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Old Jan 14th 2016, 01:20 PM   #1
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defining entropy with heat instead of work?

From what I understand, in the Carnot cycle summing qi/Ti for each step results in zero, thus indicating a new state function, entropy = qrev/T. But since dE = 0 = q+w, then q = -w, and looking at the equations derived from the cycle summing wi/Ti for each step should also result in zero. So why can't one also define entropy as wrev/T?
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Old Jan 14th 2016, 01:37 PM   #2
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I don't like using the Carnot cycle to introduce entropy as this property appears in other circumstances, even when no work is done.

However, what does a Carnot machine do?

Is it 100% efficient at this?

Answer this and you will have the Clausius statement of the second law and the answer to your own question.

It is impossible in a continuous cycle to transfer heat from a colder to a warmer body without net changes in other bodies.
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Old Feb 23rd 2016, 11:19 PM   #3
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reversible vs irreversible work for adiabatic process

Hello, I have another question:

I have a gas transitioning adiabatically between A (P1, V1) and B (P2, V2) where P1>P2 and V2>V1. The question is to determine the net work done on the gas if the gas is first expanded reversibly from A to B (w = dE = Cv(T2-T1)), and then compressed irreversibly from B to A (w = -Pext(V1-V2)) at a constant external pressure defined by A. In this scenario, simply looking at the areas under the graphs the net work should be positive.

I am trying to reconcile this with dE for the gas. For the roundtrip transition (A to B to A), dE = 0. And if we take each step as adiabatic, then dE = w for each step, but as I have described above you would end up with two different values for dE for each step, thus dE not equal to zero. My logic is flawed somewhere. If I compress irreversibly would the transition still be adiabatic? Alternatively, is the original scenario flawed: can I have an irreversible adiabatic transition?
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