Chapter 1, Problem 54RGQ

Suppose your bedroom is 18 ft long and 15 ft wide, and the distance from floor to ceiling is 8 ft 6 in. You need to know the volume of the room in metric units for some scientific calculations.

(a) What is the room's volume in cubic meters? In liters?

(b) What is the mass of air in the room in kilograms? In pounds? (Assume the density of air is 1.2 g/L and that the room is empty of furniture.)

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 $\text{65}\text{.41}{\text{×10}}^3\text{ L}$

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.