# Why is enthalpy a function of Temperature and Pressure?

by tkdiscoverer
Tags: enthalpy, pressure, thermal physics, thermodynamics
 P: 2 right now, I'm following the MIT thermodynamics video lecture. I've gone thru dU = dw + dq (d 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
 P: 24 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.

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