# State functions and their arguments

$U$ and $H$ are defined as $$U=\sum E_{pot} +\sum E_{kin} ; dU=dq+dw$$ $$H=U+pV$$

However, it was stated in a lecture note that for a fixed molar number, $U$ and $H$ can be defined using any two properties, example: $U(Temperature,Volume)$ and $H(Temperature, Pressure)$.

However, I cant figure out why only two quantities are enough, even if we consider the above example.

Can somebody show how by using just these two quantities is enough to figure out $U$ and $H$? (Just proving for the above example would be fine)

• See derivation within: en.wikipedia.org/wiki/… – Zhe Oct 19 '17 at 17:09
• – Zhe Oct 19 '17 at 17:10
• It you know temperature and pressure or temperature and density, then you know everything about the state. So, the internal energy and enthalpy per mole are determined, relative to any other state. – Chet Miller Oct 19 '17 at 17:14
• @Zhe To which derivation are you referring? I might've well missed it but didn't find it in the Wiki at a quick glance. Anyhow, this is a good question. Look forward to a proofy answer. – Linear Christmas Oct 19 '17 at 19:09
• @LinearChristmas I was thinking of the ones that refer to $dU$ in terms of $dT$ and $dV$. And likewise, $dH$ expressed as $dT$ and $dP$. – Zhe Oct 19 '17 at 20:49

This follows from the first law of thermodynamics; i.e., $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ which identifies $U$ as a function of $S$ and $V$. If in addition we have other forms of work, such as chemical work, then we add a chemical work term $\mu\mathrm{d}N$ to the first law, and introduce an additional dependence of $U$ on $N$.
In addition, we can also replace a given independent variable by its conjugate variable, since, by thermodynamic stability, the independent variable is a monotonic function of its conjugate and vice versa. That is, the following sets of independent variables can all be used to characterize $U$: $$(S, V, N), (T, V, N), (S, P, N), (S, V, \mu), (T, P, N), (T, V, \mu), (S, P, N).$$ Note in particular that $(T, P, \mu)$ cannot be used, because these three variables, all being intensive, are not independent and are related by the Gibbs-Duhem equation.