5
$\begingroup$

If you add Heat on a constant volume container, disorder will increase, so Entropy will increase. But what if I do some work on a piston-cylinder? I'll give more kinetic energy to particles, so they gain more disorder, but they also occupy a smaller volume, thus particles would be more organized (less entropy). Is this valid? Can work create entropy? I ask this because $$ dS=\frac{dq}{T}$$ By this equation it seems that entropy only depends on Heat, not in work.

$\endgroup$
1
  • $\begingroup$ $dU = TdS-pdV$... so? $\endgroup$ May 3, 2017 at 19:27

1 Answer 1

5
$\begingroup$

Reversible work transfers no entropy and generates no entropy; when you adiabatically and reversibly compress a gas, for example, the entropy increase from the higher temperature exactly offsets the entropy decrease from the smaller volume, as you intuited. However, irreversible work generates entropy because it involves a gradient in some field (e.g., pressure, force, electric field, etc.). When this gradient drives a transfer of energy, entropy is created.

$\endgroup$
3
  • $\begingroup$ thanks so much for you answer. It was helpful. So the change of entropy on the system after a reversible and adiabatic process would be zero. So the entropy production is zero. What about a diabatic reversible process? The change of entropy wouldn't be zero, right? So the entropy production wouldn't be zero as well, right? And if it was a cycle of diabatic reversible processes? The change and production of entropy would be zero? $\endgroup$ May 4, 2017 at 14:30
  • 1
    $\begingroup$ Yes, heat transfer consists of entropy transfer. Yes, reversible processes don't generate entropy (reversible heating only transfers entropy). $\endgroup$ May 4, 2017 at 18:35
  • $\begingroup$ so the entropy production during a diabatic reversible process is zero. Reversible processes, either it be a cycle or just 1 process, don't generate entropy, it just can change or not. Right? I got another question, if you would like to help me. physics.stackexchange.com/questions/330960/… $\endgroup$ May 4, 2017 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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