I need to obtain the derivative of internal energy $U$ w.r.t. pressure $p$ at a constant volume $V$. Realizing that $\mathrm{d} U = T \mathrm{d} S – p \mathrm{d} V$, I rewrite
$$
\left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial p} \right)_V
$$
The first term is equal to $T$
$$
\left( \frac{\partial U}{\partial S} \right)_V = T
$$
For the second term I used
$$
\mathrm{d} U = T \mathrm{d} S – p \mathrm{d} V
$$
again, from which
$$
\left( \frac{\partial U}{\partial S} \right)_V = T, \quad \left( \frac{\partial U}{\partial V} \right)_S = – p
$$
and therefore
$$
\left( \frac{\partial p}{\partial S} \right)_V = -\left( \frac{\partial T}{\partial V} \right)_S \quad \implies \quad \left( \frac{\partial S}{\partial p} \right)_V = \frac{1}{\left( \frac{\partial p}{\partial S} \right)_V} = – \frac{1}{\left( \frac{\partial T}{\partial V} \right)_S} = – \left( \frac{\partial V}{\partial T} \right)_S
$$
So far, we have
$$
\left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial p} \right)_V = – T \left( \frac{\partial V}{\partial T} \right)_S
$$
For $(\partial V/\partial T)_S$ I use
$$
\left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1
$$
with $x = V$, $y = T$ and $z = S$
$$
\left( \frac{\partial V}{\partial T} \right)_S \left( \frac{\partial T}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T = -1
$$
Using the definition for the heat capacity at constant volume, we get
$$
C_V = \left( \frac{\partial Q}{\partial T} \right)_V = T \left( \frac{\partial S}{\partial T} \right)_V \quad \implies \quad \left( \frac{\partial T}{\partial S} \right)_V = \frac{1}{\left( \frac{\partial S}{\partial T} \right)_V} = \frac{1}{C_V / T} = \frac{T}{C_V}
$$
For $\left( \frac{\partial S}{\partial V} \right)_T$ I use Helmholtz free energy
$$
\mathrm{d} F = – p \mathrm{d} V – S \mathrm{d} T
$$
so
$$
\left( \frac{\partial F}{\partial V} \right)_V = – p, \quad \left( \frac{\partial F}{\partial T} \right)_S = – S
$$
therefore
$$
\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V
$$
and for the last time
$$
\left( \frac{\partial p}{\partial T} \right)_V \left( \frac{\partial T}{\partial V} \right)_p \left( \frac{\partial V}{\partial p} \right)_T = -1
$$
with
$$
\alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_p \quad \implies \quad \left( \frac{\partial T}{\partial V} \right)_p = \frac{1}{\left( \frac{\partial V}{\partial T} \right)_p} = \frac{1}{\alpha V}
$$
and
$$
K_T = – \frac{1}{V} \left( \frac{\partial V}{\partial p} \right)_T \quad \implies \quad \left( \frac{\partial V}{\partial p} \right)_T = – K_T V
$$
and therefore
$$
\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V = – \frac{1}{\left( \frac{\partial T}{\partial V} \right)_p \left( \frac{\partial V}{\partial p} \right)_T} = \frac{\alpha}{K_T}
$$
going back to
$$
\left( \frac{\partial V}{\partial T} \right)_S = – \frac{1}{\left( \frac{\partial T}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T} = – \frac{1}{(T/C_V) (\alpha/K_T)} = – \frac{C_V K_T}{\alpha T}
$$
and going back to the internal energy…
$$
\left( \frac{\partial U}{\partial p} \right)_V = – T \left( \frac{\partial V}{\partial T} \right)_S = – T \left( – \frac{C_V K_T}{\alpha T} \right) = \frac{C_V K_T}{\alpha}
$$
However, the formula should be
$$
\left( \frac{\partial U}{\partial p} \right)_V = \frac{C_p K_T}{\alpha} – \alpha V T
$$
My question is: how do I obtain the second formula as quickly as possible without using Mayer's relation $C_p – C_V = (\alpha^2/K_T) V T$?
P.S. I also found a quicker way to obtain the formula with $C_V$
$$
\left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial T} \right)_V \left( \frac{\partial T}{\partial p} \right)_V = C_V \frac{K_T}{\alpha} = \frac{C_V K_T}{\alpha}
$$
Best Answer
$$dH=dU+pdV+Vdp=C_pdT+V(1-\alpha T)dP$$So, $$dU=C_pdT-pdV-V\alpha Tdp$$The rest is basically what you've already done with your second method.