When Does Helmholtz Free Energy Decrease?

wakko101
Messages
61
Reaction score
0
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. =)
 
Physics news on Phys.org
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?
 
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
<br /> 0.5 \rho g h^2\;.<br />
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
<br /> 0.5 \rho g h - \frac{k_B T}{A}\log(h)\;,<br />
where A is some parameter with dimensions of Area. I minimize this and I find a different answer
<br /> h=\sqrt{\frac{k_B T}{\rho g A}}\;.<br />
 
olgranpappy said:
...and perhaps the *free energy* (E-TS) is given by
<br /> 0.5 \rho g h - \frac{k_B T}{A}\log(h)\;,<br /> ...

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~
 
tnho said:
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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top