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 3 108 N/m2.

(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 3 103 kg/m3. Find its density at the bottom.

(c) Explain whether or when it is a good approximation to think of water as incompressible.

Short answer

a) the change in volume $=1,128.987\times {10}^5\text{N}/{\text{m}}^2$

b) B is the bulk modules of sea water is as follows:

$B=2.34\times {10}^9\text{N}/\text{m}2$

c) Bulk modules are familiar to us.$=\frac{-(1,128.987\times {10}^5\text{N}/{\text{m}}^2)\times (0.9{\text{m}}^3)}{2.34\times {10}^9\text{N}/{\text{m}}^2}$

Step-by-step solution

Step 1:

In the surface pressure is $P_s=1.013\times {10}^5\text{N}/{\text{m}}^2$

In the depth pressure is$P_d=1.13\times {10}^8\text{N}/{\text{m}}^2$

pressure variation

$\begin{array}{l}{\Delta }_p=P_S-P_d\\ =1.13\times {10}^8-1.013\times {10}^5\\ =1,128.987\times {10}^5\text{N}/{\text{m}}^2\end{array}$

B is the bulk modules of sea water is as follows:

$B=2.34\times {10}^9\text{N}/\text{m}2$

Bulk modules are familiar to us.

$\begin{array}{l}B=\frac{(-\Delta p)v_0}{\Delta v}\\ \Delta v=\frac{-\left(\Delta P\right)v_0}{B}\\ =\frac{-(1,128.987\times {10}^5\text{N}/{\text{m}}^2)\times (0.9{\text{m}}^3)}{2.34\times {10}^9\text{N}/{\text{m}}^2}\end{array}$