Question

The pressure at certain depth in a sea is 80 atm. If the density of water at the surface of the sea is $1.03\times {10}^{3}kg/m^3$ and the compressibility of water is $45.8\times {10}^{-11}P{a}^{-1}$, calculate the density of water at that depth.
$1atm=1.013\times {10}^{5}Pa$

Answer 1

$\Delta P=80atm=80\times 1.013\times {10}^{5}Pa$

Let $m$ be the mass and $\rho$ be the density of water at depth.

Volume at the surface of water $V=\frac{m}{\rho }$

And at depth $V^{\prime }=\frac{m}{{\rho }^{\prime }}$

$\Delta V=V-V^{\prime }=m(\frac{1}{\rho }-\frac{1}{{\rho }^{\prime }})$

Bulk modulus$\left(K\right)=\frac{\Delta P}{\frac{\Delta V}{V}}$

Compressibility factor $=\frac{1}{K}=\frac{\Delta V}{\Delta PV}$

$\frac{1}{K}=m(\frac{1}{\rho }-\frac{1}{{\rho }^{\prime }})\times \frac{\rho }{m}\times \frac{1}{\Delta P}=1-\left(\frac{\rho }{{\rho }^{\prime }}\right)\times \frac{1}{\Delta P}P=\frac{1.03\times {10}^3}{1-0.0037}{P}^{\prime }=1.034\times {10}^{3}kg/m^3$