Chapter 5, Problem 5.18QP

(a)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{128}\text{cm}\text{Hg}$ has to be expressed in terms of $\text{atm}$.

Concept Introduction:

Units of pressure can be converted into other units by using the conversion factor.  It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{76}\text{.0}\text{cm Hg}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{76}\text{cm}\text{Hg}}\text{or}\frac{\text{76 cm Hg}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{128}\text{cm}\text{Hg}$.  In order to convert it into $\text{atm}$ units the required conversion factors are,

    $\frac{\text{1 atm}}{\text{76}\text{cm}\text{Hg}}\text{or}\frac{\text{76 cm Hg}}{\text{1 atm}}$

As the given unit of value of pressure has $\text{cm Hg}$ as its unit, the first conversion factor has to be used.  This can be shown as,

    $\text{128}\text{cm}\text{Hg×}\frac{\text{1 atm}}{\text{76}\text{cm}\text{Hg}}\text{=}\text{1}\text{.684}\text{atm}$

Therefore, the pressure of $\text{128}\text{cm}\text{Hg}$ can be expressed in terms of $\text{atm}$ as $\text{1}\text{.684}\text{atm}$.

(b)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{255 torr}$ has to be expressed in terms of $\text{atm}$.

Concept Introduction:

Units of pressure can be converted into other units by using the conversion factor.  It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{760 torr}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{760}\text{torr}}\text{or}\frac{\text{760 torr}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{255 torr}$.  In order to convert it into $\text{atm}$ units the required conversion factors are,

    $\frac{\text{1 atm}}{\text{760}\text{torr}}\text{or}\frac{\text{760 torr}}{\text{1 atm}}$

As the given unit of value of pressure has $\text{torr}$ as its unit, the first conversion factor has to be used.  This can be shown as,

    $\text{255}\text{torr×}\frac{\text{1 atm}}{\text{760}\text{torr}}\text{=}\text{0}\text{.335}\text{atm}$

Therefore, the pressure of $\text{255 torr}$ can be expressed in terms of $\text{atm}$ as $\text{0}\text{.335}\text{atm}$.

(c)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{1405}\text{mm}\text{Hg}$ has to be expressed in terms of $\text{atm}$.

Concept Introduction:

Units of pressure can be converted into other units by using the conversion factor.  It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{760}\text{mm Hg}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{760}\text{mm}\text{Hg}}\text{or}\frac{\text{760 mm Hg}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{1405}\text{mm}\text{Hg}$.  In order to convert it into $\text{atm}$ units the required conversion factors are,

    $\frac{\text{1 atm}}{\text{760}\text{mm}\text{Hg}}\text{or}\frac{\text{760 mm Hg}}{\text{1 atm}}$

As the given unit of value of pressure has $\text{mm Hg}$ as its unit, the first conversion factor has to be used.  This can be shown as,

    $\text{1405}\text{mm}\text{Hg×}\frac{\text{1 atm}}{\text{760}\text{mm}\text{Hg}}\text{=}\text{1}\text{.849}\text{atm}$

Therefore, the pressure of $\text{1405}\text{mm}\text{Hg}$ can be expressed in terms of $\text{atm}$ as $\text{1}\text{.849}\text{atm}$.

(d)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{303}\text{kPa}$ has to be expressed in terms of $\text{atm}$.

Concept Introduction:

Units of pressure can be converted into other units by using the conversion factor.  It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{101}\text{kPa}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{101}\text{kPa}}\text{or}\frac{\text{101 kPa}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{303}\text{kPa}$.  In order to convert it into $\text{atm}$ units the required conversion factors are,

    $\frac{\text{1 atm}}{\text{101}\text{kPa}}\text{or}\frac{\text{101 kPa}}{\text{1 atm}}$

As the given unit of value of pressure has $\text{kPa}$ as its unit, the first conversion factor has to be used.  This can be shown as,

    $\text{303 kPa×}\frac{\text{1 atm}}{\text{101}\text{kPa}}\text{=}\text{3}\text{atm}$

Therefore, the pressure of $\text{303}\text{kPa}$ can be expressed in terms of $\text{atm}$ as $\text{3}\text{.0}\text{atm}$.