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 $P$. $$(\frac{\partial P}{\partial V})_S=-\frac{T}{C_v}(\frac{dP}{dT})^2.$$ 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 $$Tds=c_v(\frac{\partial T}{\partial P})_v dP+c_P(\frac{\partial T}{\partial v})_P dv$$ I get $$(\frac{\partial P}{\partial V})_S=-\frac{C_P}{C_V}(\frac{\partial T}{\partial V})_P(\frac{\partial P}{\partial T})_V$$ 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?

  • $\begingroup$ 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? $\endgroup$ – C. M. Cheng 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. $\endgroup$ – Mikael Fremling Jan 4 '16 at 15:32

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