Chapter 1, Problem 1.68P

a)

Interpretation Introduction

Interpretation:Whether evacuated ball will float or not should be identified.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

The characteristic of any substance to float or sink in another substanceis predicted by its density. Any substancefloats over another substanceif its density is less than that of other substanceand it sinks if its density is higher.

Explanation of Solution

The expression to calculate density of ball is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ball.
• $M$is mass of ball.
• $V$is volume of ball.

The value of $M$ is $0.12\text{ g}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}d=\frac{M}{V}\\ =\left(\frac{0.12\text{ g}}{560{\text{ cm}}^3}\right)\left(\frac{1{\text{ cm}}^3}{{10}^{-3}\text{ L}}\right)\\ =0.214\text{ g/L}\end{array}$

Since density of ball is less than that of air, evacuated ball will float in air.

b)

Interpretation Introduction

Interpretation:Whether ball filled with carbon dioxide will float or not should be identified.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

The characteristic of any substance to float or sink in another substance is predicted by its density. Any substance floats over another substance if its density is less than that of other substance and it sinks if its density is higher.

Explanation of Solution

The expression to calculate density of ball is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ball.
• $M$is mass of ball.
• $V$is volume of ball.

The value of $M$ is $0.12\text{ g}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}d=\frac{M}{V}\\ =\left(\frac{0.12\text{ g}}{560{\text{ cm}}^3}\right)\left(\frac{1{\text{ cm}}^3}{{10}^{-3}\text{ L}}\right)\\ =0.214\text{ g/L}\end{array}$

The formula to calculate total density of ball filled with carbon dioxide is as follows:

  $\text{Total density}=\text{Density of ball}+\text{Density of carbon dioxide}$

The density of ball is $0.214\text{ g/L}$ .

The density of carbon dioxide is $1.830\text{ g/L}$

Substitute the values in above equation.

  $\begin{array}{c}\text{Total density}=\text{Density of ball}+\text{Density of carbon dioxide}\\ =0.214\text{ g/L}+1.830\text{ g/L}\\ =2.044\text{ g/L}\end{array}$

Since density of ball filled with carbon dioxide is more than that of air, it will not float in air.

c)

Interpretation Introduction

Interpretation:Whether ball filled with hydrogen will float or not should be identified.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

The characteristic of any substance to float or sink in another substance is predicted by its density. Any substance floats over another substance if its density is less than that of other substance and it sinks if its density is higher.

Explanation of Solution

The expression to calculate density of ball is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ball.
• $M$is mass of ball.
• $V$is volume of ball.

The value of $M$ is $0.12\text{ g}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}d=\frac{M}{V}\\ =\left(\frac{0.12\text{ g}}{560{\text{ cm}}^3}\right)\left(\frac{1{\text{ cm}}^3}{{10}^{-3}\text{ L}}\right)\\ =0.214\text{ g/L}\end{array}$

The formula to calculate total density of ball filled with hydrogen is as follows:

  $\text{Total density}=\text{Density of ball}+\text{Density of hydrogen}$

The density of ball is $0.214\text{ g/L}$ .

The density of hydrogen is $0.0899\text{ g/L}$

Substitute the values in above equation.

  $\begin{array}{c}\text{Total density}=\text{Density of ball}+\text{Density of hydrogen}\\ =0.214\text{ g/L}+0.0899\text{ g/L}\\ =0.3039\text{ g/L}\end{array}$

Since density of ball filled with hydrogen is less than that of air, it will float in air.

d)

Interpretation Introduction

Interpretation:Whether ball filled with oxygen will float or not should be identified.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

The characteristic of any substance to float or sink in another substance is predicted by its density. Any substance floats over another substance if its density is less than that of other substance and it sinks if its density is higher.

Explanation of Solution

The expression to calculate density of ball is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ball.
• $M$is mass of ball.
• $V$is volume of ball.

The value of $M$ is $0.12\text{ g}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}d=\frac{M}{V}\\ =\left(\frac{0.12\text{ g}}{560{\text{ cm}}^3}\right)\left(\frac{1{\text{ cm}}^3}{{10}^{-3}\text{ L}}\right)\\ =0.214\text{ g/L}\end{array}$

The formula to calculate total density of ball filled with oxygen is as follows:

  $\text{Total density}=\text{Density of ball}+\text{Density of oxygen}$

The density of ball is $0.214\text{ g/L}$ .

The density of oxygen is $1.330\text{ g/L}$

Substitute the values in above equation.

  $\begin{array}{c}\text{Total density}=\text{Density of ball}+\text{Density of oxygen}\\ =0.214\text{ g/L}+1.330\text{ g/L}\\ =1.554\text{ g/L}\end{array}$

Since density of ball filled with oxygen is more than that of air, it will not float in air.

e)

Interpretation Introduction

Interpretation:Whether ball filled with nitrogen will float or not should be identified.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

The characteristic of any substance to float or sink in another substance is predicted by its density. Any substance floats over another substance if its density is less than that of other substance and it sinks if its density is higher.

Explanation of Solution

The expression to calculate density of ball is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ball.
• $M$is mass of ball.
• $V$is volume of ball.

The value of $M$ is $0.12\text{ g}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}d=\frac{M}{V}\\ =\left(\frac{0.12\text{ g}}{560{\text{ cm}}^3}\right)\left(\frac{1{\text{ cm}}^3}{{10}^{-3}\text{ L}}\right)\\ =0.214\text{ g/L}\end{array}$

The formula to calculate total density of ball filled with nitrogen is as follows:

  $\text{Total density}=\text{Density of ball}+\text{Density of nitrogen}$

The density of ball is $0.214\text{ g/L}$ .

The density of nitrogen is $1.165\text{ g/L}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Total density}=\text{Density of ball}+\text{Density of nitrogen}\\ =0.214\text{ g/L}+1.165\text{ g/L}\\ =1.379\text{ g/L}\end{array}$

Since density of ball filled with nitrogen is more than that of air, it will not float in air.

f)

Interpretation Introduction

Interpretation:Weight added to ball to make it sink in case it already floats in it is to be determined.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

The characteristic of any substance to float or sink in another substance is predicted by its density. Any substance floats over another substance if its density is less than that of other substance and it sinks if its density is higher.

Answer to Problem 1.68P

  $\text{0}\text{.55 g}$ of gas and $0.50\text{ g}$ of hydrogen gas should be added to ball so as to make it sink.

Explanation of Solution

The ball floats in air when it is evacuated and when filled with hydrogen.

The expression to calculate density of ball is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ball.
• $M$is mass of ball.
• $V$is volume of ball.

Rearrange above equation for $M$ .

  $M=dV$

The value of $d$ is $1.189\text{ g/L}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}M=dV\\ =\left(1.189\text{ g/L}\right)\left(560{\text{ cm}}^3\right)\left(\frac{{10}^{-3}\text{ L}}{1{\text{ cm}}^3}\right)\\ =0.66584\text{ g}\end{array}$

Therefore $0.66584\text{ g}$ is required for ball to sink in air.

The mass of added gas can be calculated as follows:

  $\begin{array}{c}\text{Mass of added gas}=\left(0.66584-0.12\right)\text{ g}\\ =0.54584\text{ g}\\ \approx \text{0}\text{.55 g}\end{array}$

Hence, $\text{0}\text{.55 g}$ of gas should be added to ball so as to make it sink in air.

The value of $d$ is $0.0899\text{g/L}$ .

The value of $V$ is $560{\text{ cm}}^3$ .

Substitute the value in above equation for mass of hydrogen.

  $\begin{array}{c}M=dV\\ =\left(0.0899\text{ g/L}\right)\left(560{\text{ cm}}^3\right)\left(\frac{{10}^{-3}\text{ L}}{1{\text{ cm}}^3}\right)\\ =0.05034\text{ g}\end{array}$

The formula to calculate total mass of ball filled with hydrogen is as follows:

  $\text{Total mass}=\text{Mass of ball}+\text{Mass of hydrogen}$

The mass of ballis $0.12\text{ g}$ .

The mass of hydrogen is $0.05034\text{ g}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Total mass}=\text{Mass of ball}+\text{Mass of hydrogen}\\ =\left(0.12+0.05034\right)\text{ g}\\ =0.17034\text{ g}\end{array}$

But $0.66584\text{ g}$ is required for ball to sink in air. Therefore weight added to ball can be calculated as follows:

  $\begin{array}{c}\text{Added weight}=\left(0.66584-0.17034\right)\text{ g}\\ =0.4955\text{ g}\\ \approx \text{0}\text{.50 g}\end{array}$

Hence, $0.50\text{ g}$ of hydrogen is added to ball to make it sink in air.