Chapter 9, Problem 30QP

(a)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The conversion factor which is used to convert the pressure from $\text{atm}$ to $\text{torr}$ is as follows:

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

For $1.15\text{ atm}$

$\begin{array}{c}1.15\text{ atm}=1.15\text{ atm}\times \frac{760\text{ torr}}{1\text{ atm}}\\ =\text{874 torr}\end{array}$

(b)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The conversion factor which is used to convert the pressure from $\text{torr}$ to $\text{atm}$ is as follows:

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

For $968\text{ torr}$ ,

$\begin{array}{c}\text{968 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{968 torr}\\ =1.27\text{ atm}\end{array}$ .

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The conversion factor that is used to convert the pressure value from $\text{Pa}$ to $\text{atm}$ is as follows:

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

For $2.50\times {10}^5\text{ Pa}$ ,

$\begin{array}{c}2.50\times {10}^5\text{ Pa}=\frac{1\text{ atm}}{101325\text{ Pa}}\times 2.50\times {10}^5\text{ Pa}\\ =2.47\text{ atm}\end{array}$

(d)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The conversion factor that is used to convert the pressure value is as follows:

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

For $695\text{ mm Hg}$ ,

$\begin{array}{c}695\text{ mm Hg}=\frac{1\text{ torr}}{1\text{ mm Hg}}\times 695\text{ mm Hg}\\ =695\text{ torr}\end{array}$

(e)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The conversion factor that is used to convert pressure values from $\text{atm}$ to $\text{Pa}$ is as follows:

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

For $0.953\text{ atm}$

$\begin{array}{c}\text{0}\text{.953 atm}=\frac{101325\text{ Pa}}{1\text{ atm}}\times 0.953\text{ atm}\\ =96562.72\text{ Pa}\\ =9.66\times {\text{10}}^4\text{ torr}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The conversion of the pressure values from $\text{torr}$ to $\text{mm Hg}$

Explanation of Solution

The conversion factor that is used to convert pressure values from $\text{torr}$ to $\text{mm Hg}$ is as follows:

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

For $653\text{ torr}$ ,

$\begin{array}{c}653\text{ torr}=\frac{1\text{ mm Hg}}{1\text{ torr}}\times 653\text{ torr}\\ =653\text{ mm Hg}\end{array}$

(g)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The conversion factor that is used to convert of pressure values from $\text{torr}$ to $\text{mm Hg}$ is as follows:

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

For $1545\text{ mm Hg}$ ,

$\begin{array}{c}\text{1545 mm Hg}=\frac{\text{101325}}{760}\text{ Pa}\times \text{1545}\\ =205983.06\text{ Pa}\end{array}$

(h)

Interpretation Introduction

Interpretation:

The conversion of the pressure values from $\text{kPa}$ to $\text{atm}$ .

Explanation of Solution

The conversion factor that is used to convert of pressure values from $\text{kPa}$ to $\text{atm}$ is as follows:

$\begin{array}{c}101.325\text{ kPa}=1\text{ atm}\\ \text{1}=\frac{1\text{ atm}}{101.325\text{ kPa}}\end{array}$

For $3.73\text{ kPa}$ ,

$\begin{array}{c}3.73\text{ kPa}=\frac{1\text{ atm}}{101.325\text{ kPa}}\times 3.73\text{kPa}\\ =0.0368\text{ atm}\end{array}$