Van der Waals gas - free energy

[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?

 

[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?