Chapter 13, Problem 5A

(a)

Interpretation Introduction

Interpretation:

The unit of pressure has to be converted into mm Hg.

Concept Introduction:

The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimeters of mercury) are most commonly used by chemists.

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

Answer to Problem 5A

  $0.9975\mathrm{a}\mathrm{t}\mathrm{m}=758.1\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}$

Explanation of Solution

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

  $\begin{array}{c}0.9975\mathrm{a}\mathrm{t}\mathrm{m}=0.9975\times 760\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =758.1\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The unit of pressure has to be converted into mm Hg.

Concept Introduction:

The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimetres of mercury) are most commonly used by chemists.

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

Answer to Problem 5A

  $225400\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}=1690.64\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}$

Explanation of Solution

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

  $\begin{array}{c}225400\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}=\frac{225400}{101325}\mathrm{a}\mathrm{t}\mathrm{m}\\ =2.224\mathrm{a}\mathrm{t}\mathrm{m}\\ =2.224\times 760\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =1690.64\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The unit of pressure has to be converted into mm Hg.

Concept Introduction:

The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimetres of mercury) are most commonly used by chemists.

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

Answer to Problem 5A

  $99.7\mathrm{k}\mathrm{P}\mathrm{a}=747.8\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}$

Explanation of Solution

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

  $\begin{array}{c}99.7\mathrm{k}\mathrm{P}\mathrm{a}=99700\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}\\ =\frac{99700}{101325}\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.984\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.984\times 760\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =747.8\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The unit of pressure has to be converted into mm Hg.

Concept Introduction:

The SI unit for pressure is the pascal (Pa) and the units such as atm, torr and mm Hg (millimetres of mercury) are most commonly used by chemists.

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

Answer to Problem 5A

  $1.078\mathrm{a}\mathrm{t}\mathrm{m}=819.3\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}$

Explanation of Solution

  $\begin{array}{c}1\text{ standard atmosphere }=1.00\text{ atm }\\ \text{= 760}\text{.0 mm Hg }\\ \text{= 760}\text{.0 torr }\\ \text{= 101,325 Pa}\end{array}$

  $\begin{array}{c}1.078\mathrm{a}\mathrm{t}\mathrm{m}=1.078\times 760\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =819.3\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\end{array}$