Van der Waals gas - free energy

1. Oct 5, 2010

Quelsita

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=-$$\Delta$$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:

$$\int$$(dQ/T) $$\leq$$S(B)-S(A)

and since T is constant throughout I can say that:
Q= $$\int$$(dQ) $$\leq$$ 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$$\leq$$ -$$\Delta$$U+ T{S(B)-S(A)}
L$$\leq$$ 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. Oct 5, 2010

Quelsita

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?