Chapter 9, Problem 86QP

(a)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and mass of $\text{Xe}$ 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 change in accordance with the units of pressure, volume and temperature.

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

$°\text{C}=273.15+\text{K}$

For $54°\text{C}$ ,

$\begin{array}{c}°\text{C}=54°\text{C}+\text{273}\text{.15}\\ =327.15\text{ K}\end{array}$

The conversion factor that is used to convert pressure value from $\text{torr}$ to $\text{atm}$ is as follows:

$\begin{array}{c}760\text{ torr}=1\text{ atm}\\ \text{1}=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

For $790\text{ torr}$ ,

$\begin{array}{c}\text{790 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{790 torr}\\ =1.04\text{ atm}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $0.15\text{ L}$ for $V$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $327.15\text{ K}$ for $T$ and $1.04\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}1.04\text{ atm}\times 0.15\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 327.15\text{ K}\\ n=5.8\times {10}^{-3}\text{ mol}\end{array}$

The mass of gas is calculated using the given formula:

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

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

Substitute $5.8\times {10}^{-3}\text{ mol}$ for $n$ and $131{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}5.8\times {10}^{-3}\text{ mol}=\frac{mass}{131{\text{ g mol}}^{-1}}\\ mass=0.76\text{ g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and mass of $\text{CO}$ gas.

Explanation of Solution

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

$°\text{C}=273.15+\text{K}$

For $54°\text{C}$ ,

$\begin{array}{c}°\text{C}=54°\text{C}+\text{273}\text{.15}\\ =327.15\text{ K}\end{array}$

The conversion factor that is used to convert pressure value from $\text{torr}$ to $\text{atm}$ is as follows:

$\begin{array}{c}760\text{ torr}=1\text{ atm}\\ \text{1}=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

For $790\text{ torr}$ ,

$\begin{array}{c}\text{790 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{790 torr}\\ =1.04\text{ atm}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $38.1\text{ L}$ for $V$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $327.15\text{ K}$ for $T$ and $1.04\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}1.04\text{ atm}\times 38.1\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 327.15\text{ K}\\ n=1.5\text{ mol}\end{array}$

The mass of gas is calculated using the given formula:

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

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

Substitute $1.5\text{ mol}$ for $n$ and $28{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}1.5\text{ mol}=\frac{mass}{28{\text{ g mol}}^{-1}}\\ mass=42\text{ g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and mass 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 $54°\text{C}$ ,

$\begin{array}{c}°\text{C}=54°\text{C}+\text{273}\text{.15}\\ =327.15\text{ K}\end{array}$

The conversion factor that is used to convert pressure value from $\text{torr}$ to $\text{atm}$ is as follows:

$\begin{array}{c}760\text{ torr}=1\text{ atm}\\ \text{1}=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

For $790\text{ torr}$ ,

$\begin{array}{c}\text{790 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{790 torr}\\ =1.04\text{ atm}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $2.5\text{ L}$ for $V$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $327.15\text{ K}$ for $T$ and $1.04\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}1.04\text{ atm}\times 2.5\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 327.15\text{ K}\\ n=0.097\text{ mol}\end{array}$

The mass of gas is calculated using the given formula:

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

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

Substitute $0.097\text{ mol}$ for $n$ and $32{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}0.097\text{ mol}=\frac{mass}{{\text{32 g mol}}^{-1}}\\ mass=3.1\text{ g}\end{array}$