Irreversible process with $dS=0$ (isolated system)

Question

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

Chris · Accepted Answer · 2018-02-19T21:53:36.040

2

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

edited Feb 19 '18 at 21:53

answered Feb 16 '18 at 20:25

Chris

17,377

fxmarty · Answer 2 · 2018-02-16T21:08:20.133

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).

Irreversible process with $dS=0$ (isolated system)

2 Answers2