Derivation of a equation in closed, phases coexisting system 
I'm trying to show that for a closed system consisting of two phases coexisting in equilibrium at a temperature $T$ and under a pressure $P$. $$\left(\frac{\partial P}{\partial V}\right)_S=-\frac{T}{C_v}\left(\frac{dP}{dT}\right)^2$$ Here $\frac{dP}{dT}$ is the slope of the phase equilibrium curve.

I tried to perform some calculation at both sides of the equation. For the left side, using the relation $$Tds=c_v\left(\frac{\partial T}{\partial P}\right)_v dP+c_P\left(\frac{\partial T}{\partial v}\right)_P dv$$
I get $$\left(\frac{\partial P}{\partial V}\right)_S=-\frac{C_P}{C_V}\left(\frac{\partial T}{\partial V}\right)_P\left(\frac{\partial P}{\partial T}\right)_V$$
For the right side, I use Clausius-Clapeyron equation $$\frac{dP}{dT}=\frac{l}{T\Delta v}$$
where $l$ is the latent heat. But I have difficulty to relate two sides. Maybe I should head for another direction?
 A: This question is problem 10.7 in Adkins, "Equilibrium Thermodynamics." I was going to ask a question about this problem, since my solution, detailed below, is not quite the same as Adkins', but then I noticed the exact same question had been asked already.
For a system of one component in two phases, the equilibrium condition is that the chemical potentials are equal $\mu^{I}(P,T)=\mu^{II}(P,T)$. In principle, this equation could be solved to get $P=P(T)$. This is the co-existence curve of the two phases. So, $P(T)$ is supposed to be known and so is the slope $dP/dT$. So, in whatever way we change the system, $P,T$ are locked in step as $P=P(T)$. The volume V of the system and the mole fractions of the component in each phase will change depending on the details of the change.
Since $\left(\frac{\partial P}{\partial V}\right)_{S}$ appears in the problem, it seems worthwhile to consider a reversible adiabatic change in which the entropy of the complete system is constant. Furthermore, the problem contains the heat capacity at constant volume, so $S=S(V,T)$ seems appropriate.
\begin{equation}
0=dS=\left(\frac{\partial S}{\partial V}\right)_{T}dV+\left(\frac{\partial S}{\partial T}\right)_{V}dT
\end{equation}
Dividing by $dT$ and bringing in the heat capacity at constant volume
$C_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V}$ gives,
\begin{equation}
0=\left(\frac{\partial S}{\partial V}\right)_{T}\frac{dV}{dT}+\frac{C_{V}}{T}
\end{equation}
Now use Maxwell's relation $\left(\frac{\partial S}{\partial V}\right)_{T}
=\left(\frac{\partial P}{\partial T}\right)_{V}$. This gives,
\begin{equation}
0=\left(\frac{\partial P}{\partial T}\right)_{V}\frac{dV}{dT}+\frac{C_{V}}{T}
\end{equation}
It remains to get a formula for the volume change $dV/dT$ in terms of the slope $dP/dT$ of the co-existence curve. Since we need to bring in $dP/dT$ and $S$ is constant, it seems sensible to use $V=V(S,P)$.
\begin{equation}
\frac{dV}{dT}=\left(\frac{\partial V}{\partial S}\right)_{P}\frac{dS}{dT}+\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}
=\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}
\end{equation}
The last equality is because the change is adiabatic. Substituting in the penultimate equation,
\begin{equation}
0=\left(\frac{\partial P}{\partial T}\right)_{V}\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}+\frac{C_{V}}{T}
\end{equation}
A re-arrangement of this result gives something close to the desired answer,
\begin{equation}
\left(\frac{\partial P}{\partial V}\right)_{S}=-\frac{T}{C_{V}}\left(\frac{\partial P}{\partial T}\right)_{V}\frac{dP}{dT}
\end{equation}
This would be the desired answer to Adkins' question if $\left(\frac{\partial P}{\partial T}\right)_{V}=\frac{dP}{dT}$. However,
\begin{equation}
\frac{dP}{dT}=\left(\frac{\partial P}{\partial T}\right)_{V}+\left(\frac{\partial P}{\partial V}\right)_{T}\frac{dV}{dT}
=\left(\frac{\partial P}{\partial T}\right)_{V}+\left(\frac{\partial P}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}
\end{equation}
The last equality used the earlier result for $dV/dT$. Now, the usual gymnastics with partials can be used to obtain the well known result (see equation (8.7) of Adkins, for example),
\begin{equation}
\left(\frac{\partial P}{\partial V}\right)_{S}=\frac{C_{P}}{C_{V}}\left(\frac{\partial P}{\partial V}\right)_{T}=\gamma\left(\frac{\partial P}{\partial V}\right)_{T}
\end{equation}
The coefficient of $dP/dT$ on the RHS of the penultimate equation is $1/\gamma$. We can now re-arrange things to get the relation between $dP/dT$ and  $\left(\frac{\partial P}{\partial T}\right)_{V}$ for an adiabatic process.
\begin{equation}
\left(\frac{\partial P}{\partial T}\right)_{V}=\frac{\gamma-1}{\gamma}\frac{dP}{dT}
\end{equation}
Substituting, in my partial result,
\begin{equation}
\left(\frac{\partial P}{\partial V}\right)_{S}=-\frac{T}{C_{V}}\left(\frac{\gamma-1}{\gamma}\right)\left(\frac{dP}{dT}\right)^{2}
\end{equation}
So, this result differs from that of Adkins by the factor containing the ratios of heat capacities. I would appreciate it if someone more adept at thermodynamics than myself would figure out if this factor is erroneous or not.
