Question

Calculate the pressure due to the ocean at the bottom of the Marianas Trench near the Philippines, given its depth is 11.0 km and assuming the density of sea water is constant all the way down. (b) Calculate the percent decrease in volume of sea water due to such a pressure, assuming its bulk modulus is the same as water and is constant. (c) What would be the percent increase in its density? Is the assumption of constant density valid? Will the actual pressure be greater or smaller than that calculated under this assumption?

Short answer

(a) The pressure is obtained as:\(P = 1.105 \times 1{0^8} N/{m^2}\).

(b) The percentage decrease in volume is obtained as:\(5\% \).

(c) The percentage increase in density is obtained as:\(5.26\% \). The assumption of constant density is said not be valid. As we descend, the density of the ocean increases, increasing the real pressure.

Step-by-step solution

Conceptual Introduction

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

Given Data

The depth is:

\(\begin{array}{c}{\rm{h = 11 km}}\\{\rm{ = 11,000 m}}\end{array}\)

The density of the seawater is\(\rho = 1.025 \times 1{0^3} kg/{m^3}\).

Evaluating the pressure

The pressure due to the ocean at the bottom of the Marianas Trench near the Philippines is given by:

\(\begin{array}{c}P = \rho gh\\P = 1.025 \times 1{0^3}*9.8*11000\\P = 1.105 \times 1{0^8} N/{m^2}\end{array}\)

Therefore, the pressure is: \(P = 1.105 \times 1{0^8} N/{m^2}\).

Percentage decrease in volume

The percent decrease in volume of seawater is given by

\(\Delta V/V = - \Delta P/B\)

Here, \(B = 2.2 \times 1{0^9} N/{m^2}\) is the bulk modulus of water.

\(\begin{array}{c}\Delta V/V = - 1.105 \times 1{0^8}/2.2 \times 1{0^9}\\\Delta V/V = - 0.05\\\Delta V/V\% = - 0.05*100\\ = - 5\% \end{array}\)

Therefore, the percentage decrease is: \(5\% \).

Percentage increase in density

The ratio of the final to initial density is:

\(\begin{array}{l}\frac{{{\rho _f}}}{{{\rho _i}}} = \frac{{m/95}}{{m/100}}\\\frac{{{\rho _f}}}{{{\rho _i}}} = 1.0526\end{array}\)

The percent increase in the density is:

\(\begin{array}{c}\% increase = (1.0526 - 1) \times 100\% \\\% increase = 5.26\% \end{array}\)

The assumption is not valid.

The actual pressure will be greater as the density of the ocean increases as we go down.

Therefore, the increase in density is: \(5.26\% \).