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I think your definition of enthalpy is not wrong.

My ranting is how most of thermo starts with U(S, V, N) whereas I would prefer with S(U, V, N). Either way the differential reads:

dU = T dS - p dV + mu dN

which, if we follow the standard way, is just saying that

T = (dU/dS)_{V, N}

- p = (dU/dV)_{S, N}

mu = (dU/dN)_{S, V}

where the derivatives are really just partial derivatives so I really should have written them with a curly d. Mathematically enthalpy is the Legendre transform where we eliminate V in favor of p:

H(S, p, N) = (U(S, V, N) + p V)_{V = V*(p)}

where V* extremizes the term in parentheses, i.e.

(dU/dV) (S, V*(p), N) + p = 0

Of course this equation is just the above definition of the pressure at constant V, but now we are meant to solve this equation to find V* as a function of p (and S and N), and then plug that back in to get H as a function of p.

The Legendre transform property ensures that:

(dH/dp)_{S,N} = V*(p)

You should note that taking the derivative of H wrt p would in principle also induce a term proportional to dV* / dp since in the definition of H there is V* which depends on p. But that term cancels out because V* is an extremum! So that is why dH/dp gives just V*(p) which is the inverse function of p(V) that you get from (minus) dU/dV. This is the "inverse of derivatives" property of the Legendre transform mentioned in the original post.

Clearly the enthalpy is the nicer gadget to have if you work at constant pressure since then one of its arguments is simply held constant.




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