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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?

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    $\begingroup$ 1. You should do your homework on your own 2. Start with understanding the problem formulation, specifically: What does "two phases coexisting in equilibrium" mean? Once you have that, the rest is fiddling around with differentials $\endgroup$
    – Bort
    Jan 4 '16 at 14:21
  • $\begingroup$ I asked the question because I can't solve it. I tried to prove this by using Clausius-Clapeyron equation but didn't have much progress. Would you answer this with more detail? $\endgroup$ Jan 4 '16 at 14:33
  • $\begingroup$ Give some details of you attempt with the Clausius-Clapeyron equation and we can help. Then it won't be discarded as homework question as easily. $\endgroup$ Jan 4 '16 at 15:32
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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.

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