First Law of Thermodynamics

Going from the above equations, using simple algebra, only PV will give the solution W = FD, which is what I understand to be the definition of work. However, I'm not interested in the algebra; I'm interested in the calculus behind this equation. In calculus, W = FD isn't really W. It's dW. Which leads to my original question, why the VdP term is not in the equation for dW.

W is energy, dW is work.

In an ideal gas, PV = nRT
Since in a closed system, n, R are constant, you can replace one of the other 2 variables with this equality. (eg. P = T/V) and thus eliminate one of the 3 partial derivatives (dT, dP, dV)



I think this table is another clue to the issue:
http://en.wikipedia.org/wiki/Table_of_th...d_concepts
U = TS - pV + sum(...)
H = U + pV

If you scroll down further, you see
dU = TdS - pdV + d(sum)
dH = TdS + Vdp + d(sum)

So the Vdp term is going into enthalpy for some reason.