Question

1.    Eight kilometers below the surface of the ocean the pressure is 82Mpa. Determine the density of the seawater at this depth if the density at the surface is 1025kg/m³ and the average bulk modulus of elasticity is 2.3GPa.

Answer 1

Gives

h=8km

Bulk modulus

$\beta=\frac{∆pV}{∆V}$

$\frac{∆p}{\beta}=\frac{∆V}{V}$

We know that

$\frac{∆V}{V}=\frac{\frac{m}{\rho}-\frac{m}{\rho'}}{\frac {m}{\rho}}$

$\frac{∆V}{V}={1}-{\frac{\rho}{\rho'}}$

$\frac{∆p}{\beta}=1-\frac{\rho}{\rho'}$

$\rho'=\frac{\rho}{1-\frac{∆p}{\beta}}$

Put value

$\rho'=\frac{1025}{1-\frac{80.36\times10^6-82\times10^6}{2.3\times10^9}}$

$\rho'=\frac{1025}{1-\frac{1.64\times10^6}{2.3\times10^9}}$

$\rho'=\frac{1025}{1-7.13\times10^{-4}}$

$\rho'=1025.73kg/m^3$