Chapter 9, Problem 29QP

(a)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{torr}$ to $\text{atm}$ .

Explanation of Solution

The conversion factor required to convert the value of pressure is:

$\begin{array}{c}760\text{ torr}=1\text{ atm}\\ 1=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

For $745\text{ torr}$ ,

$\begin{array}{c}\text{745 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{745 torr}\\ =0.98\text{ atm}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{atm}$ to $\text{torr}$ .

Explanation of Solution

The conversion factor required to convert the value of pressure is:

$\begin{array}{c}1\text{ atm}=760\text{ torr}\\ 1=\frac{760\text{ torr}}{1\text{ atm}}\end{array}$

For $1.23\text{ atm}$ ,

$\begin{array}{c}1.23\text{ atm}=\frac{760\text{ torr}}{1\text{ atm}}\times \text{1}\text{.23 atm}\\ =934.8\text{ torr}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{mm Hg}$ to $\text{atm}$ .

Explanation of Solution

The conversion factor used to convert the value of pressure is:

$90.1\text{ mm Hg}$ to $\text{atm}$ :

$\begin{array}{c}760\text{ mmHg}=1\text{ atm}\\ 1=\frac{1\text{ atm}}{760\text{ mmHg}}\end{array}$

For $\text{90}\text{.1 mm Hg}$ ,

$\begin{array}{c}\text{90}\text{.1 mm Hg}=\frac{1}{760}\text{ atm}\times \text{90}\text{.1}\\ =0.119\text{ atm}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{kPa}$ to $\text{Pa}$ .

Explanation of Solution

The conversion factor used to convert the value of pressure is:

$\begin{array}{c}1\text{ kPa}=1000\text{ Pa}\\ 1=\frac{1000\text{ Pa}}{1\text{ kPa}}\end{array}$

For $0.643\text{ kPa}$ ,

$\begin{array}{c}\text{0}\text{.643 kPa}=\frac{1000\text{ Pa}}{1\text{ kPa}}\times \text{0}\text{.643 kPa}\\ =643\text{ Pa}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{Pa}$ to $\text{mm Hg}$ .

Explanation of Solution

The conversion factor used to convert the value of pressure is:

$1.35\times {10}^5\text{ Pa}$ to $\text{mm Hg}$ :

$\begin{array}{c}101325\text{ Pa}=760\text{ mm Hg}\\ 1=\frac{760\text{ mm Hg}}{101325\text{ Pa}}\end{array}$

For $1.35\times {10}^5\text{ Pa}$ ,

$\begin{array}{c}1.35\times {10}^5\text{ Pa}=\frac{760\text{ mm Hg}}{101325\text{ Pa}}\times 1.35\times {10}^5\text{ Pa}\\ =1012.58\text{ mm Hg}\\ \text{=1}\text{.01}\times {\text{10}}^3\text{ mm Hg}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{Pa}$ to $\text{torr}$ .

Explanation of Solution

The conversion factor used to convert the value of pressure is:

$\begin{array}{c}101325\text{ Pa}=760\text{ torr}\\ 1=\frac{760\text{ torr}}{101325\text{ Pa}}\end{array}$

For $7.51\times {10}^4\text{ Pa}$ ,

$\begin{array}{c}7.51\times {10}^4\text{ Pa}=\frac{760\text{ torr}}{101325\text{ Pa}}\times 7.51\times {10}^4\text{ Pa}\\ =563.296\text{ torr}\end{array}$

(g)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{torr}$ to $\text{Pa}$ .

Explanation of Solution

The conversion factor used to convert the value of pressure is:

$\begin{array}{c}760\text{ torr}=\text{101325 Pa}\\ \text{1}=\frac{\text{101325 Pa}}{760\text{ torr}}\end{array}$

For $798\text{ torr}$ ,

$\begin{array}{c}798\text{ torr}=\frac{\text{101325 Pa}}{760\text{ torr}}\times 798\text{ torr}\\ =106391.25\text{ Pa}\\ \text{=1}\text{.06}\times {\text{10}}^5\text{ Pa}\end{array}$

(h)

Interpretation Introduction

Interpretation:

The conversion of pressure from $\text{cm Hg}$ to $\text{mm Hg}$ .

Explanation of Solution

The conversion factor used to convert the value of pressure is:

$29.3\text{ cm Hg}$ to $\text{mm Hg}$ :

$\begin{array}{c}1\text{ cm Hg}=10\text{ mm Hg}\\ 1=\frac{10\text{ mm Hg}}{1\text{ cm Hg}}\end{array}$

For $29.3\text{ cm Hg}$ ,

$\begin{array}{c}\text{29}\text{.3 cm Hg}=\frac{10\text{ mm Hg}}{1\text{ mm Hg}}\times 29.3\\ =293\text{ mm Hg}\end{array}$