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Van der Waals gas - free energy

  1. Oct 5, 2010 #1
    For this problem, I'm really jst trying to figure out everything that is going on and then I can simply follow through with the derivatives once I know what I'm working with.

    Q: For the van der Waals gas, introduce the free energy as F = U – TS and verify that its derivatives over V and T give the correct expressions for p and S.

    I know the eqaution for free energy is the Helmholz free energy where
    F=( Hem.) free energy
    U= Internal energy of the system
    T= absolute temperature (K)
    S= Entropy

    From the first law of thermo. :
    L=-[tex]\Delta[/tex]U +Q (L being the external work)

    since the system is in thermal contact w/ an environment at constant temperature and transforming from a state A to B:

    [tex]\int[/tex](dQ/T) [tex]\leq[/tex]S(B)-S(A)

    and since T is constant throughout I can say that:
    Q= [tex]\int[/tex](dQ) [tex]\leq[/tex] T{S(B)-S(A)}

    my question is, where did the inequality come from??

    Then, I can plug in Q to the First law?:
    L[tex]\leq[/tex] -[tex]\Delta[/tex]U+ T{S(B)-S(A)}
    L[tex]\leq[/tex] U(A)-U(B) + T{S(B)-S(A)}
    again, why is an inequality used?
    where does the volume and pressure come into play?
  2. jcsd
  3. Oct 5, 2010 #2
    ok, is the inequality used because the process is not reversible?
    can L=pdV be applied somewhere?

    also, I found in our text that deltaF=(dF/dV) dV and it states that this was found using:
    L< F(A)-F(B)= -deltaF....how is this so?
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