Chapter 5, Problem 5.15P

(a)

Interpretation Introduction

Interpretation:

The mass of the atmosphere above $1{\text{ m}}^{\text{2}}$ of ocean is to be calculated.

Concept introduction:

The expression to calculate the force is as follows:

$F=ma$        (1)

Here, $m$ is the mass, $a$ is the acceleration and $F$ is the force.

The relation between $\text{atm}$ and $\text{kPa}$ is as follows:

$1\text{ atm}=101.325\text{ kPa}$

The relation between $\text{atm}$ and $\text{kPa}$ is as follows:

$1\text{ kPa}=7.500617\text{ mmHg}$

(b)

Interpretation Introduction

Interpretation:

The height at which pressure becomes equal to the atmospheric pressure is to be calculated.

Concept introduction:

The pressure can be defined as the force per unit area. The atmospheric pressure is also called barometric pressure.

The expression to calculate the density of substance is as follows:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$