Chapter 8, Problem 11QRT

(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 11QRT

Pressure in $atm$ is $\text{0}\text{.158 atm}.$

Explanation of Solution

Given that pressure is $120mm\text{Hg}\text{.}$

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

  $\begin{array}{c}P=\text{ 120 mmHg }\times \frac{1\text{ atm}}{\text{760 mmHg}}\\ \\ =\text{ 0}\text{.158 atm}\end{array}$

Hence, pressure in $atm$ is $0.158atm.$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a)

Answer to Problem 11QRT

Pressure in $mm\text{Hg}$ is $1.52\times 10^{3}mmHg.$

Explanation of Solution

Given that pressure is $2.00\text{ atm}\text{.}$

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

  $\begin{array}{c}P=2.00\text{ atm }\times \frac{\text{760 mmHg}}{1\text{ atm}}\\ \\ =\text{ 1}\text{.52}\times {\text{10}}^3\text{ mmHg}\end{array}$

Hence, pressure in $mm\text{Hg}$ is $1.52\times 10^{3}mmHg.$

(c)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $kPa$ to $mmHg$ where the given pressure is $\text{100}\text{.0 kPa}.$

Concept Introduction:

Refer to part (a)

Answer to Problem 11QRT

Pressure in $mm\text{Hg}$ is $750.0mmHg.$

Explanation of Solution

Given that pressure is $\text{100}\text{.0 kPa}.$

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

  $\begin{array}{c}P=\text{ 100}.0\text{ kPa}\times \frac{\text{1 atm}}{\text{101}\text{.325 kPa}}\times \frac{\text{760 mmHg}}{1\text{ atm}}\\ \\ =\text{ 750}\text{.0 mmHg}\end{array}$

Hence, pressure in $mm\text{Hg}$ is $750.0mmHg.$

(d)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $kPa$ to $atm$ where the given pressure is $200.0\text{ kPa}\text{.}$

Concept Introduction:

Refer to part (a)

Answer to Problem 11QRT

Pressure in $atm$ is $1.97atm.$

Explanation of Solution

Given that pressure is $\text{200}\text{.0 kPa}.$

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

  $\begin{array}{c}P=\text{ 200}.0\text{ kPa}\times \frac{\text{1 atm}}{\text{101}\text{.325 kPa}}\\ \\ =\text{ 1}\text{.97 atm}\end{array}$

Hence, pressure in $atm$ is $1.97atm.$

(e)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $kPa$ to $atm$ where the given pressure is $\text{36 kPa}\text{.}$

Concept Introduction:

Refer to part (a)

Answer to Problem 11QRT

Pressure in $atm$ is $0.36atm.$

Explanation of Solution

Given that pressure is $\text{36 kPa}.$

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

  $\begin{array}{c}P=36\text{ kPa}\times \frac{\text{1 atm}}{\text{101}\text{.325 kPa}}\\ \\ =\text{ 0}\text{.36 atm}\end{array}$

Hence, pressure in $atm$ is $0.36atm.$

(e)

Interpretation Introduction

Interpretation:

Unit of pressure has to be converted from $kPa$ to $mmHg$ where the given pressure is $\text{600}\text{.0 kPa}.$

Concept Introduction:

Refer to part (a)

Answer to Problem 11QRT

Pressure in $mm\text{Hg}$ is $4.50\times 10^{3}mmHg.$

Explanation of Solution

Given that pressure is $\text{600}\text{.0 kPa}.$

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

  $\begin{array}{c}P=\text{ 600}.0\text{ kPa}\times \frac{\text{1 atm}}{\text{101}\text{.325 kPa}}\times \frac{\text{760 mmHg}}{1\text{ atm}}\\ \\ =\text{ 4}\text{.50}\times {\text{10}}^3\text{ mmHg}\end{array}$

Hence, pressure in $mm\text{Hg}$ is $4.50\times 10^{3}mmHg.$