Chapter 8, Problem 10QRT

(a)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $mm\text{Hg}$ to $atm.$

Concept Introduction:

Pressure of gas is the force applied on a unit area.

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

Pressure can be expressed in different units.

General unit conversions:

  $\begin{array}{l}\text{1}\text{Pa}\text{=}{\text{1Nm}}^{-2}{\text{ = 1 kgm}}^{-1}{s}^{-2};{\text{ 1 kPa = 10}}^{\text{3}}\text{ Pa}\\ \\ \text{1}\text{bar}\text{=}{\text{10}}^{\text{5}}{\text{Pa = 10}}^{\text{3}}{\text{ kPa; mbar = 10}}^{\text{2}}\text{ Pa}\\ \\ \text{1}\text{atm}\text{=}1.01325\times {10}^5\text{ Pa = 101}\text{.325 kPa}\\ \\ 1\text{ atm = 760 Torr = 760}\text{mm}\text{Hg}\\ \\ 1\text{ atm = 14}{\text{.7 lb/in}}^{\text{2}}\left(psi\right)\text{ = 1}\text{.01325 bar}\end{array}$

Answer to Problem 10QRT

Pressure in $atm$ is $0.947atm.$

Explanation of Solution

Given that pressure is $720mm\text{Hg}\text{.}$

Convert the unit of pressure from $mm\text{Hg}$ to $atm$ as follows,

  $\begin{array}{c}P=\text{ 720 mmHg }\times \frac{1\text{ atm}}{\text{760 mmHg}}\\ \\ =\text{ 0}\text{.947 atm}\end{array}$

Hence, pressure in $atm$ is $0.947atm.$

(b)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $atm$ to $mmHg.$

Concept Introduction:

Refer to part (a)

Answer to Problem 10QRT

Pressure in $mm\text{Hg}$ is $950.0mmHg.$

Explanation of Solution

Given that pressure is $1.25\text{ atm}\text{.}$

Convert the unit of pressure from $atm$ to $mm\text{Hg}$ as follows,

  $\begin{array}{c}P=1.25\text{ atm }\times \frac{\text{760 mmHg}}{1\text{ atm}}\\ \\ =\text{ 950}\text{.0 mmHg}\end{array}$

Hence, pressure in $mm\text{Hg}$ is $950.0mmHg.$

(c)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $mm\text{Hg}$ to $torr.$

Concept Introduction:

Refer to part (a)

Answer to Problem 10QRT

Pressure in $torr$ is $542torr.$

Explanation of Solution

Given that pressure is $\text{542 mmHg}.$

Convert the unit of pressure from $mm\text{Hg}$ to $torr$ as follows,

  $\begin{array}{c}P=\text{ 542 mmHg }\times \frac{\text{760 torr}}{\text{760 mmHg}}\\ \\ =\text{ 542 torr}\end{array}$

Hence, pressure in $torr$ is $542torr.$

(d)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $mm\text{Hg}$ to $\text{kPa}.$

Concept Introduction:

Refer to part (a)

Answer to Problem 10QRT

Pressure in $kPa$ is $98.7kPa.$

Explanation of Solution

Given that pressure is $\text{740 mmHg}.$

Convert the unit of pressure from $mm\text{Hg}$ to $kPa$ as follows,

  $\begin{array}{c}P=\text{ 740}\text{.0 mmHg }\times \frac{\text{1 atm}}{\text{760 mmHg}}\times \frac{\text{101}\text{.325 kPa}}{\text{1 atm}}\\ \\ =\text{ 98}\text{.7 kPa}\end{array}$

Hence, pressure in $kPa$ is $98.7kPa.$

(d)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $kPa$ to $\text{atm}.$

Concept Introduction:

Refer to part (a)

Answer to Problem 10QRT

Pressure in $atm$ is $6.91atm.$

Explanation of Solution

Given that pressure is $\text{700}\text{.0 kPa}.$

Convert the unit of pressure from $kPa$ to $atm$ as follows,

  $\begin{array}{c}P=\text{ 700}\text{.0 kPa }\times \frac{\text{101}\text{.325 kPa}}{\text{1 atm}}\\ \\ =\text{ 6}\text{.91 atm}\end{array}$

Hence, pressure in $atm$ is $6.91atm.$