Clausius-Clapeyron relation for high pressures I can see that the Clausius-Clapeyron relation depends on the change in specific volume $\Delta v = v_g (1 - \frac{v_c}{v_g}) = v_g - v_c$, with $v_c$ equal to the volume of the condensed phase and $v_g$ equal to the volume of the gas phase. One approximation is to replace $v_g = \frac{RT}{P}$ and $\frac{v_c}{v_g} = 0$ when $v_g >> v_c$. This only works for low pressures and temperatures (much lower than critical pressure and temperature).
For high pressures/temperatures how can I calculate $\Delta v$ (and so $v_c$, and $v_g$)?
 A: We can use a cubic equation such as the van der Waals equation,
$$P = \frac{RT}{v-b}-\frac{a}{v^2}$$
to calculate the properties of a fluid according to the procedure below.

*

*Solve the van der Waals equation for $v_l$ and $v_v$ at temperature $T$ and pressure $P^\text{sat}(T)$. The equation has three real solutions, the smallest is $v_l$, the largest is $v_v$ and the middle root is discarded.


*Calculate the heat of vaporization as follows:
$$\Delta h_\text{vap} = h^R(v_v)-h^R(v_l)$$
with
$$
   h^R(v) = Pv - RT - \frac{a}{v}
$$
where $a$ is the parameter in the van der Waals equation.
Comments

*

*The van der Waals is not accurate enough for engineering calculations. Instead we use the Soave-Redlich-Kwong (SRK) or the Peng-Robinson (PR) equations, both of which are modifications of the van der Waals but a bit more complicated in algebra. Even these are more appropriate for small non polar molecules. Nonetheless, the van der Waals ought to give a better aproximation than the Clausius-Clapeyron.


*The procedure described here assumes that we know the saturation pressure at any temperature. If we don't, we can get it from the van der Waals equation, but this involves another layer of iterations. I'll be happy to explain if that might be useful.
