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.
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$\begingroup$ $dU = TdS-pdV$... so? $\endgroup$– Damian SowinskiCommented May 3, 2017 at 19:27
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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.
answered May 3, 2017 at 19:40
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$\begingroup$ 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? $\endgroup$ Commented May 4, 2017 at 14:30
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1$\begingroup$ Yes, heat transfer consists of entropy transfer. Yes, reversible processes don't generate entropy (reversible heating only transfers entropy). $\endgroup$ Commented May 4, 2017 at 18:35
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$\begingroup$ 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/… $\endgroup$ Commented May 4, 2017 at 18:40