0
$\begingroup$

When calculating entropy change for a irreversible process,do I assume a reversible path and then integrated it?

$\endgroup$
  • $\begingroup$ That will give you a lower bound, but not the actual entropy change. $\endgroup$ – probably_someone Feb 11 at 19:25
0
$\begingroup$

Once you have determined the final state of the system for the irreversible process (say using the 1st law of thermodynamics), you devise an alternate reversible process path between the same initial and final end states and integrate along that path. For further details on how to do this, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

$\endgroup$
  • $\begingroup$ Thanks!The example#2 in the reference is so much helpful! $\endgroup$ – XYC Feb 11 at 21:35
  • $\begingroup$ So why don't you accept my answer? $\endgroup$ – Chet Miller Feb 12 at 13:18
1
$\begingroup$

If the endpoints of the irreversible path are equilibrium points and they can also be connected with a reversible path then the answer is yes, because the entropy is a state function. Complications ensue if there is no reversible path connecting the endpoints of the irreversible path or the irreversible process does not start/end in thermodynamic equilibrium. An example for the latter is ferromagnetic hysteresis or plastic deformation, see Bridgman: "The Thermodynamics of Plastic Deformation and Generalized Entropy", REV. MOD. PHYS. VOL. 22. NUMBER 1 JANUARY, 1950. and Tolman & Fine: "On the Irreversible Production of Entropy" REV. MOD. PHYS. VOL. 20, NUMBER 1 JANUARY, 1948

$\endgroup$

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.