# Why is enthalpy a function of Temperature and Pressure?

1. Apr 28, 2012

### 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 = ($\delta$H/$\delta$ T)dT + ($\delta$H/$\delta$V)dV ???

Thanks! : D

Last edited: Apr 28, 2012
2. Apr 29, 2012

### mSSM

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

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:
$$\mathrm{d}Q = \mathrm{d}E + P\mathrm{d}V = \mathrm{d}(E + PV) = \mathrm{d}W$$

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:
$$\mathrm{d}W = T\mathrm{d}S + V\mathrm{d}P$$

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.

Last edited: Apr 29, 2012