Q: What is the density of water at a depth where the pressure is 79.0 atm, given its density at the surface is 1.03 × 10^{3} kg/m^{3} ? Compressibility of water = 45.8 × 10^{-10} Pa^{-1}

Sol: Bulk Modulus of Elasticity , $\large K = \frac{\Delta P}{\Delta V/V} $

$\large \frac{\Delta V}{V} = \frac{\Delta P}{K}$

$\large \frac{\Delta V}{V} = ( \Delta P ) C$ ; (Since C= 1/K )

$\large \frac{\Delta V}{V} = ( 79 \times 1.01 \times 10^5 ) (45.8 \times 10^{-10}) $

= 36.65 × 10^{-3}

$\large \frac{\Delta V}{V} = \frac{m/\rho – m/\rho’}{m/\rho} = (1-\frac{\rho}{\rho’})$

$\large \frac{\rho}{\rho’} = 1- \frac{\Delta V}{V} $

$\large \rho’ = \frac{\rho}{1-\frac{\Delta V}{V}}$

On putting the known values ,

$\large \rho’ = 1.07 \times 10^3 kg/m^3$