Chapter 5, Problem 5.49P

(a)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  $d=\frac{m}{V}$ …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is $0{}^o\text{C}$ or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  $PV=nRT$ …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  $PV=\left(\frac{m}{M}\right)RT$

Or,

  $V=\frac{mRT}{PM}$

Putting the value of volume in equation (1) thus,

  $d=\frac{m}{\frac{mRT}{PM}}$

Or,

  $d=\frac{PM}{RT}$

Answer to Problem 5.49P

The density of ${\text{SO}}_{\text{2}}$ is $2.86\text{ g/L}$ which is denser than air.

Explanation of Solution

The given gas is ${\text{SO}}_{\text{2}}$ .

The molar mass of the gas is 64.066 g/mol. The pressure and temperature at STP will be 1 atm and 298.15 K thus, density can be calculated using the following relation:

  $d=\frac{PM}{RT}$

Putting the values,

  $\begin{array}{c}d=\frac{\left(1\text{ atm}\right)\left(64.066\text{ g/mol}\right)}{\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(273.15\text{ K}\right)}\\ =2.86\text{ g/L}\end{array}$

The density of air at STP is 1.29 g/L. The value of density of ${\text{SO}}_{\text{2}}$ is more than air thus, it is denser than air.

(b)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  $d=\frac{m}{V}$ …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is $0{}^o\text{C}$ or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  $PV=nRT$ …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the value of number of moles from equation (3) to (2) thus,

  $PV=\left(\frac{m}{M}\right)RT$

Or,

  $V=\frac{mRT}{PM}$

Putting the value of volume in equation (1) thus,

  $d=\frac{m}{\frac{mRT}{PM}}$

Or,

  $d=\frac{PM}{RT}$

Answer to Problem 5.49P

The density of ${\text{CH}}_4$ is $0.716\text{ g/L}$ which is not denser than air.

Explanation of Solution

The given gas is ${\text{CH}}_4$ .

The molar mass of the gas is 16.04 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  $d=\frac{PM}{RT}$

Putting the values,

  $\begin{array}{c}d=\frac{\left(1\text{ atm}\right)\left(16.04\text{ g/mol}\right)}{\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(273.15\text{ K}\right)}\\ =0.716\text{ g/L}\end{array}$

The density of air at STP is 1.29 g/L. The value of density of ${\text{CH}}_4$ is less than air thus, the air is denser than ${\text{CH}}_4$ .

(c)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  $d=\frac{m}{V}$ …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is $0{}^o\text{C}$ or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  $PV=nRT$ …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  $PV=\left(\frac{m}{M}\right)RT$

Or,

  $V=\frac{mRT}{PM}$

Putting the value of volume in equation (1) thus,

  $d=\frac{m}{\frac{mRT}{PM}}$

Or,

  $d=\frac{PM}{RT}$

Answer to Problem 5.49P

The density of ${\text{H}}_2$ is $0.09\text{ g/L}$ which is not denser than air.

Explanation of Solution

The given gas is ${\text{H}}_2$ .

The molar mass of the gas is 1.008 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  $d=\frac{PM}{RT}$

Putting the values,

  $\begin{array}{c}d=\frac{\left(1\text{ atm}\right)\left(2.016\text{ g/mol}\right)}{\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(273.15\text{ K}\right)}\\ =0.09\text{ g/L}\end{array}$

The density of air at STP is 1.29 g/L. The value of density of ${\text{H}}_2$ is less than air thus, the air is denser than ${\text{H}}_2$ .

(d)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  $d=\frac{m}{V}$ …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is $0{}^o\text{C}$ or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  $PV=nRT$ …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  $PV=\left(\frac{m}{M}\right)RT$

Or,

  $V=\frac{mRT}{PM}$

Putting the value of volume in equation (1) thus,

  $d=\frac{m}{\frac{mRT}{PM}}$

Or,

  $d=\frac{PM}{RT}$

Answer to Problem 5.49P

The density of $\text{He}$ is $0.1787\text{ g/L}$ which is not denser than air.

Explanation of Solution

The given gas is $\text{He}$ .

The molar mass of the gas is 4.002 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  $d=\frac{PM}{RT}$

Putting the values,

  $\begin{array}{c}d=\frac{\left(1\text{ atm}\right)\left(\text{4}\text{.002 g/mol}\right)}{\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(273.15\text{ K}\right)}\\ =0.1787\text{ g/L}\end{array}$

The density of air at STP is 1.29 g/L. The value of density of $\text{He}$ is less than air thus, the air is denser than $\text{He}$ .

(e)

Interpretation Introduction

Interpretation: The density of gas at STP should be calculated. If it is greater than the density of air or not should be determined.

Concept Introduction: The density of a substance is its mass per unit volume. It is mathematically represented as follows:

  $d=\frac{m}{V}$ …… (1)

Here, m is mass and V is volume.

At STP, the value of temperature is $0{}^o\text{C}$ or 273.15 K and that of pressure is 1 atm. From ideal gas law, pressure, volume, number of moles and temperature are related to each other as follows:

  $PV=nRT$ …… (2)

Here, P is pressure, V is volume, n is the number of moles, R is Universal gas constant and T is temperature.

Also, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the value of the number of moles from equation (3) to (2) thus,

  $PV=\left(\frac{m}{M}\right)RT$

Or,

  $V=\frac{mRT}{PM}$

Putting the value of volume in equation (1) thus,

  $d=\frac{m}{\frac{mRT}{PM}}$

Or,

  $d=\frac{PM}{RT}$

Answer to Problem 5.49P

The density of ${\text{CO}}_2$ is $1.96\text{ g/L}$ which is denser than air.

Explanation of Solution

The given gas is ${\text{CO}}_2$ .

The molar mass of the gas is 44.01 g/mol. The pressure and temperature at STP will be 1 atm and 273.15 K thus, density can be calculated using the following relation:

  $d=\frac{PM}{RT}$

Putting the values,

  $\begin{array}{c}d=\frac{\left(1\text{ atm}\right)\left(\text{44}\text{.01 g/mol}\right)}{\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(273.15\text{ K}\right)}\\ =1.96\text{ g/L}\end{array}$

The density of air at STP is 1.29 g/L. The value of density of ${\text{CO}}_2$ is more than air thus, it is denser than air.