- #1
Zag
- 49
- 9
Hello everyone,
I've been reviewing some concepts on Thermodynamics and, even though I feel like I am gaining a level of comprehension about the subject that I could not have achieved before as an undergraduate, I am also running into some situations in which some thermodynamic concepts seem to contradict each other. Here is one of these situations involving Entropy of reversible and irreversible processes:
From Clausius' Theorem, we know that: [itex]\oint\mathrm{\frac{dQ}{T}} \leq 0[/itex] (the equality holding for reversible processes)
Now, following the text-book that I'm using (K. Huang - Statistical Mechanics), we can consider a transformation between two states [itex]A[/itex] and [itex]B[/itex] as being either reversible [itex](R)[/itex] or irreversible [itex](I)[/itex]:
Now considering a closed path given by [itex]I + (-R)[/itex], we can rewrite Clausis' theorem in the following way:
[itex]\int_{I}\mathrm{\frac{dQ}{T}} + \int_{-R}\mathrm{\frac{dQ}{T}} \leq 0[/itex]
But since for a reversible path the integral yields the variation in the Entropy: [itex]\int_{R}\mathrm{\frac{dQ}{T}} = \Delta S[/itex], we get:
[itex]\int_{I}\mathrm{\frac{dQ}{T}} \leq S(B) - S(A)[/itex]
Now comes the tricky part that has been confusing me. If the system under consideration is isolated, not heat exchange occurs and [itex]\mathrm{dQ} = 0[/itex], from which we conclude that:
[itex]S(B) - S(A) \geq 0[/itex]
Again with the equality holding for a reversible process, and the inequality holding for irreversible processes. Even though I get the idea, for me this seems to somehow contradict other aspects of the Entropy (probably because I'm missing something). The point is: how come [itex]S(B) - S(A) \geq 0[/itex] for two fixed states [itex]A[/itex] and [itex]B[/itex] if the entropy is a function of state? Given two fixed states [itex]A[/itex] and [itex]B[/itex], I could in principle go from the first to the second either reversibly or irreversibly, and because the Entropy depends only on the initial and final state, [itex]S(B) - S(A)[/itex] should be the same for both cases! How come this expression tells us it must be different? What am I missing?
Thanks a lot guys!
I've been reviewing some concepts on Thermodynamics and, even though I feel like I am gaining a level of comprehension about the subject that I could not have achieved before as an undergraduate, I am also running into some situations in which some thermodynamic concepts seem to contradict each other. Here is one of these situations involving Entropy of reversible and irreversible processes:
From Clausius' Theorem, we know that: [itex]\oint\mathrm{\frac{dQ}{T}} \leq 0[/itex] (the equality holding for reversible processes)
Now, following the text-book that I'm using (K. Huang - Statistical Mechanics), we can consider a transformation between two states [itex]A[/itex] and [itex]B[/itex] as being either reversible [itex](R)[/itex] or irreversible [itex](I)[/itex]:
Now considering a closed path given by [itex]I + (-R)[/itex], we can rewrite Clausis' theorem in the following way:
[itex]\int_{I}\mathrm{\frac{dQ}{T}} + \int_{-R}\mathrm{\frac{dQ}{T}} \leq 0[/itex]
But since for a reversible path the integral yields the variation in the Entropy: [itex]\int_{R}\mathrm{\frac{dQ}{T}} = \Delta S[/itex], we get:
[itex]\int_{I}\mathrm{\frac{dQ}{T}} \leq S(B) - S(A)[/itex]
Now comes the tricky part that has been confusing me. If the system under consideration is isolated, not heat exchange occurs and [itex]\mathrm{dQ} = 0[/itex], from which we conclude that:
[itex]S(B) - S(A) \geq 0[/itex]
Again with the equality holding for a reversible process, and the inequality holding for irreversible processes. Even though I get the idea, for me this seems to somehow contradict other aspects of the Entropy (probably because I'm missing something). The point is: how come [itex]S(B) - S(A) \geq 0[/itex] for two fixed states [itex]A[/itex] and [itex]B[/itex] if the entropy is a function of state? Given two fixed states [itex]A[/itex] and [itex]B[/itex], I could in principle go from the first to the second either reversibly or irreversibly, and because the Entropy depends only on the initial and final state, [itex]S(B) - S(A)[/itex] should be the same for both cases! How come this expression tells us it must be different? What am I missing?
Thanks a lot guys!