Chapter 9, Problem 85QP

(a)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and mass of ${\text{CH}}_4$ 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 conversion factor of temperature from Celsius to Kelvin is as follows:

$°\text{C}=273.15+\text{K}$

For $87.5°\text{C}$ ,

$\begin{array}{c}°\text{C}=87.5°\text{C}+\text{273}\text{.15}\\ =360.7\text{ K}\end{array}$

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 $722\text{ torr}$ ,

$\begin{array}{c}\text{722 torr}=\frac{1}{760}\text{ atm}\times \text{722}\\ =0.95\text{ atm}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $7.62\text{ L}$ for $V$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $360.7\text{ K}$ for $T$ and $0.95\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}0.95\text{ atm}\times 7.62\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 360.7\text{ K}\\ n=0.245\text{ mol}\end{array}$

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 $0.245\text{ mol}$ for $n$ and $16{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}0.245\text{ mol}=\frac{mass}{16{\text{ g mol}}^{-1}}\\ mass=3.92\text{ g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and mass of ${\text{H}}_2$ gas.

Explanation of Solution

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

$°\text{C}=273.15+\text{K}$

For $87.5°\text{C}$ ,

$\begin{array}{c}°\text{C}=87.5°\text{C}+\text{273}\text{.15}\\ =360.7\text{ K}\end{array}$

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 $722\text{ torr}$ ,

$\begin{array}{c}\text{722 torr}=\frac{1}{760}\text{ atm}\times \text{722}\\ =0.95\text{ atm}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $0.135\text{ L}$ for $V$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $360.7\text{ K}$ for $T$ and $0.95\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}0.95\text{ atm}\times 0.135\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 360.7\text{ K}\\ n=\text{4}\text{.33}\times {\text{10}}^{-3}\text{ mol}\end{array}$

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 $4.33\times {10}^{-3}\text{ mol}$ for $n$ and $2{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}4.33\times {10}^{-3}\text{ mol}=\frac{mass}{{\text{ 2 g mol}}^{-1}}\\ mass=8.74\times {10}^{-3}\text{ g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The calculation of the number of moles and mass of ${\text{N}}_2$ gas.

Explanation of Solution

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

$°\text{C}=273.15+\text{K}$

For $87.5°\text{C}$ ,

$\begin{array}{c}°\text{C}=87.5°\text{C}+\text{273}\text{.15}\\ =360.7\text{ K}\end{array}$

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 $722\text{ torr}$ ,

$\begin{array}{c}\text{722 torr}=\frac{1}{760}\text{ atm}\times \text{722}\\ =0.95\text{ atm}\end{array}$

Recall equation (1),

$PV=nRT\text{.}$

Substitute $7.62\text{ L}$ for $V$ , $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ for $\text{R}$ , $360.7\text{ K}$ for $T$ and $0.95\text{ atm}$ for $P$ in the above equation:

$\begin{array}{c}0.95\text{ atm}\times 8.96\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\times 360.7\text{ K}\\ n=0.288\text{ mol}\end{array}$

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 $0.288\text{ mol}$ for $n$ and $28{\text{ g mol}}^{-1}$ for $M$ in the above equation.

$\begin{array}{c}0.288\text{ mol}=\frac{mass}{28{\text{ g mol}}^{-1}}\\ mass=8.06\text{ g}\end{array}$