Chapter 8.4, Problem 8.6P

Interpretation Introduction

Interpretation:

The pressure given should be converted into the units of $\text{atm,}\text{mm}\text{Hg,}\text{and}\text{pascals}$

Concept Introduction:

Pressure is given by the ratio of unit of forces that applied into a unit area.

$\text{Pressure}\text{=}\frac{\text{Force}}{\text{Area}}$

Pressure can be expressed in different units. But it is normally reported in $\text{mm}\text{Hg}\left(\text{millimeters}\text{of}\text{mercury}\right)$

General unit conversions:

$\begin{array}{l}\text{1}\text{atm}\text{=}\text{760}\text{mm}\text{Hg}\text{=}\text{14}\text{.7}\text{psi}\text{=}\text{101,325}\text{Pa}\\ \text{1}\text{pascal}\text{=}\text{1}\text{newton}/{\text{meter}}^{\text{2}}\\ \text{1}\text{atm}\text{=}\text{760}\text{mm}\text{Hg}\text{=}\text{101}\text{.325}\text{kilopascals}\text{=}\text{1}\text{.01325}\text{bar}\\ \text{1}\text{bar}\text{=}{\text{1×10}}^{\text{5}}\text{pascal}\end{array}$