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Thermodynamics: Small Question On Gibb's Free Energy

  1. Apr 27, 2006 #1
    Hi to everyone at the forum.
    I've attached the question in word format so you can see it properly but
    it is basically this:

    The Gibb's Free Energy is:

    G(T, P, N) = U - TS + PV

    Where N is the number of particles. (Only one type of particle is present).

    Explain why the N-dependence of G is given by:

    G(T, P, N) = Ng(T, P)

    Where g(T, P) is independent of N.


    So far the only thing I can think is that the number of particles, N, is not dependent on anything else and can be factored out. But to be honest I don't understand the question very well and can see how my answer (if it is an answer) is rather flimsy.

    Thank you for any help you can give,


    Attached Files:

  2. jcsd
  3. Apr 27, 2006 #2

    Physics Monkey

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    Hi Ben,

    Welcome to Physics Forums! To understand what's going on here, ask yourself the following question: what would happen to the Gibbs free energy if I doubled the size of the system while keeping all the intensive variables (like T and P) fixed?
  4. Apr 27, 2006 #3
    I thought that pressure was an extensive variable? In Gibb's free energy is it assumed that pressure is always constant?
    I guess that if you doubled the size of the system, keeping pressure constant, then you would need twice as many particles. The Gibb's free energy would double as well?
  5. Apr 27, 2006 #4

    Physics Monkey

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    Nope, pressure is an intensive variable. And yes, the Gibbs free energy describes a system where you control the pressure. You are right that the Gibbs potential doubles, but I sense some uncertainty on your part. To understand what's going on a little better, look at your expression for the Gibbs potential in terms energy, entropy, and volume. Doubling the size of the system will clearly double the volume (if you keep pressure and temperature fixed). What does such a doubling do to the entropy and energy?

    Once you're satisfied that the Gibbs potential doubles when you double N and keep T and P fixed, it's easy to show that the Gibbs potential can depend on N only linearly.
  6. Apr 28, 2006 #5
    God that was stupid of me! Of course pressure is extensive. I had (unwisely) consulted wikipedia where the entry is wrong for intensive/extensive.
    Right well as far I can see I would say that by doubling the volume while keeping T and P fixed would result in a doubling of entropy and a halving of internal energy? Although now I'm feeling quite confused again; internal energy doesn't depend on anything but temperature, is that right?
    In that case shouldn't it just remain constant if T is fixed?
  7. Apr 28, 2006 #6

    Physics Monkey

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    I think you just typed the wrong word, but just to be sure, pressure is intensive.

    You're right about entropy, but why do you think the energy would be cut in half? That doesn't make any sense. The internal energy of an ideal gas depends on T and N, but the dependence on N is simple. Think of it this way, if you had a monoatomic ideal gas at temperature T, the internal energy per particle is 3/2 kT. So if I have another system with twice as many particles at the same temperature, what would happen to the total internal energy?
  8. Apr 28, 2006 #7
    Ok I see, so internal energy will double for twice the number of particles.
    So overall will the - S and + V doubling sort of cancel out leaving the Gibb's energy doubled?
    Also, why does the entropy double for double the size of the system? Is it because there are now twice as many particles in twice the volume and so they are more disordered?
    Thanks for the continued help,
  9. Apr 28, 2006 #8
    Oops, just noticed I typed extensive again in that post before! Yep, pressure's intensive.
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