# 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?

• 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
– Bort
Jan 4 '16 at 14:21
• 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? Jan 4 '16 at 14:33
• 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. Jan 4 '16 at 15:32

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. $$$$0=dS=\left(\frac{\partial S}{\partial V}\right)_{T}dV+\left(\frac{\partial S}{\partial T}\right)_{V}dT$$$$ Dividing by $$dT$$ and bringing in the heat capacity at constant volume $$C_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V}$$ gives, $$$$0=\left(\frac{\partial S}{\partial V}\right)_{T}\frac{dV}{dT}+\frac{C_{V}}{T}$$$$ Now use Maxwell's relation $$\left(\frac{\partial S}{\partial V}\right)_{T} =\left(\frac{\partial P}{\partial T}\right)_{V}$$. This gives, $$$$0=\left(\frac{\partial P}{\partial T}\right)_{V}\frac{dV}{dT}+\frac{C_{V}}{T}$$$$ It remains to get a formula for the volume change $$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)$$. $$$$\frac{dV}{dT}=\left(\frac{\partial V}{\partial S}\right)_{P}\frac{dS}{dT}+\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT} =\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}$$$$ The last equality is because the change is adiabatic. Substituting in the penultimate equation, $$$$0=\left(\frac{\partial P}{\partial T}\right)_{V}\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}+\frac{C_{V}}{T}$$$$ A re-arrangement of this result gives something close to the desired answer, $$$$\left(\frac{\partial P}{\partial V}\right)_{S}=-\frac{T}{C_{V}}\left(\frac{\partial P}{\partial T}\right)_{V}\frac{dP}{dT}$$$$ This would be the desired answer to Adkins' question if $$\left(\frac{\partial P}{\partial T}\right)_{V}=\frac{dP}{dT}$$. However, $$$$\frac{dP}{dT}=\left(\frac{\partial P}{\partial T}\right)_{V}+\left(\frac{\partial P}{\partial V}\right)_{T}\frac{dV}{dT} =\left(\frac{\partial P}{\partial T}\right)_{V}+\left(\frac{\partial P}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial P}\right)_{S}\frac{dP}{dT}$$$$ The last equality used the earlier result for $$dV/dT$$. Now, the usual gymnastics with partials can be used to obtain the well known result (see equation (8.7) of Adkins, for example), $$$$\left(\frac{\partial P}{\partial V}\right)_{S}=\frac{C_{P}}{C_{V}}\left(\frac{\partial P}{\partial V}\right)_{T}=\gamma\left(\frac{\partial P}{\partial V}\right)_{T}$$$$ The coefficient of $$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}$$ for an adiabatic process. $$$$\left(\frac{\partial P}{\partial T}\right)_{V}=\frac{\gamma-1}{\gamma}\frac{dP}{dT}$$$$ Substituting, in my partial result, $$$$\left(\frac{\partial P}{\partial V}\right)_{S}=-\frac{T}{C_{V}}\left(\frac{\gamma-1}{\gamma}\right)\left(\frac{dP}{dT}\right)^{2}$$$$ 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.