Help! This is a question in elementary thermodynamics.
I was deriving the expression for Gibb’s free energy today:
delta G = delta H - T delta S
that is
delta G = delta E + P delta V - T delta S
But is it not true that the first law of thermodynamics is
delta E = T delta S - P delta V ?
That would make delta G = 0.
What am I missing here?
I suspect this is a homework question… nonetheless, I wrestled with an inadequate textbook in Thermo as well. I always found http://scienceworld.wolfram.com/physics/ to be a great help in looking up the “basic” or “standard” derivations of thermodynamic equations. If it’s not homework, I can come back and help you finish this puppy off (I ended up with a B+ in Thermo).
You are doing fine. The first law is dE=dQ-dW. dW=PdV, and for reversible processes, dQ=TdS, so you can write dE=TdS-PdV for reversible processes. You concluded dG=0, which is true for reversible processes. For irreversible processes, you have to allow dQ<TdS.
Out of respect for your professor, I probably shouldn’t have given you an answer that explicit. Please read the above post as “Hint: It involves reversibility”.
Or maybe you are just deriving the Gibbs Free Energy on your own time, for fun 
Actually I am a high school teacher, and I had this problem bring my class to a screeching halt this morning. I did not know there was a caveat to the equation
dE = T dS - P dV
Anyway, thanks.
Let me just say that the fact you are teaching about Gibbs Free Energy in high school is awesome. I don’t think we ever got beyond the proverbial “pith balls and inclined planes.”
I’m pretty sure we learned something about Gibbs’ free energy in high school, but in chemistry class, not physics. That helped me to get an A in university thermo 1 and 2.
Hyperelastic already explained the answer. Kudos for teaching this early.