Chapter 14, Problem 17E

Interpretation Introduction

(a)

Interpretation:

The density of air at STP having “effective” molar mass of $29\text{g}/\text{mol}$ is to be stated.

Concept Introduction:

The ideal gas law is the compilation of Charles’s Law $\left(\text{V}\propto \text{T}\right)$, Boyle’s Law $\left(\text{V}\propto \frac{\text{1}}{\text{P}}\right)$ and Avogadro’s Law $\left(\text{V}\propto \text{n}\right)$.

The ideal gas equation resultant of all the ideal gas law’s is shown below.

$\text{PV}=\text{nRT}$

Effective molar mass means the average molar mass of all the gases present in the air.

Answer to Problem 17E

The density of air at STP having “effective” molar mass of $29\text{g}/\text{mol}$ is $\text{1}\text{.3}\text{g}/\text{L}$.

Explanation of Solution

The ideal gas equation is shown below.

$\text{PV}=\text{nRT}$ …(1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{n}$ is the number of moles.

• $\text{R}$ is gas constant.

• $\text{T}$ is temperature.

Density of any substance is defined as the mass per unit volume. That is, mass of the substance divided by the volume that mass is occupying.

$\text{Density}=\frac{\text{Mass}\left(\text{M}\right)}{\text{Volume}\left(\text{V}\right)}$ …(2)

Conversion of ideal gas equation in terms of density of gas is shown below.

Divide the equation (1) by $\text{V}$ on both sides and simplify.

$\begin{array}{rll}\text{PV}& =& \text{nRT}\\ \frac{\text{PV}}{\text{V}}& =& \frac{\text{nRT}}{\text{V}}\\ \text{P}& =& \frac{\text{nRT}}{\text{V}}\end{array}$

The number of moles of gas is related to its molar mass by the formula shown below.

$\text{n}=\frac{\text{M}}{\text{MM}}$

Where,

• $\text{M}$ is the mass of gas.

• $\text{MM}$ is the molar mass of gas.

Substitute the value of $\text{n}$ in the above simplified expression.

$\begin{array}{l}\text{P}=\frac{\text{nRT}}{\text{V}}\\ \text{P}=\frac{\frac{\text{M}}{\text{MM}}\text{RT}}{\text{V}}\\ \text{P}=\frac{\text{MRT}}{\left(\text{MM}\right)\text{V}}\end{array}$

Using equation (2) and simplify.

$\begin{array}{rll}\text{P}& =& \frac{\text{MRT}}{\left(\text{MM}\right)\text{V}}\\ \text{P}& =& \frac{\text{DRT}}{\text{MM}}\\ \frac{\text{P}\left(\text{MM}\right)}{\text{RT}}& =& \text{D}\end{array}$

The formula of density of gas in terms of molar mass, temperature, and pressure is given below.

$\text{D}=\frac{\text{P}\left(\text{MM}\right)}{\text{RT}}$ …(3)

Under STP conditions temperature is $\text{273 K}$ and pressure is $\text{0}\text{.987 atm}$.

The molar mass of gas is provided as $29\text{g}/\text{mol}$.

Substitute the values of P, R, T and MM in the equation (3).

$\begin{array}{rll}\text{D}& =& \frac{\text{0}\text{.987 atm}\left(29\text{g}/\text{mol}\right)}{\left(\text{0}\text{.0821}{\text{L atm mol}}^{-1}{\text{ K}}^{-1}\right)\text{273}\text{K}}\\ & =& \frac{\text{29}{\text{g atm mol}}^{-1}}{\text{22}\text{.4}{\text{L atm mol}}^{-1}}\\ & =& \text{1}\text{.3 }\text{g}/\text{L}\end{array}$

The density of gas at STP having molar mass of $29\text{g}/\text{mol}$ is $\text{1}\text{.3 }\text{g}/\text{L}$.

Conclusion

The density of air at STP having “effective” molar mass of $29\text{g}/\text{mol}$ is $\text{1}\text{.3}\text{g}/\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The density of air at temperature of $\text{20}°\text{C}$ and pressure of $\text{751 torr}$ having “effective” molar mass of $29\text{g}/\text{mol}$ is to be stated.

Concept Introduction:

The ideal gas law is the compilation of Charles’s Law $\left(\text{V}\propto \text{T}\right)$, Boyle’s Law $\left(\text{V}\propto \frac{\text{1}}{\text{P}}\right)$ and Avogadro’s Law $\left(\text{V}\propto \text{n}\right)$.

The ideal gas equation resultant of all the ideal gas law’s is shown below.

$\text{PV}=\text{nRT}$

Effective molar mass means the average molar mass of all the gases present in the air.

Answer to Problem 17E

The density of air at temperature of $\text{20}°\text{C}$ and pressure of $\text{751 torr}$ having “effective” molar mass of $29\text{g}/\text{mol}$ is $\text{1}\text{.2}\text{g}/\text{L}$.

Explanation of Solution

The ideal gas equation is shown below.

$\text{PV}=\text{nRT}$ …(1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{n}$ is the number of moles.

• $\text{R}$ is gas constant.

• $\text{T}$ is temperature.

Density of any substance is defined as the mass per unit volume. That is, mass of the substance divided by the volume that mass is occupying.

$\text{Density}=\frac{\text{Mass}\left(\text{M}\right)}{\text{Volume}\left(\text{V}\right)}$ …(2)

Conversion of ideal gas equation in terms of density of gas is shown below.

Divide the equation (1) by $\text{V}$ on both sides and simplify.

$\begin{array}{rll}\text{PV}& =& \text{nRT}\\ \frac{\text{PV}}{\text{V}}& =& \frac{\text{nRT}}{\text{V}}\\ \text{P}& =& \frac{\text{nRT}}{\text{V}}\end{array}$

The number of moles of gas is related to its molar mass by the formula shown below.

$\text{n}=\frac{\text{M}}{\text{MM}}$

Where,

• $\text{M}$ is the mass of gas.

• $\text{MM}$ is the molar mass of gas.

Substitute the value of $\text{n}$ in the above simplified expression.

$\begin{array}{l}\text{P}=\frac{\text{nRT}}{\text{V}}\\ \text{P}=\frac{\frac{\text{M}}{\text{MM}}\text{RT}}{\text{V}}\\ \text{P}=\frac{\text{MRT}}{\left(\text{MM}\right)\text{V}}\end{array}$

Using equation (2) and simplify.

$\begin{array}{rll}\text{P}& =& \frac{\text{MRT}}{\left(\text{MM}\right)\text{V}}\\ \text{P}& =& \frac{\text{DRT}}{\text{MM}}\\ \frac{\text{P}\left(\text{MM}\right)}{\text{RT}}& =& \text{D}\end{array}$

The formula of density of gas in terms of molar mass, temperature, and pressure is given below.

$\text{D}=\frac{\text{P}\left(\text{MM}\right)}{\text{RT}}$ …(3)

Molar mass of oxygen gas is ${\text{32 g mol}}^{-1}$.

The temperature of air is $\text{20}°\text{C}$ and pressure is $\text{751 torr}$.

The conversion of temperature in degree celsius to kelvin units is shown below.

$\text{T}\left(\text{K}\right)=\text{T}\left({}^{\text{ο}}\text{C}\right)+273$ …(4)

Where,

• $\text{T}\left(\text{K}\right)$ is the temperature in Kelvin units.

• $\text{T}\left(°\text{C}\right)$ is the temperature in degree Celsius units.

Temperature of air in degree Celsius units is $20°\text{C}$.

Substitute the value $\text{T}\left(°\text{C}\right)$ and in the equation 4.

$\begin{array}{rll}\text{T}\left(\text{K}\right)& =& \text{T}\left(°\text{C}\right)+273\\ & =& 20+\text{273}\\ & =& \text{293}\end{array}$

The conversion of pressure in torr to atm is shown below.

$\text{760 torr}=\text{1 atm}$

The pressure of $\text{751 torr}$ in atm units is calculated below.

$\begin{array}{l}\text{760 torr}=\text{1 atm}\\ \text{1 torr}=\frac{\text{1 atm}}{\text{760 torr}}\\ \text{751 torr}=\frac{\text{1 atm}}{\text{760 torr}}\times \text{751 torr}\\ \text{751 torr}=\text{0}\text{.988 atm}\end{array}$

Substitute the values of P, R, T and MM in the equation (3).

$\begin{array}{rll}\text{D}& =& \frac{\text{0}\text{.988 atm}\left({\text{29 g mol}}^{-1}\right)}{\left(\text{0}{\text{.0821 L atm mol}}^{-1}{\text{ K}}^{-1}\right)\text{293 K}}\\ & =& \frac{{\text{29 g atm mol}}^{-1}}{\text{24}{\text{.1 L atm mol}}^{-1}}\\ & =& \text{1}\text{.2 }\text{g}/\text{L}\end{array}$

The density of air at temperature of $\text{20}°\text{C}$ and pressure of $\text{751 torr}$ having “effective” molar mass of $29\text{g}/\text{mol}$ is $\text{1}\text{.2 }\text{g}/\text{L}$.

Conclusion

The density of air at temperature of $\text{20}°\text{C}$ and pressure of $\text{751 torr}$ having “effective” molar mass of $29\text{g}/\text{mol}$ is $\text{1}\text{.2 }\text{g}/\text{L}$.