Why is the helmholtz free energy minimized?

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kidoceanoe · Accepted Answer

Consider the fact that by the second law of thermodynamics, the total entropy S of the universe must increase. That is, the sum of all entropies has the relationship

d S_{u n i v e r s e} = d S_{s y s} + d S_{s u r r} \geq 0,

where 'sys' and 'surr' represent ours system and surrounds respectively. Hold this result for the moment. Recall the thermodynamic identity

d U = T d S - P d V + μ d N

, and recognize that for constant volume and number of particles,

d S_{s u r r} = \frac{d U_{s u r r}}{T} = - \frac{d U_{s y s}}{T} .

Note that the last equality comes from the first law: energy is conserved, so

d U_{s u r r} = - d U_{s y s}

. Then, plugging this into our earlier result and multiplying by T, we have

d S_{u n i v e r s e} = d S_{s y s} + d S_{s u r r} = d S_{s y s} - \frac{d U_{s y s}}{T}

Now the proper definition for Helmholtz free energy is

F = U - T S

, so for constant temperature,

d F = d U - T d S = - (T d S - d U)

. We can plug this into our last result as

T d S_{u n i v e r s e} = - d F_{s y s}

and finally

d S_{u n i v e r s e} = - \frac{d F_{s y s}}{T} .

Note that in order to maximize the energy of the universe, you have make the Helmholtz free energy as negative as possible (negative, but large magnitude).

Why is the helmholtz free energy minimized?

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