Thermodynamic potentials and equilibrium


by mrandersdk
Tags: equilibrium, potentials, thermodynamic
mrandersdk
mrandersdk is offline
#1
Dec10-07, 03:06 PM
P: 230
If I have some arbitrary Thermodynamic system, and I want to find the equilibrium state what do I do?

Lets say that I know

Internal energy: U(S,V)
Helmholtz free energy: F(T,V)
Enthalpy: H(S,P)
Gibbs free energy: G(T,P)

(I omitted the dependence on particles N)

I've been discussing this with some friends. They say that just minimize one of them and you have the state. You can then just transform back and forth between them to get the other state variables, meaning if I minimize U and find the S and V that minimizes the system, you can then just find T and V from them (from some equation of state), and they are the values of T and V that minimizes F(T,V).

I think this is wrong I understood it as, I need to take a look at my system. If I choose that my system have a fixed S and V when the system is going towards equilibrium, then I can minimize U and I find the S and V, for the system in equilibrium. But if I have minimized H for example, and found S and P they wouldn't be the correct values for my system in equilibrium (because I assumed fixed S and V, not fixed S and P), that is the two S's found wouldn't agree, and if I could determine V from the S and P found from minimizing H, that wouldn't agree with the one from minimizing U.

I hope you understand what i'm asking. You have to minimize the correct potential depending one the situation you are describing, not just some random one. I'm not meaning what would be the easiest, but that you simply can't choose any potential.

by the way, when we are talking about a system at fixed S and V, and then say that U is a function of S and V, seems a bit strange, I guess that if we take some general system U is a function of those, we then determine U as a function of those, we then say that we keep S and V fixed for our system, so that we have to minimize U (assuming that I was right in my first quaestion), but saying that we keep S and V fixed doesn't change that U is a function of S and V, the only important thing is that when saying this, we say we have to minimize U. Is this right or am I completely confused?
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vivesdn
vivesdn is offline
#2
Dec10-07, 03:26 PM
P: 104
If you fix a given variable, you will find the equilibrium for such a situation. A different equilibrium state will be reached if you allow that variable free.

But anyway, the equilibrium is found at the minimum of G. There is an expression that relates variation on G to the equilibrium constant.
mrandersdk
mrandersdk is offline
#3
Dec10-07, 03:38 PM
P: 230
so you say that you always have to minimize G, independent of the situation? Because I my point is that I think that minimizing G, gives only the equilibrium if the process is going on at fixed Pressure and Temperature

to quote wikipedia:

Gibbs energy is also the chemical potential that is minimized when a system reaches equilibrium at constant pressure and temperature

vivesdn
vivesdn is offline
#4
Dec10-07, 04:50 PM
P: 104

Thermodynamic potentials and equilibrium


Well, yes, my chemical background seems to be still there, and chemical reactions tend to take place at constant pressure.
Anyway, whatever the equilibrium conditions will be, the equilibrium itself must be a local minimum of G. And depending on your system, the equilibrium can be open: liquid water an vapour can be in equilibrium at a range of temperatures and pressures (T and P linked by a phase curve), but if equilibrium must include solid water, then only one combination of T an P is possible.

Just on a chemical example: you can know the equilibrium state through G variation at constant P and T. But if the system is adiabatic, this will produce a temperature change as enthalpy usually is not zero. You have to realize that the minimum of G now will be achieved at different composition as T will be different, which in turn will modify the total enthalpy change.

Of course, if you work on a constant volume system, you should minimize F. But again, as equilibrium is characterized by a constant T, P and V among others, both F and G will be local minima.
mrandersdk
mrandersdk is offline
#5
Dec10-07, 05:58 PM
P: 230
okay so you say that I have to minimize the correct potential depending on what is constant in my system, but in some cases the minimas of different potentials agree, or are you saying because in equilibrium, all termodynamic variables have to be constant, I could vary either of the potentials?

Because i'm not sure, that all termodynamic variables are constant in equilibrium (they are of cause constant by definition), implies that the different potentials is minima for all processes, i'm reading it as the importent thing is how the system is coming in equilibrium, that is there is a differens in keeping the volume and entropy constant in your expirement, and that they are constant in equilibrium. I thought that when you keep them constant in you experiment you need to minimize U, but if you made an experiment where you kept T and P constant you need to minimize G, but minimizing U in that case would yield the wrong equilibrium conditions, do you see the difference?
vivesdn
vivesdn is offline
#6
Dec10-07, 06:27 PM
P: 104
Depending on your system and which variables are kept constant you will choose F or G to be minimized. However, it is easy to keep P constant (atmosphere) but if you do not refrigerate the system to keep T, it will increase its temperature (again I am assuming a chemical reaction, which will usually be exothermic). Then, the temperature increase will have an impact on the equilibrium. You can repeat the process until the result is consistent.

If you keep the volume, you will proceed minimizing F and evaluating the impact of U on the final temperature.

The equilibrium will be a local minimum for both F and G anyway.
Count Iblis
Count Iblis is offline
#7
Dec10-07, 06:32 PM
P: 2,159
No, what you always have to do, in principle, is maximize S for the entire system. This amounts to minimizing F for a system that is kept at a constant volume and temperature when you take into account the change in entropy of the heath bath that is required to keep the system at constant T. If you keep the pressure constant instead of the volume, then maximizing the entropy of the system plus heath bath amounts to minimizing G for the system.
mrandersdk
mrandersdk is offline
#8
Dec11-07, 01:29 AM
P: 230
Count Iblis, that sounds more like the way i understood it, so it is importent what potential you use, depending on you situation and you can't just minimize whatever potential you like. So if I fx. know F, but I want to keep T and P constant in my experiment i have to make the a legendre transformation of F to find G and then minimize G to find the equilibrium?
Count Iblis
Count Iblis is offline
#9
Dec11-07, 08:58 AM
P: 2,159
Quote Quote by mrandersdk View Post
Count Iblis, that sounds more like the way i understood it, so it is importent what potential you use, depending on you situation and you can't just minimize whatever potential you like. So if I fx. know F, but I want to keep T and P constant in my experiment i have to make the a legendre transformation of F to find G and then minimize G to find the equilibrium?
That's right. It is actually not difficult to prove. In case of constant T and P, you just apply the relation:

dE = TdS - P dV (1)

to the heath bath (e.g. the environment) that keeps the system at constant T and P. The assumption is that the heath bath has a huge heat capacity, so that its temperature indeed stays constant. This means that even if the system undergoes rapid changes so that the assumptions of quasistatic changes are not valid and equation (1) does not apply to the system (or actually the more general relation dE = TdS - P dV + mu1 dN1 + mu2 dN2 + ...), it still applies to the heath bath.

Suppose the system is initially in a state which can be described using the usual thermodynamical variables. E.g. the entropy, and pressure of two compunds kept separated in the system but that react violently when combined is well defined, even though it is not strictly in thermodynamical equilibrium.

If the intitial internal energy is E1, entropy is S1 and volume is V1 and the final values (e.g. after the chemicals are mixed and the reaction is completed and equilibrium thermodynamical variables can again be defined) are E2, S2, and V2 then, of course, the entropy increase of the system is:

Delta S = S2 - S1


The entropy increase of the heath bath follows from equation (1):

dS' = [dE' + P dV']/T

Where the prime indicates that the variable refers to the heath bath.
During the change in the system, the heath bath remains in thermodynamical equilibrium so we can use this equation. If you integrate this from the start to the fintish of the change of the sstem, you get:

Delta S' = [Delta E' + P Delta V']/T

This follows from the fact that P and T are constant during any changes of the system. Because of conservation of energy:

Delta E' =- Delta E

Also, the system plus heath bath is assumed to have a fixed volume (to keep the pressure constant then means that the volume of te heath bath must be must larger than the system, just like the heat capacity of the heath bath needs to be much larger than that of the system in order that the temperature is kept constant). So we have:

Delta V' = -Delta V

Therefore:

Delta S' = - [Delta E + P Delta V]/T

And the total entropy change is:

Delta S + Delta S' =

Delta S - [Delta E + P Delta V]/T =

- Delta G/T

Total entropy change must always be positive, so Delta G can only be negative.


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