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Quelsita

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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?