# Why is the helmholtz free energy minimized?

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.

• The wikipedia page has $\max_S$, not the maximum of the expression.
– user195162
Commented Jun 13, 2018 at 19:30
• @Quantumness I know, but I still don't even know what's that and why is that. Commented Jun 13, 2018 at 19:33
• It is referring to maximum entropy, the second law of thermodynamics.
– user195162
Commented Jun 13, 2018 at 19:43
• @Quantumness I still don't get the derivation :( Commented Jun 13, 2018 at 19:50
• Could you refer to the specific aspect where you are confused? I'm not sure I understand what you are looking for.
– user195162
Commented Jun 13, 2018 at 20:05

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).

• Tell me one thing you first assumed the process to be Isochoric and then at last you further assumed isothermal path, combining the two should mean no process happening at all, right !? Then how can you claim that F should be minimised ? Commented Apr 22 at 14:32