Chapter 15, Problem 103A

(a)

Interpretation Introduction

Interpretation: To determine the volume of each balloon.

Concept Introduction: Ideal gas equation is given as,

By using the ideal gas equation:

  $\text{PV=nRT}$

Where, P = pressure, V = volume, T = temperature, R is gas constant

Answer to Problem 103A

22.4 L

Explanation of Solution

The volume of 1 mol of any gas is 22.4 L.

All the balloons have 1 mol of gas, thus, the volume of each balloon is 22.4 L.

(b)

Interpretation Introduction

Interpretation: To determine the mass of each balloon.

Concept Introduction: The molar mass of a compound could be estimated by combining the standard atomic masses (in g/mol) of such constituent atoms.

Answer to Problem 103A

The mass He is 4 g, the mass of methane is 16 and the mass of oxygen is 32 g.

Explanation of Solution

The mass of Helium = 4 g/mol

  $\begin{array}{c}\text{Mass}\text{of}\text{}{\text{CH}}_{\text{4}}\text{=mass}\text{of}\text{C+mass}\text{of}\text{H}\\ =12g+4(1)g\\ =16g\\ \end{array}$

  $\begin{array}{c}\text{The}\text{mass}\text{of}{\text{O}}_{\text{2}}\text{=2×16g}\\ \text{=32g}\end{array}$

(c)

Interpretation Introduction

Interpretation: To determine the density of each balloon.

Concept Introduction: Density can be. determined by the volume and mass of the gas. It can be represented by the symbol d which can be expressed as $\text{d=}\frac{\text{m}}{\text{v}}$

Answer to Problem 103A

The density of helium gas will be $\text{0}\text{.178 g/L}$ , the density of methane gas will be $\text{0}\text{.714}\text{g/L}$ , and the density of oxygen gas will be $\text{1}\text{.42}\text{g/L}$ .

Explanation of Solution

It is known that,

The volume of the gas at STP 22.4 L

The mass of the helium gas = 4 g

The mass of the methane gas = 16 g

The mass of oxygen gas = 32g

Density can be determined by using the formula:

  $\begin{array}{c}\text{d=}\frac{\text{m}}{\text{v}}\\ \text{=}\frac{\text{4}\text{g}}{\text{22}\text{.4 L}}\\ =\text{0}\text{.178 g/L}\end{array}$

The density of methane gas can be calculated by as:

  $\begin{array}{c}\text{d=}\frac{\text{m}}{\text{v}}\\ \text{=}\frac{\text{16}\text{g}}{\text{22}\text{.4 L}}\\ =\text{0}\text{.714}\text{g/L}\end{array}$

The density of oxygen gas can be calculated as:

  $\begin{array}{c}\text{d=}\frac{\text{m}}{\text{v}}\\ \text{=}\frac{32\text{g}}{\text{22}\text{.4 L}}\\ \text{=1}\text{.42}\text{g/L}\end{array}$

(d)

Interpretation Introduction

Interpretation: To predict which balloon will rise and which will sink.

Concept Introduction: Density can be determined by the volume and mass of the gas. It can be represented by the symbol d which can be expressed as $\text{d=}\frac{\text{m}}{\text{v}}$

Answer to Problem 103A

Each balloon will rise.

Explanation of Solution

The density of helium gas is $\text{0}\text{.178 g/L}$ , the density of methane gas is $\text{0}\text{.714}\text{g/L}$ , and the density of oxygen gas is $\text{1}\text{.42}\text{g/L}$ .

Density of air is $\text{1}\text{.2}\text{g/mL or 1200 g/L}$ .

As the density of air is much higher than the density of each balloon, thus, each balloon will rise.