Why is the helmholtz free energy minimized?

Question

I don't quite understand the principle of minimum energy despite having read the derivation on Wikipedia.

I think I got lost when the free energy was defined as $A= \max_S{\left(U-TS\right)}$, because I don't know why is the max there.

Answer 1

Here's an alternative derivation to show that Helmholtz energy will be minimized.

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

$$ dS_{\rm universe} = dS_{\rm sys} + dS_{\rm surr} \geq 0,$$

where 'sys' and 'surr' represent ours system and surrounds respectively. Hold this result for the moment. Recall the thermodynamic identity $dU = TdS-PdV+\mu dN$, and recognize that for constant volume and number of particles,

$$ dS_{\rm surr} = \frac{dU_{\rm surr}}{T} = - \frac{dU_{\rm sys}}{T}. $$

Note that the last equality comes from the first law: energy is conserved, so $dU_{\rm surr} = -dU_{\rm sys}$. Then, plugging this into our earlier result and multiplying by $T$, we have

$$ dS_{\rm universe} = dS_{\rm sys} + dS_{\rm surr} = dS_{\rm sys} - \frac{dU_{\rm sys}}{T}$$ $$ \rightarrow TdS_{\rm universe} = TdS_{\rm sys} - dU_{\rm sys}.$$

Now the proper definition for Helmholtz free energy is $F = U - TS$, so for constant temperature, $dF = dU - TdS = -(TdS-dU)$. We can plug this into our last result as

$$ T dS_{\rm universe} = - dF_{\rm sys} $$

and finally

$$ dS_{\rm universe} = - \frac{dF_{\rm sys}}{T}.$$

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