# Compressibility of Saturated Liquid Water

I'd appreciate some help in squaring this thermodynamic circle (which may exist only in my mind, but I digress ...)

Let's consider pure water in a two-phase(liquid/vapour) state. Within the vapour dome, the density of water can be determined from mass and volumetric balances according to

$$\rho = \Bigg[\frac{x}{\rho_V(P)}+\frac{1-x}{\rho_L(P)}\Bigg]^{-1}$$ where $x$ is the vapour mass (not volume) fraction (which we knuckle-dragging engineers call the "steam quality"), $P$ is the pressure, and where $\rho_L(P)$ and $\rho_V(P)$ are the saturated liquid and saturated vapour densities to the left and right of the critical point (apex of the vapour dome) respectively. Both $\rho_L$ and $\rho_V$ are functions of pressure $P$, alone. The classical definition of the compressibility (really the isenthalpic compressibility within the vapour dome since the vapour mass fraction is a proxy for enthalpy here) is

$$\beta=-\frac{1}{V}\frac{\partial V}{\partial P}=\rho\frac{\partial \rho}{\partial P}$$

Now along the saturated liquid curve $x=0$ (to the left of the vapour dome apex), the density drops with pressure (because the dominant effect is that of the increase in the boiling point with an increase in pressure), so that for a saturated liquid, the compressibility is

$$\beta_{x=0}=\rho_L\frac{\partial \rho_L}{\partial P}$$

which would be negative. This defies not only the physics, but is also inconsistent with the (positive) compressibility that would be evaluated form a simple finite difference between a point on the saturated liquid line and a point at the same pressure in the subcooled region just above and to the left of the vapour dome boundary. I suspect that I'm missing some key concept here.