Helmholtz free energy decreases

Helmholtz free energy decreases....

Hello...

I'm having trouble getting my head around something. I'm not exactly sure what is meant by the Helmholtz free energy (F) decreasing in a system that's in thermal equilibrium with it's environment. Under what circumstances does it decrease?

Cheers,
W. =)

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Mapes
Homework Helper
Gold Member
The free energy could decrease if the system lost matter to the surrounding environment, if it did work on the environment, if the temperature of the environment decreased slowly to remain in thermal equilibrium with the system, or if a spontaneous internal change occurred. Does this help?

olgranpappy
Homework Helper
The free energy is just a useful quantity to have for determining how a system at a given temperature behaves. the minimization principle of the free energy is a generalization of of the minimization prinicple of the energy (which you are probably used to) to the case of finite temperature.

For example (I'm just making this up off the top of my head to illustrate the point, so don't take it too literally), perhaps I have a system at density $$\rho$$ that rises up to some height $$h$$ in a tube. And so the energy is
$$0.5 \rho g h^2\;.$$
And so, I can minimize the energy to find that the height h should be zero... the fluid doesn't rise up in the tube...

But, what if I'm at finite temperature, and perhaps the *free energy* (E-TS) is given by
$$0.5 \rho g h - \frac{k_B T}{A}\log(h)\;,$$
where $$A$$ is some parameter with dimensions of Area. I minimize this and I find a different answer
$$h=\sqrt{\frac{k_B T}{\rho g A}}\;.$$

...and perhaps the *free energy* (E-TS) is given by
$$0.5 \rho g h - \frac{k_B T}{A}\log(h)\;,$$ ...
I bet it should be $$0.5\rho g h^2 - \frac{k_B T}{A}\log(h)$$. =P

It is a very good example~thank you~

olgranpappy
Homework Helper
I bet it should be $$0.5\rho g h^2 - \frac{k_B T}{A}\log(h)$$. =P

It is a very good example~thank you~
whoopsy daisy, you're right--that's what I meant.