Question

The deepest point in the ocean is in the Mariana Trench, about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 x 10^8 Pa. (a) Calculate the change in volume of 1.00 m3 of seawater carried from the surface to this deepest point. (b) The density of seawater at the surface is 1.03 x 10^3 kg/m3. Find its density at the bottom. (c) Explain whether or when it is a good approximation to think of water as incompressible.

Answer 1

Solution;

Given;

Depth =11km

Pressure=1.13×10⁸Pa

Density of seawater=1.03×10³kg/m³

Volume is 1.00m³

(a)

Change in volume;

$B=-V(\frac{\Delta P}{\Delta V})$

$2.1×10^9=-1(\frac{1.13×10^8-1.01×10^5}{\Delta V})$

$\Delta V=-0.053m^3$

(Negative means volume reduced)

(b)

Change in density is calculated as;

$\Delta \rho=\frac{m\Delta \rho}{V}=\frac{0.053×10^3}{1}$

$\Delta \rho=0.053×10^3kg/m^3$

Density at the bottom is calculated as;

$\Delta \rho=\rho_2-\rho_1$

$\rho_2=(1.03+0.053)×10^3kg/m^3$

$\rho_2=1.083kg/m^3$

(c)

Yes it is a good approximation because in the 11km the density of water changed by only (5-6)%.