# 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 $$\begin{split} dU &= dH - pdV - Vdp\\ &= \left(\left(\frac{\partial H}{\partial T}\right)_pdT + \left(\frac{\partial H}{\partial p}\right)_T dp\right) - pdV - Vdp. \end{split}$$

Equating the two expressions for $dU$ and requiring that the coefficient of $dp$ should vanish we obtain the relations $$\begin{split} &\left(\frac{\partial U }{\partial T}\right)_V = \left(\frac{\partial H}{\partial T}\right)_p, \quad \quad -p = \left(\frac{\partial U}{\partial V}\right)_T, \quad \quad V = \left(\frac{\partial H}{\partial p}\right)_T. \end{split}$$

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 $H = H(S,p)$.

• Are you sure about that relation between specific heats? – J. Murray Aug 27 '18 at 17:42
• Yes, it follows from the first law of thermodynamics $\delta Q = dU + pdV$. If we now say that $U = U(T,V)$ and calculate $dU$ and substitute it in into the 1st law we can obtain the above relation by dividing through by $dT$ while keeping the pressure constant. – MOOSE Aug 27 '18 at 17:51
• But $U \neq U(T,V)$. It is a function of $S$ and $V$. You can't just swap out variables like that without performing the requisite legendre transforms. – J. Murray Aug 27 '18 at 18:01
• I think this might be what I'm misunderstanding, but let me play the devils advocate. The internal energy is the sum of the kinetic and potential energy of the particles in the gas. The kinetic energy is proportional to the temperature and the potential energy depends on the volume. Hence I would say that we can treat the internal energy as a function of both $T$ and $V$. – MOOSE Aug 27 '18 at 18:10
• the issue is not that whether U can be or cannot be taken as function of T,V or not. The issue is that the function B, U=B(S,V), is not he same function D as U = D(T,V) although they both give the same value U. – hyportnex Aug 27 '18 at 19:05

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

$$-\left(\frac{\partial U}{\partial V}\right)_T = p + \frac{\beta}{\alpha}T$$