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Q & A: thermodynamics equilibrium conditions

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Most recent answer: 02/26/2014
In thermodynamics class, we derived the equilibrium correlations from a constant entropy systems. In this case, we assumed, that the wall of the system is closed, fixed size and isentropic (adiabatic) and we concluded that in this case, the internal energy will be minimized in the equilibrium state. My question is, where does the extra energy go, if we minimize the internal energy, when it can't escape neither through heat (wall is adiabatic) nor through work (walls are fixed sized)?
- Adam (age 18)

That's an excellent question. You're right to be puzzled. If the system can neither exchange energy with the outside through mechanical work nor through heat flow, how can it make sense to talk about energy being minimized?

There is, however, one broad class of problems for which it can still exchange energy. Those would be cases where a long-range field (electrical or magnetic) couples the inside to the outside. Then work can be done without any changes in the container walls.  In such cases, however, the thermally insulating rigid walls don't make the system isentropic.

Think of a bunch of spins in a fixed external magnetic field.  As the spins partially line up, the entropy they lose will be more than compensated by the entropy gained by other degrees of freedom inside the box from the energy released untill equilibrium is reached. 

I'm not sure how this would be described in terms of the energy of the system. As I've described it (with no changes in any spins outsie the system) it doesn't sound like any work is being done on the outside, so the magnetic energy lost stays inside the system. Can you send us a more complete description of what was actually claimed in class?

Mike W.

(published on 02/26/2014)

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