kislotd

2022-07-14

Can we show that the helmholtz free energy at equilibrium is minimized from its second derivative?

Damarion Pierce

Expert

The first derivative at an extremum point ${x}_{0}$ is 0 only when evaluated at ${x}_{0}$. Same goes with the second derivative. So you first have to evaluate the second derivative, then you plug in the specific $x={x}_{0}$
For example, take $y={x}^{2}$
$\frac{\mathrm{d}y}{\mathrm{d}x}=2x$ and $\frac{{\mathrm{d}}^{2}y}{{\mathrm{d}}^{2}x}=2$
At $x=0$, the minimum, $\frac{\mathrm{d}y}{\mathrm{d}x}{|}_{0}=0$, and $\frac{{\mathrm{d}}^{2}y}{{\mathrm{d}}^{2}x}{|}_{0}=2$
If we were to apply your logic however, we would have $\frac{{\mathrm{d}}^{2}y}{{\mathrm{d}}^{2}x}=\frac{\mathrm{d}}{\mathrm{d}x}\underset{=0}{\underset{⏟}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}}=0.$
So, you have to expand your last line.
${\mathrm{d}}^{2}F=-\mathrm{d}P\mathrm{d}V-P{\mathrm{d}}^{2}V-\mathrm{d}S\mathrm{d}T-S{\mathrm{d}}^{2}T.$
Now you apply the condition that you are at equilibrium, so $\mathrm{d}T{|}_{\mathrm{e}\mathrm{q}}=\mathrm{d}V{|}_{\mathrm{e}\mathrm{q}}=\mathrm{d}P{|}_{\mathrm{e}\mathrm{q}}=\mathrm{d}S{|}_{\mathrm{e}\mathrm{q}}=0$, so that:
${\mathrm{d}}^{2}F{|}_{\mathrm{e}\mathrm{q}}=-P{\mathrm{d}}^{2}V{|}_{\mathrm{e}\mathrm{q}}-S{\mathrm{d}}^{2}T{|}_{\mathrm{e}\mathrm{q}}.$
Then I guess that if pressure is positive, then the volume is a maximum so ${\mathrm{d}}^{2}V<0$ so that the first term is positive.

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