Chapter 9, Problem 161QP

(a)

Interpretation Introduction

Interpretation:

The gas which has the greatest density is to be identified.

Explanation of Solution

The ideal gas equation states that pressure $\left(P\right)$ and volume of a gas $\left(V\right)$ is directly proportional to number of moles $\left(n\right)$ and temperature $\left(T\right)$ of the gas.

$PV=nRT$ …… (1)

Here, $R$ is the universal gas constant having value $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ .

Density $\left(\rho \right)$ of a gas is equal to the ratio of mass $\left(m\right)$ of the gas to the volume $\left(V\right)$ of the gas.

$\rho =\frac{m}{V}$ …… (2)

The number of moles $\left(n\right)$ is equal to the ratio of mass $\left(m\right)$ of the gas to the molar mass $\left(MM\right)$ of the gas.

$n=\frac{m}{MM}$ …… (3)

The gas is kept in $2.50\text{ L}$ container with pressure of $1.12\text{ atm}$ and temperature $25°\text{C}$ . Convert temperature units from degree Celsius to kelvin as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Convert temperature $25°\text{C}$ units from degree Celsius to kelvin as follows:

$\begin{array}{c}T=25+273.15\\ =298.15\text{ K}\end{array}$

Substitute $P$ as $1.12\text{ atm}$ , $V$ as $2.50\text{ L}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ , and $T$ as $298.15\text{ K}$ in equation (1):

$\begin{array}{c}1.12\text{ atm}\times 2.50\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 298.15\text{ K}\\ n=\frac{1.12\text{ atm}\times 2.50\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 298.15\text{ K}}\\ =0.114\text{ mol}\end{array}$

The molar mass of ${\text{N}}_2$ gas is as follows:

$\begin{array}{c}{\text{N}}_2=2\times 14{\text{ g mol}}^{-1}\\ =28{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $0.114\text{ mol}$ and $MM$ as $28{\text{ g mol}}^{-1}$ in equation (3):

$\begin{array}{c}0.114\text{ mol}=\frac{m}{28{\text{ g mol}}^{-1}}\\ m=3.192\text{ g}\end{array}$

Substitute $m$ as $3.192\text{ g}$ and $V$ as $2.50\text{ L}$ in equation (2):

$\begin{array}{c}\rho =\frac{3.192\text{g}}{2.50\text{ L}}\\ =1.278{\text{ g L}}^{-1}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The gas which has the greatest density is to be identified.

Explanation of Solution

The molar mass of ${\text{O}}_2$ gas is as follows:

$\begin{array}{c}{\text{O}}_2=2\times 16{\text{ g mol}}^{-1}\\ =32{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $0.114\text{ mol}$ and $MM$ as $32{\text{ g mol}}^{-1}$ in equation (3):

$\begin{array}{c}0.114\text{ mol}=\frac{m}{32{\text{ g mol}}^{-1}}\\ m=3.648\text{ g}\end{array}$

Substitute $m$ as $3.648\text{ g}$ and $V$ as $2.50\text{ L}$ in equation (2):

$\begin{array}{c}\rho =\frac{3.648\text{ g}}{2.50\text{ L}}\\ =1.46{\text{ g L}}^{-1}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The gas which has the greatest density is to be identified.

Explanation of Solution

The molar mass of ${\text{CO}}_2$ gas is as follows:

$\begin{array}{c}{\text{CO}}_2=12{\text{ g mol}}^{-1}+2\times 16{\text{ g mol}}^{-1}\\ =44{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $0.114\text{ mol}$ and $MM$ as $44{\text{ g mol}}^{-1}$ in equation (3):

$\begin{array}{c}0.114\text{ mol}=\frac{m}{44{\text{ g mol}}^{-1}}\\ m=5.016\text{ g}\end{array}$

Substitute $m$ as $5.016\text{ g}$ and $V$ as $2.50\text{ L}$ in equation (2):

$\begin{array}{c}\rho =\frac{5.016\text{ g}}{2.50\text{ L}}\\ =2.0064{\text{ g L}}^{-1}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The gas which has the greatest density is to be identified.

Explanation of Solution

The molar mass of $\text{He}$ gas is as follows:

$\text{He}=4{\text{ g mol}}^{-1}$

Substitute $n$ as $0.114\text{ mol}$ and $MM$ as $4{\text{ g mol}}^{-1}$ in equation (3):

$\begin{array}{c}0.114\text{ mol}=\frac{m}{4{\text{ g mol}}^{-1}}\\ m=0.456\text{ g}\end{array}$

Substitute $m$ as $0.456\text{ g}$ and $V$ as $2.50\text{ L}$ in equation (2):

$\begin{array}{c}\rho =\frac{0.456\text{ g}}{2.50\text{ L}}\\ =0.227{\text{ g L}}^{-1}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The gas which has the greatest density is to be identified.

Explanation of Solution

The molar mass of ${\text{SO}}_2$ gas is as follows:

$\begin{array}{c}{\text{SO}}_2=32{\text{ g mol}}^{-1}+2\times 16{\text{ g mol}}^{-1}\\ =64{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $0.114\text{ mol}$ and $MM$ as $64{\text{ g mol}}^{-1}$ in equation (3):

$\begin{array}{c}0.114\text{ mol}=\frac{m}{64{\text{ g mol}}^{-1}}\\ m=7.296\text{ g}\end{array}$

Substitute $m$ as $7.296\text{ g}$ and $V$ as $2.50\text{ L}$ in equation (2):

$\begin{array}{c}\rho =\frac{7.296\text{ g}}{2.50\text{ L}}\\ =2.918{\text{ g L}}^{-1}\end{array}$

Therefore, sulfur dioxide is the gas having the highest density.

The arrangement in the increasing order of the densities of gases are:

${\text{SO}}_2>{\text{CO}}_2>{\text{O}}_2>{\text{N}}_2>\text{He}$ .