# Integrated vs. Differential Form of the Enthalpy

I have been under the impression that the definition of enthalpy as

$$H = U + PV$$

is purely mathematical, the result of taking the Legendre transform of the internal energy $U$, and merely provides a convenient expression to find the differential form of the enthalpy,

$$dH = dU +d(PV) = TdS + VdP + \sum_i\mu_i dN_i$$

which is a useful expression for performing experiments at constant pressure, which I take to be the main reason for even defining the enthalpy in the first place.

Because the natural variable $P$ in the enthalpy is intensive, it cannot be integrated using an Euler integral. But in several places I have seen the expression

\begin{align*} H &= U + PV \\ &= TS - PV +\sum_i\mu_i N_i + PV \\ &= TS + \sum_i\mu_i N_i \end{align*}

defined as the Euler integral of the enthalpy. It seems we must have

$$dH = dU + d(PV) = d(TS) + \sum_i d(\mu_iN_i)$$

But I do not immediately see why

$$TdS + VdP + \sum_i\mu_i dN_i = d(TS) + \sum_i d(\mu_iN_i)$$

If I expand the differentials, will I eventually be able to use some Maxwell relation to arrive at an identity? Is there a meaningful physical interpretation of the term

$$U + PV$$

or is it, as I suspect, purely mathematical?

• Not a Maxwell relation, no. But you are essentially discovering the Gibbs-Duhem relation, which relates to extensivity and intensivity of variables. This is already discussed at physics.stackexchange.com/q/60084 and you can find a general discussion on Wikipedia. – user197851 Sep 16 '18 at 19:28
• @LonelyProf I see what you mean now. Gibbs-Duhem provides a constraint on the intensive variables such that $VdP = SdT + Nd\mu$ which results in the identity above. Thank you for the insight. – nguzman Sep 16 '18 at 19:51