# Spontaneity / Free Energy of Non-Isothermal Process

I'm trying to determine a lower bound for the work input necessary to make an entropy-reducing process "spontaneous" in the sense that the 2nd law is not violated.

For a constant temperature and volume process, I'd calculate the change in Helmholtz free energy and say $\Delta A = -W$. When considering a non-isothermal process (suppose we have no thermal reservoir), I can still calculate $dA = dU - TdS - SdT$, but can I say anything about what that $dA$ means for the spontaneity of such a (infinitesimal) process?

EDIT: I think the following is equivalent, fundamentally -- How do we determine how much work we can get out of a system undergoing an entropy-increasing process, when that process is non-isothermal?

• If you are doing work, can you really call the process "spontaneous"? – Floris Sep 4 '14 at 3:21
• Spontaneous probably isn't the best word choice -- I've edited the question above. – Will Sep 4 '14 at 3:31
• The second law of thermodynamics is a measured physical fact. How are you going to invalidate a measured fact? As for the question of "spontaneity", are you asking for methods to determine the thermodynamic noise in systems? That noise doesn't violate the second law, either. – CuriousOne Sep 4 '14 at 4:06
• Isn't this just asking for the efficiency of the Carnot cycle? That's the maximum amount of work you can extract from the heat flow between a source and sink at different temperatures (although admittedly there the heat flow has to be infinitely slow). Or am I missing the point of your question? – Floris Sep 4 '14 at 4:16
• I am getting the feeling, that you are essentially trying to resurrect Laplace's Demon en.wikipedia.org/wiki/Laplace's_demon? I am sorry to inform you, but that little fella has been dead on arrival since 1814. Fluctuations do not violate the second law and they can't be used to generate any energy. Neither is there any "extra" energy needed to have fluctuations. What is needed, however, is a second temperature bath (usually held at a lower temperature) to measure them against, which is basically equivalent to the second law. – CuriousOne Sep 4 '14 at 6:13

## 1 Answer

In a general process one always has $dU=\delta Q + \delta W$ and $dQ \le TdS$ between two equilibrium points. If you define $A = U-TS$ then $d(A+TS) = \delta Q + \delta W \le TdS + \delta W$ always and therefore $dA + TdS +SdT \le TdS + \delta W$ or $$dA + SdT \le \delta W$$ for any process. This shows how expending work $\delta W$ on the system distirbutes itself between temperature change $dT$ and free energy change $dA$. This is true irrespective of whether the process is spontaneous or not. If there is no external work done, $\delta W=0$, then $dA \le -SdT$ shows how much the free energy must decrease for a given amount temperature change.