Why is enthelpy a function of Temperature and Pressure?

[tkdiscoverer]

right now, I'm following the MIT thermodynamics video lecture.

I've gone thru

dU = [STRIKE]d[/STRIKE]w + [STRIKE]d[/STRIKE]q
([STRIKE]d[/STRIKE] for "is path dependent")

and

H = U + pV

But why is enthalpy a function of temperature and pressure?
is it because pV = nRT and thus, V = nRT/p, so we only need p and T to get V?

dH = ( \delta H/ \delta )TdT + ( \delta H/ \delta p)dp

but why not:
dH = (\deltaH/\delta T)dT + (\deltaH/\deltaV)dV ?

Thanks! : D

 

[mSSM]

You have (I won't write the path-dependent parts explicitly; just assume they are there :D ):
<br /> \mathrm{d}E = T\mathrm{d}S - P\mathrm{d}V<br />

We know that \mathrm{d}Q = T\mathrm{d}T, so that for a process at constant pressure we can rewrite the quantity of heat as the differential:
<br /> \mathrm{d}Q = \mathrm{d}E + P\mathrm{d}V = \mathrm{d}(E + PV) = \mathrm{d}W <br />

of some quantity
W = E+PV

If you now want to find the total differential of the heat function itself (now no longer assuming constant pressure), you can easily get:
<br /> \mathrm{d}W = T\mathrm{d}S + V\mathrm{d}P<br />

For your notation: look at the definition of your U. When you make the total differential of the heat function, you have to write explicitly the terms for the energy, where -P\mathrm{d}V cancels with one of the differential you get from \mathrm{d}(PV) in the heat function.