Chapter 9, Problem 83QP

(a)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and volume of ${\text{NH}}_3$ gas.

Explanation of Solution

Ideal gas law gives a relation between pressure $\left(P\right)$ , volume $\left(V\right)$ , temperature $\left(T\right)$ , and the number of moles $\left(n\right)$ of gas. The gas law equation is:

$\begin{array}{l}PV\propto nT\\ PV=nRT\mathrm{......}\left(1\right)\end{array}$

Where, R is the universal gas constant and its values changes in accordance with the units of pressure, volume and temperature.

The given mass of ${\text{NH}}_3$ gas is $5.8\text{ g}$ . Pressure and temperature of ${\text{NH}}_3$ gas are $15\text{ atm}$ and $100°\text{C}$ , respectively. The conversion factor of temperature from Celsius to Kelvin is as follows:

$°\text{C}=273.15+\text{K}$

For $100°\text{C}$ ,

$\begin{array}{c}°\text{C}=100°\text{C}+\text{273}\text{.15}\\ =373.15\text{ K}\end{array}$

The number of moles of gases is calculated using the given formula:

$n=\frac{m}{M}$

Here, $m$ is the mass and $M$ is the molar mass.

Substitute $5.8\text{ g}$ for $m$ and $17{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}n=\frac{5.8\text{ g}}{17{\text{ g mol}}^{-1}}\\ =0.34\text{mol}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $0.34\text{ mol}$ for $n$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $373.15\text{ K}$ for $T$ and $15\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}15\text{ atm}\times V=0.34\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 373.15\text{ K}\\ V=0.69\text{ L}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and volume of ${\text{O}}_2$ gas.

Explanation of Solution

The conversion factor of temperature from Celsius to Kelvin is as follows:

$°\text{C}=273.15+\text{K}$

For $100°\text{C}$ ,

$\begin{array}{c}°\text{C}=100°\text{C}+\text{273}\text{.15}\\ =373.15\text{ K}\end{array}$

The number of moles of gases is calculated using the given formula:

$n=\frac{m}{M}$

Here, $m$ is the mass and $M$ is the molar mass.

Substitute $48\text{ g}$ for $m$ and $32{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}n=\frac{48\text{ g}}{{\text{32 g mol}}^{-1}}\\ =1.5\text{ mol}\end{array}$

Recall equation (1)

$PV=nRT\text{.}$

Substitute $1.5\text{ mol}$ for $n$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $373.15\text{ K}$ for $T$ and $15\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}15\text{ atm}\times V=1.5\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 373.15\text{ K}\\ V=3.1\text{ L}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and volume of $\text{He}$ gas.

Explanation of Solution

The conversion factor of temperature from Celsius to Kelvin is as follows:

$°\text{C}=273.15+\text{K}$

For $100°\text{C}$ ,

$\begin{array}{c}°\text{C}=100°\text{C}+\text{273}\text{.15}\\ =373.15\text{ K}\end{array}$

The number of moles of gases is calculated using the given formula:

$n=\frac{m}{M}$

Here, $m$ is the mass and $M$ is the molar mass.

Substitute $10.8\text{ g}$ for $m$ and $4{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}=\frac{10.8\text{ g}}{{\text{4 g mol}}^{-1}}\\ n=2.7\text{ mol}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $2.7\text{ mol}$ for $n$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $373.15\text{ K}$ for $T$ and $15\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}15\text{ atm}\times V=2.7\text{ mol}\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 373.15\text{ K}\\ V=5.51\text{ L}\end{array}$