Using a Legendre transformation to convert enthalpy to internal energy Let us consider the enthalpy as a function of temperature and pressure, such that $H = H(T,p).$
Now, say we want to exchange the pressure $p$ by the volume $V$ by performing a Legendre transformation. Let's call the result $U = U(T,V)$ (the internal energy). Thus the Legendre transformation should take the form
$$U(T,V) = H(T,p) - pV.$$
If we calculate $dU$ we obtain
$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$
or alternatively
\begin{equation}
\begin{split}
dU &= dH - pdV - Vdp\\
&= \left(\left(\frac{\partial H}{\partial T}\right)_pdT + \left(\frac{\partial H}{\partial p}\right)_T dp\right) - pdV - Vdp.
\end{split}
\end{equation}
Equating the two expressions for $dU$ and requiring that the coefficient of $dp$ should vanish we obtain the relations
\begin{equation}
\begin{split}
&\left(\frac{\partial U }{\partial T}\right)_V = \left(\frac{\partial H}{\partial T}\right)_p, \quad \quad -p = \left(\frac{\partial U}{\partial V}\right)_T, \quad \quad V = \left(\frac{\partial H}{\partial p}\right)_T.
\end{split}
\end{equation}
But the relation $-p = \left(\frac{\partial U}{\partial V}\right)_T$ can't be correct because we have the very general relation
$$C_p - C_v = \left(p + \left(\frac{\partial U}{\partial V}\right)_T\right)\left(\frac{\partial V}{\partial T}\right)_p \neq 0.$$
So there are two options:


*

*Something is wrong about the Legendre transformation I started out with. If this is the case, can anyone see what it is?

*The pressure $-p = \left(\frac{\partial U}{\partial V}\right)_T$ is not the pressure of the gas. If this is the case, what is the meaning of this pressure?


I know that the natural variables of $U$ is the entropy $S$ and the volume $V$, and that the natural variables of $H$ is entropy $S$ and pressure $p$. But I do not see why I should get something unphysical by treating $U = U(T,V)$ and $H = H(T,p)$ instead of $U = U(S,V)$ and $H = H(S,p)$.
 A: Okay I see the problem - I misread the question (more than once).  
The issue is that 
$$p \neq - \left(\frac{\partial U}{\partial V}\right)_T$$
but rather
$$p = -\left(\frac{\partial U}{\partial V}\right)_S$$
This is very different.  From the first law,
$$dU = TdS - pdV$$
but $T=T(S,V)$ and $p=p(S,V)$ are functions, not independent variables.  
In other words, $\left(\frac{\partial U}{\partial V}\right)_S$ is equal to the change in $U$ divided by the change in $V$ while $S$ is held constant.  On the other hand, $\left(\frac{\partial U}{\partial V}\right)_T$ is equal to the change in $U$ divided by the change in $V$ while $S$ and $V$ both change in such a way as to hold $T(S,V)$ constant.

Explicitly,
$$dU = TdS - pdV$$
$$dT = \left(\frac{\partial T}{\partial S}\right)_V dS + \left(\frac{\partial T}{\partial V}\right)_S dV$$
where we've defined $\alpha \equiv \left(\frac{\partial T}{\partial S}\right)_V$ and $\beta \equiv \left(\frac{\partial T}{\partial V}\right)_S$ for convenience.  We then have that
$$dS = \frac{1}{\alpha} dT - \frac{\beta}{\alpha} dV$$
so
$$ dU = T \left(\frac{1}{\alpha} dT - \frac{\beta}{\alpha} dV\right) - p dV = \frac{T}{\alpha}dT - \left(p + \frac{\beta}{\alpha}T\right)dV$$
and so
$$-\left(\frac{\partial U}{\partial V}\right)_T = p + \frac{\beta}{\alpha}T$$
