Answers 2

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No. An irreversible process is, by definition, one in which entropy increases.

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  • $\begingroup$ Thanks, I was trying to show it only using the (definition (?)) that a process is reversible iff its reversed process is also an valid one. $\endgroup$
    – minum
    Feb 16, 2018 at 21:16
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2nd principle says that for a given system $dS = \delta S_e + \delta S_c$. Therefore if $dS = 0$ it is still possible to have $\delta S_c > 0$ (i.e. an irrevesible transformation) if $S_e = \frac{Q}{T_{ext}} < 0$.

For example, the Joule-Thompson expansion is an isentropic transformation though irreversible.

What is always positive when we think about entropy is the entropy created $S_c$, not the difference of total enthalpy of your system, which can exchange entropy with other systems (if not a closed system).

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  • $\begingroup$ Thanks, but I am considering an closed system S_e = 0 and it seems that my problem stems from the definition of reversibility. $\endgroup$
    – minum
    Feb 16, 2018 at 21:20

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