Chapter 13, Problem 6A

(a)

Interpretation Introduction

Interpretation:

The indicated pressure conversions have to be made.

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 6A

  $1.5\times 10^5\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}=1690.64\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}$

Explanation of Solution

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

Thus,

  $\begin{array}{c}1.5\times 10^5\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}=\frac{150000}{101325}\mathrm{a}\mathrm{t}\mathrm{m}\\ =1.48\mathrm{a}\mathrm{t}\mathrm{m}\\ =2.224\times 760\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =1690.64\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The indicated pressure conversions have to be made.

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 6A

  $1.21\mathrm{a}\mathrm{t}\mathrm{m}=122603.25\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}$

Explanation of Solution

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

Thus,

  $\begin{array}{c}1.21\mathrm{a}\mathrm{t}\mathrm{m}=1.21\times 101325\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\\ =122603.25\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The indicated pressure conversions have to be made.

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 6A

  $97345\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}=730.15\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}$

Explanation of Solution

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

Thus,

  $\begin{array}{c}97345\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}=\frac{97345}{101325}\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.961\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.961\times 760\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =730.15\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The indicated pressure conversions have to be made.

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 6A

  $1.32\mathrm{k}\mathrm{P}\mathrm{a}=1320\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}$

Explanation of Solution

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

Thus,

  $\begin{array}{c}1.32\mathrm{k}\mathrm{P}\mathrm{a}=1.32\times 1000\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\\ =1320\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\end{array}$