Chapter 13, Problem 4A

(a)

Interpretation Introduction

Interpretation:

The unit of pressure have to be converted into atmospheres.

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 4A

  $105.2\mathrm{k}\mathrm{P}\mathrm{a}=1.038\mathrm{a}\mathrm{t}\mathrm{m}$

Explanation of Solution

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

  $\begin{array}{c}105.2\mathrm{k}\mathrm{P}\mathrm{a}=105200\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{}\\ =\frac{105200}{101325}\mathrm{a}\mathrm{t}\mathrm{m}\\ =1.038\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The unit of pressure have to be converted into atmospheres.

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 4A

  $75.2\mathrm{c}\mathrm{m}\mathrm{H}\mathrm{g}=0.989\mathrm{a}\mathrm{t}\mathrm{m}$

Explanation of Solution

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

  $\begin{array}{c}75.2\mathrm{c}\mathrm{m}\mathrm{H}\mathrm{g}=752\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}\\ =\frac{752}{760}\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.989\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The unit of pressure have to be converted into atmospheres.

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 4A

  $752\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}=0.989\mathrm{a}\mathrm{t}\mathrm{m}$

Explanation of Solution

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

  $\begin{array}{c}752\mathrm{}\mathrm{m}\mathrm{m}\mathrm{}\mathrm{H}\mathrm{g}=\frac{752}{760}\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.989\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The unit of pressure have to be converted into atmospheres.

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 4A

  $767\mathrm{}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{}=1.01\mathrm{a}\mathrm{t}\mathrm{m}$

Explanation of Solution

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

  $\begin{array}{c}767\mathrm{}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{}=\frac{767}{760}\mathrm{a}\mathrm{t}\mathrm{m}\\ =1.01\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$