Chapter 5, Problem 5.19QP

(a)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{54}\text{.0}\text{cm}\text{Hg}$ has to be expressed in terms of $\text{psi}$.

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,

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

It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{14}{\text{.7 lb/in}}^{\text{2}}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{14}\text{.7}\text{psi}}\text{or}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{54}\text{.0}\text{cm}\text{Hg}$.  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,

    $\text{54}\text{.0}\text{cm}\text{Hg×}\frac{\text{1 atm}}{\text{76}\text{cm}\text{Hg}}\text{=}\text{0}\text{.71}\text{atm}$

The converted value of pressure has $\text{atm}$ as its unit; the second conversion factor has to be used.  This can be shown as,

    $\text{0}\text{.71}\text{atm×}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}\text{=}\text{10}\text{.437}\text{psi}$

Therefore, the pressure of $\text{54}\text{.0}\text{cm}\text{Hg}$ can be expressed in terms of $\text{psi}$ as $\text{10}\text{.437}\text{psi}$.

(b)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{155 torr}$ has to be expressed in terms of $\text{psi}$.

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,

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

It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{14}{\text{.7 lb/in}}^{\text{2}}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{14}\text{.7}\text{psi}}\text{or}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{155 torr}$.  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,

    $\text{155}\text{torr×}\frac{\text{1 atm}}{\text{760}\text{torr}}\text{=}\text{0}\text{.204}\text{atm}$

The converted value of pressure has $\text{atm}$ as its unit; the second conversion factor has to be used.  This can be shown as,

    $\text{0}\text{.204}\text{atm×}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}\text{=}2.9988\text{psi}$

Therefore, the pressure of $\text{155 torr}$ can be expressed in terms of $\text{psi}$ as $\text{2}\text{.9988}\text{psi}$.

(c)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{800}\text{mm}\text{Hg}$ has to be expressed in terms of $\text{psi}$.

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,

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

It is known that $\text{1}\text{atm}$ of pressure is equal to $\text{14}{\text{.7 lb/in}}^{\text{2}}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{14}\text{.7}\text{psi}}\text{or}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{800}\text{mm}\text{Hg}$.  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,

    $\text{800}\text{mm}\text{Hg×}\frac{\text{1 atm}}{\text{760}\text{mm}\text{Hg}}\text{=}\text{1}\text{.052}\text{atm}$

The converted value of pressure has $\text{atm}$ as its unit; the second conversion factor has to be used.  This can be shown as,

    $\text{1}\text{.052}\text{atm×}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}\text{=}\text{15}\text{.464}\text{psi}$

Therefore, the pressure of $\text{800}\text{mm}\text{Hg}$ can be expressed in terms of $\text{psi}$ as $\text{15}\text{.464}\text{psi}$.

(d)

Interpretation Introduction

Interpretation:

Given pressure unit of $\text{1}\text{.50}\text{atm}$ has to be expressed in terms of $\text{psi}$.

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 $\text{14}{\text{.7 lb/in}}^{\text{2}}$.  Conversion factors can be given as,

    $\frac{\text{1 atm}}{\text{14}\text{.7}\text{psi}}\text{or}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}$

Explanation of Solution

Given unit of pressure is $\text{1}\text{.50 atm}$.  In order to convert it into $\text{psi}$ units the required conversion factors are,

    $\frac{\text{1 atm}}{\text{14}\text{.7}\text{psi}}\text{or}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}$

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

    $\text{1}\text{.50}\text{atm×}\frac{\text{14}\text{.7 psi}}{\text{1 atm}}\text{=}22.05\text{psi}$

Therefore, the pressure of $\text{1}\text{.50}\text{atm}$ can be expressed in terms of $\text{psi}$ as $\text{22}\text{.05 psi}$.