1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics: Helmholtz free energy

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Lets take a look at the Helmholtz free energy. It is defined as the amount of energy one has to put in to create a system once the spontaneous energy transfer to the sytem from the environment is accounted for.

    Now lets look at it once we have created the system. Now F = U - TS is the energy we get from annihilating system. Why is it we have to subtract TS? Is it because we are talking about reversible processes, or what?

    Thansk in advance.
  2. jcsd
  3. Dec 16, 2008 #2
    You start with a system that has an entropy of S in contact with a heat bath at temperature T. We assume that whatever the system does, the heat bat always stays in internal thermal equilibrium.

    If you annihilate the system and convert the system into pure energy and then convert that energy as much as possible into work or smuggle as much of that energy outside the universe (it is best to think of the latter possibility as energy leaving the universe in the form of work, the "creation out of nothing scenarios in some textbooks" are only consistent if the thing is created at absolute zero and zero entropy), then the total entropy increase is -S + Q/T, where Q is the heat energy taken from the total energy U dumped in the heat bath. The energy that you can take away (in the form of work, or in the textbook scenario smuggle out of the universe) is thus U - Q

    Since by the second law -S + Q/T >= 0, we have:

    Q >= T S ------------>

    U - Q <= U - T S

    So, the maximum amount of energy that you can take away after annihilating the system is U - T S.
  4. Dec 16, 2008 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    First law: dQ = dU + dW = dU + PdV

    dS = dQ/T so dQ = TdS

    So, the first law can be written as:

    TdS = dU + PdV

    where PdV = W is the work done BY the system. So, work done to system = - PdV = dU - TdS

    So if you define a state function F = U - TS then, while holding temperature constant, the difference between two states F1 and F2 is:

    F1 - F2 = dF = U1-U2 - T(S1-S2) = dU - TdS

    In other words, the difference in F is the same as the work done on the system.

    dF = dU - TdS = - PdV

    So reversing the process gives you the work that is done by the system in returning to its original state (holding T constant):

    PdV = - dF

    In other words, the change in F (dF) represents the Potential Work that can be done by the system in returning to its original state by an isothermal process.

  5. Dec 16, 2008 #4
    I don't agree with Andrew's derivation. You have to clearly distinguish the system which during the change is not in thermal equilibrium and the heat bath which is always in thermal equilibrium. Also, if you allow the volume to change, then the work is done against (or by) the heat bath, but that is then not the useful work we want to extract from the system. In that case you are led the the Gibbs energy as giving you the total amount of work you can extract from the system at best.

    So, yes, if you have a system at constant temperature in thermal equilibrium and allow the volume to change quasi staticly, minus the change in F is the work done against the heat bath, but this is not the general result in thermodynamics that says that the maximum work that you can extract from the system plus heat bath if the system is kept at constant volume, so it doesn't do any pressure volume work against the heat bath, is given by the drop in F.

    We are then talking about a finite change in F as the system is in a state of thermal equilibrium and then changes into another state of thermal equilibrium. E.g. you start with a mixture of oxygen and hydrogen and end with water. But the initial state and the final states are at the same temperature and volume. But the intermediary state may be far from thermal equilibrium.

    If we let the process proceed quasistatically, then thermodymamics could be used to describe the process as it proceeds, but then you would have to add extra variables that describe the change in the system. In this case you have to add the chemical potentials for water, hydrogen and oxygen.
  6. Dec 16, 2008 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    I am not sure what you mean by "against the heat bath". And I am not sure what your point is. These are state functions (F, S, U). The changes in S, U and F are independent of any process. Although F = Work potential only for an isothermal process, the function F depends only upon the state of the system.

    The various thermodynamic potentials are defined in ways that make them useful for particular processes. The Helmholtz potential provides a measure of the amount of work that can be done by a system in changing between two states, and for that reason is useful.

    Last edited: Dec 16, 2008
  7. Dec 16, 2008 #6
    It's explained in detail here

    See also the discussion page were similar questions were raised:


    So, of course, the change in F is indepependent of a process, so you can consider a quasistatic change from one state to another state, but you then have to enlarge the thermodynamic state space by adding new parameters and generalized forces, otherwise you are simply looking at volume pressure work done by a single system in thermal equilibrium, and that's not what the general statement relating drop in free energy to maximum work that can be extracted is all about.

    Basically, the important point (stressed in all the theoretical physics textbooks) is that the heat bath is always in thermal equilibrium, all changes here are quasi static. But the system is not necessarily in thermal equilibrium during the change. So, e.g., it is not correct to assume that dS = dQ/T holds for the system.
  8. Dec 16, 2008 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    dS = dQ/T holds ONLY for a reversible (quasi static) process. [itex]\Delta F[/itex] is the maximum work you can get out of a system in moving between two states. If the process is not reversible, you will not get the maximum work.

  9. Dec 17, 2008 #8
    Thanks guys. I get it now.

    I have a question quite similar to this one, so I'll just ask it here, instead of creating a new thread. I hope this is OK.

    Today at class we looked at a system, which is able to be in 4 different energy states E1, E2, E3 and E4, where E1 denotes the state with the lowest energy, and E4 denotes the state with the highest energy.

    My teacher said that when the temperature is 0, then the system is in the lowest energy state. His reasoning for this was that the total entropy of "system + surroundings" is maximal in the lowest energy state when T=0, and he used the following identity to account for this:

    \frac{\partial S}{\partial U} = \frac{1}{T}.

    If the system is not in the lowest energy state (e.g. if it is in E2), then going to the state E1 would make the entropy of the surroundings go to infinity because of the equation above, which makes sense.

    Now my question is: Who says that the surroundings must have a temperature T=0 as well? Because if the temperature of the surrounds T >> 1, then why can the system be in E2?

    Thanks in advance guys! I really appreciate this.
  10. Dec 17, 2008 #9
    Actually, the total entropy is always maximal if the system is in the lowest energy state. If the temperature is zero, then the probability that the system is in the lowest energy state is equal to 1. If the temperatutre is higher, then the proabability for the system to be in the higher energy state increases, but the lower energy states always have higher probability.

    The entropy is defined as:

    S = k Log(Omega)

    where Omega is the number of energy eigenstates of the system that are within some small energy resolution dE around the specified internal energy.

    The equation dS/dU = 1/T should be interpreted as defining the temperature for a closed system:

    d Log(Omega)/dU = 1/(k T)

    For a closed system, you can, in principle, compute the energy levels, and then compute Omega. Then this fixes T as a function of the internal energy .

    In your problem you should think of the system with four energy levels in contact with a large heat bath at some temperature T. The thermodynamical equations apply for the heat bath only.

    The heat bath plus system is a closed system with constant energy U. Then if the fpur level system is in state j containing energy Ej, the heat bath will have an energy of U - Ej. We can then compute the total number of states for the heat bath can be in:

    Log[Omega(U - Ej)] = Log[Omega(U)] - Ej dLog[Omega]/dU + higher order terms.


    dLog[Omega]/dU = 1/(k T)

    The higher order terms thus involve the change in temperature of the heat bath because some energy has flowed to the four level system. Un the llimit that the heat bath is much larger than the system, these terms become zero.

    Therefore, we have:

    Omega(U - Ej) = Omega(U)Exp[-Ej/(kT)]

    So, when the four level system is in some fixed microstate j, the heat bath can be in Omega(U)Exp[-Ej/(kT)] possible microstates. All possible microstates are equally likely, so the probability for the four level state to be in some microstate j is thus proportiaonal to
    Exp[-Ej/(kT)]. This means that the higher the energy of the system, the lower the probability will be. At T = 0, the system can only be in the ground state (assuming E1 < E2).
  11. Dec 17, 2008 #10

    1: Can this also be seen from the following propability?

    p_i = \frac{1}{Z}\exp(-\beta E_i),

    where Z is the partition function. So for T -> 0, all pi's -> 0 expect p1, which will be Z(-1)=1?

    2: Ok, so from your explanation, the T is the formula is the temperature of the heat bath. But I thought of something new just now: Let's say that we are at a very low temperature (i.e. T ~ 0, so the probability of being in higher states is not entirely zero), and the system is not in the lowest state, i.e. it is in e.g. E2. Then it suddenly goes to state E1, because the above propability "tells" the system to do that. Then the entropy of the surroundings will almost increase with an infinite rate, because of dS/dU = beta, and hence the system+surroundings will have a very high entropy (perhaps even maximum?). Why can't one use this reasoning?
  12. Dec 18, 2008 #11
    Yes, you can also derive it from the partition function.

    Point 2: That's actually not that much different from the partiton function derivation, because if you study how the formula for the partition function is derived, you see that it is exactly by considering the Omega function of the bath....
  13. Dec 18, 2008 #12
    Thanks for being so patient with me.

    Merry Christmas.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Thermodynamics: Helmholtz free energy
  1. Helmholtz Free Energy (Replies: 0)

  2. Helmholtz free energy (Replies: 10)