Chapter 1, Problem 54RGQ

(a)

Interpretation Introduction

Interpretation: The volume of the room in cubic meters and liters has to be given.

Concept Introduction:

Density: is a characteristic property of a substance.  The density of a substance is the relationship between the mass of the substance and how much space it takes up (volume).

  $\text{Density = }\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 54RGQ

Volume in liters is $\text{65}\text{.41}{\text{×10}}^3\text{ L}$

Explanation of Solution

The volume of the room in cubic meters and litres is given,

  $\begin{array}{l}\underset{\_}{\text{Calculate the volume of room:}}\\ \text{Volume of a room is l×h×w }\\ \text{Given: length = 18ft = 5}\text{.4864m ,}\\ \text{ wide = 15 ft = 4}\text{.572 m,}\\ \text{ height = 8 ft 6 in = 2}\text{.5904 m}\\ \text{Volume = 5}\text{.49 m }\times \text{4}\text{.6 m }\times \text{2}\text{.59 m = 65}\text{.41}{\text{m}}^{\text{3}}\\ \underset{\_}{\text{Conversion of units:}}\\ \text{Given: }\text{Volume = 65}\text{.41}{\text{m}}^{\text{3}}\\ {\text{Known: 1 m}}^{\text{3}}{\text{ = 10}}^3\text{ L}\\ \text{Whereas,}\\ \text{65}\text{.41}{\text{m}}^{\text{3}}\text{ is}\\ \text{ = }\frac{\text{65}{\text{.41m}}^{\text{3}}{\text{×10}}^3\text{ L}}{{\text{1 m}}^{\text{3}}}\text{=}\text{65}\text{.41}{\text{×10}}^3\text{ L}\\ \text{ =}\text{65}\text{.41}{\text{×10}}^3\text{ L}.\end{array}$

The volume in cubic liters is $\text{65}\text{.41}{\text{×10}}^3\text{ L}$

(b)

Interpretation Introduction

Interpretation: The volume of the room in cubic meters and liters has to be given.

Concept Introduction:

Density: is a characteristic property of a substance.  The density of a substance is the relationship between the mass of the substance and how much space it takes up (volume).

  $\text{Density = }\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 54RGQ

Mass of air in kilogram is $78.492Kg$ and in pounds is $0.1730pounds.$

Explanation of Solution

The mass of air is calculated as,

  $\begin{array}{l}\underset{\_}{\text{Calculate the mass of air:}}\\ \text{Given: }\text{Density = 1}\text{.2 g/L ; Volume = 65}\text{.41}{\text{×10}}^{\text{3}}\text{ L}\\ \text{Known: Density = }\frac{\text{Mass}}{\text{Volume}}\\ \text{Whereas,}\\ \text{Mass = Density×Volume }\\ \text{ = 1}\text{.2 g/L× 65}\text{.41}{\text{×10}}^{\text{3}}\text{ L}\\ \text{ =}\text{78}{\text{.492×10}}^{\text{3}}\text{g}\text{.}\\ \underset{\_}{\text{Conversion of units:}}\\ \text{Known:1g = 0}\text{.001 kg}\\ \text{Whereas, }\\ \text{78}{\text{.492×10}}^{\text{3}}\text{g is,}\\ \text{= }\frac{\left(\text{78}{\text{.492×10}}^{\text{3}}\text{g}\right)\left(\text{0}\text{.001 kg}\right)}{\text{1g}}\text{ = }78.492Kg\text{.}\\ \underset{\_}{\text{Conversion of units:}}\\ \text{Known:1g = 0}\text{.00220462 pound}\\ \text{Whereas, }\\ \text{78}{\text{.492×10}}^{\text{3}}\text{g is,}\\ \text{= }\frac{\left(\text{78}{\text{.492×10}}^{\text{3}}\text{g}\right)\left(\text{0}\text{.00220462 pound}\right)}{\text{1g}}\text{ = }0.1730pounds\text{.}\end{array}$

The obtained mass in kilograms and pounds are shown above.