13
$\begingroup$

So I'm preparing for my Thermodynamics undergrad exam, and I just can't wrap my head around the significance of reversibility vs. irreversibility of a process in relation to entropy. I mean if entropy is a state function, and a system in state A has S(A) entropy, and a system in state B has S(B), then what do we care whether the path between them is reversible or irreversible?

Also, my professor has stated that in an irreversible cycle the change in entropy is not zero. How can that be if a cycle is defined by having the exact same state as start and end, and entropy is a state function?

All this confuses me a lot, and I'd really appreciate some clarification.

Improve this question
$\endgroup$
2
  • $\begingroup$ Here is what solved the same confusion to me when I was at your point: "A state function is NOT the same thing as a mechanical potential.". It does look pretty much like one, does it not, with the definition of different paths that the system can take and all? So what's the difference? The difference is that in potential problems the potential is the only relevant physical quantity. It completely determines the dynamics of the system. In thermodynamics that's not the case. The relevant physics in thermodynamics is the heat transfer between temperature baths, which is an irreversible process. $\endgroup$ – CuriousOne Jun 6 '15 at 20:35
  • $\begingroup$ Hmm, thanks for the comment. I'm not sure I entirely get this, but you are totally right that in my head I treated these two, namely state functions and mathematical potentials as the same. I'm quite surprised they are not the same, actually. I will try and understand what you are saying here. $\endgroup$ – Benjamin Márkus Jun 7 '15 at 15:41
2
$\begingroup$

You need to consider the surroundings as well. If you go from state A to state B via a reversible process, the change in system's entropy exactly cancels out the opposite change in entropy for the surroundings; so overall there is no change in entropy. On the other hand, if it were an irreversible process, entropy change of the system (though same as the reversible case as it's a state function), doesn't cancel out the entropy change for surroundings. And overall there is a positive change in universe's entropy.

Further, in the reversible case, you can directly relate the change in system's entropy with heat transferred reversibly divided by temperature of the surrounding. But in irreversible case, the heat transferred irreversibly can't be used to evaluate entropy change and you would need to use an equivalent reversible process with same equilibrium end points.

Improve this answer
$\endgroup$
3
  • $\begingroup$ How should this work? The reversible process changes the system entropy by dS_sys^irr=dQ/T, while the irreversible process changes it by dS_sys^rev > dQ/T. So there is a difference in the 2 values. The environment contribution should be the same, because the environment is very large and takes up energy reversibly (so always dS_env=-dQ/T)? $\endgroup$ – Guiste Sep 7 '20 at 15:40
  • $\begingroup$ The environment contribution should be the same, because the environment is very large and takes up energy reversibly (so always dS_env=-dQ/T)? The change in total entropy (sys+env) is also different (=0 for reversible process and >0 for irreversible), so how can the entropy be a state function then? $\endgroup$ – Guiste Sep 7 '20 at 16:45
  • $\begingroup$ Total entropy(sys + env) is NOT a state function. But the entropy of the system ALONE is a state function, which means it does not depend on how you reached a state, be it reversible or irreversible path, the entropy of system in that state will be same. $\endgroup$ – zixtor Sep 8 '20 at 16:26
1
$\begingroup$

  1. In case of $ S(A)=S(B) $ there are many paths to connect $ A $ to $ B $ but only a single one that is reversible, i.e. can be traveled both ways without increasing the total entropy in the universe. This result is very important in cyclic processes.

  2. Your proffesor talked about the total entropy in the universe, not the entropy of the system that undergoes a cyclic process.

See Chapter 4 of "Thermodynamics & Introduction to Thermostatistics" of Herbert Callen, which has an excellent explanation for this topic.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Actually, there are an infinite number of reversible paths, all of which give exactly the same change in entropy. $\endgroup$ – Chet Miller Feb 4 '16 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.