Chapter 10, Problem 1PS

Pressure

(See Section 10.1 and Example 10.1.)

The pressure of a gas is 440 mm Hg. Express this pressure in units of (a) atmospheres, (b) bars, and (c) kilopascals.

Interpretation Introduction

Interpretation:

The pressure given should be converted into the units of atmosphere.

Concept Introduction:

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

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:

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

Answer to Problem 1PS

  $\text{440}\text{mm}\text{Hg}\text{=}\text{0}\text{.578}\text{atm}$

Explanation of Solution

The given pressure are in the units of atmosphere

Given the value of pressure as $440\text{mm}\text{Hg}$

Conversion into atm:

  $\begin{array}{l}\text{1}\text{atm}\text{=}\text{760}\text{mm}\text{Hg}\\ \text{440}\text{mm}\text{Hg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mm}\text{Hg}}\text{=}\text{0}\text{.578}\text{atm}\end{array}$

Interpretation Introduction

Interpretation:

The pressure given should be converted into the units of bars.

Concept Introduction:

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

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:

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

Answer to Problem 1PS

  $\text{440}\text{mm}\text{Hg}\text{=0}\text{.586}\text{bar}$

Explanation of Solution

The given pressure are in the units of bar

Given the value of pressure as $440\text{mm}\text{Hg}$

Conversion into bar:

  $\text{440}\text{mm}\text{Hg}\text{×}\frac{\text{1}\text{.013}\text{bar}}{\text{760}\text{mm}\text{Hg}}\text{=}\text{0}\text{.586}\text{bar}$

Interpretation Introduction

Interpretation:

The pressure given should be converted into the units of kilopascals.

Concept Introduction:

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

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:

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

Answer to Problem 1PS

  $\text{440}\text{mm}\text{Hg}\text{=}\text{58}\text{.6}\text{k}\text{Pa}$

Explanation of Solution

The given pressure are in the units of kilopascals.

Given the value of pressure as $440\text{mm}\text{Hg}$

Conversion into kilopascals:

  $\text{440}\text{mm}\text{Hg}\text{×}\frac{\text{101}\text{.3}\text{kPa}}{\text{760}\text{mm}\text{Hg}}\text{=}\text{58}\text{.6}\text{k}\text{Pa}$