Chapter 14, Problem 41E

Interpretation Introduction

(a)

Interpretation:

In the reaction, $\text{S}+{\text{O}}_{\text{2}}\to {\text{SO}}_{\text{2}}$, the volume of ${\text{O}}_{\text{2}}$ needed to form $\text{35}\text{.2}\text{L}$ of ${\text{SO}}_{\text{2}}$, when both gases are measured at $\text{741}\text{torr}$ and ${\text{26}}^{\text{o}}\text{C}$, is to be calculated.

Concept introduction:

Ideal gas is defined as the gas in which the collisions between the molecules and the atoms are perfectly elastic and there are no intermolecular attractive forces found between them. The ideal gas equation is given by the expression as shown below

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

Answer to Problem 41E

The volume of ${\text{O}}_{\text{2}}$ needed to form $\text{35}\text{.2}\text{L}$ of ${\text{SO}}_{\text{2}}$, when both the gases are measured at $\text{741}\text{torr}$ and ${\text{26}}^{\text{o}}\text{C}$, is $\text{35}\text{.2}\text{L}$.

Explanation of Solution

In the reaction, $\text{S}+{\text{O}}_{\text{2}}\to {\text{SO}}_{\text{2}}$, the pressure of both the gases is $741\text{torr}$ and the temperature of both the gases is ${\text{26}}^{\text{o}}\text{C}$.

The mole ratio of the reactants and products is $\left(1:1:1\right)$ and at same temperature and pressure the mole ratio for both the sides is same as the volume ratio. Therefore, the volume ratio of ${\text{O}}_2$ and ${\text{SO}}_{\text{2}}$ can be expressed as shown below.

$\frac{1\text{L}{\text{O}}_{\text{2}}}{\text{1}\text{L}{\text{SO}}_{\text{2}}}$

The volume of ${\text{SO}}_{\text{2}}$ produced by $1\text{L}{\text{O}}_{\text{2}}$ is $1\text{L}$.

Therefore, the volume of ${\text{O}}_{\text{2}}\left({\text{V}}_{{\text{O}}_{\text{2}}}\right)$ which will produce $\text{35}\text{.2}\text{L}{\text{SO}}_{\text{2}}$ is calculated as shown below.

$\begin{array}{rll}{\text{V}}_{{\text{O}}_{\text{2}}}& =& \frac{\text{1}\text{L}{\text{O}}_{\text{2}}}{\text{1}\text{L}{\text{SO}}_{\text{2}}}\times 35.2\text{L}{\text{SO}}_{\text{2}}\\ & =& 35.2\text{L}{\text{O}}_{\text{2}}\end{array}$

Therefore, the volume of ${\text{O}}_{\text{2}}$ needed is $\text{35}\text{.2}\text{L}$.

Conclusion

The volume of ${\text{O}}_{\text{2}}$ needed to form $\text{35}\text{.2}\text{L}$ of ${\text{SO}}_{\text{2}}$, when both the gases are measured at $\text{741}\text{torr}$ and $\text{26}{}^{\text{o}}\text{C}$, is $\text{35}\text{.2}\text{L}$.

Interpretation Introduction

(b)

Interpretation:

In the reaction, $\text{S}+{\text{O}}_{\text{2}}\to {\text{SO}}_{\text{2}}$, the volume of ${\text{O}}_{\text{2}}$ needed to form $\text{35}\text{.2}\text{L}$ of ${\text{SO}}_{\text{2}}$, when only the ${\text{SO}}_{\text{2}}$ is at $\text{741}\text{torr}$ and ${\text{26}}^{\text{o}}\text{C}$, but the ${\text{O}}_{\text{2}}$ is at ${\text{17}}^{\text{o}}\text{C}$ and $\text{847}\text{torr}$, is to be calculated.

Concept introduction:

Ideal gas is defined as the gas in which the collisions between the molecules and the atoms are perfectly elastic and there are no intermolecular attractive forces found between them. The ideal gas equation is given by the expression as shown below

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

Answer to Problem 41E

The volume of ${\text{O}}_{\text{2}}$ needed to form $\text{35}\text{.2}\text{L}$ of ${\text{SO}}_{\text{2}}$, when only the ${\text{SO}}_{\text{2}}$ is at $\text{741}\text{torr}$ and $\text{26}{}^{\text{o}}\text{C}$, but the ${\text{O}}_{\text{2}}$ is at $\text{17}{}^{\text{o}}\text{C}$ and $\text{847}\text{torr}$, is $\text{29}\text{.2}\text{L}$.

Explanation of Solution

In the reaction, $\text{S}+{\text{O}}_{\text{2}}\to {\text{SO}}_{\text{2}}$, the initial pressure and temperature of ${\text{SO}}_{\text{2}}$ are $\text{741}\text{torr}$ and $\text{26}{}^{\text{o}}\text{C}$ respectively and the initial volume of ${\text{SO}}_{\text{2}}$ is $\text{35}\text{.2}\text{L}$. The final temperature and pressure of ${\text{SO}}_{\text{2}}$ are $\text{17}{}^{\text{o}}\text{C}$ and $\text{847}\text{torr}$ respectively.

Conversion of temperature from Celsius to Kelvin can be done as shown below.

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

Therefore, $\text{26}{}^{\text{o}}\text{C}$ and $\text{17}{}^{\text{o}}\text{C}$ can be calculated to Kelvin as shown below.

$\begin{array}{rll}{\text{T}}_{\text{final}}\left(\text{K}\right)& =& \left(\text{17}+273\right)\text{K}\\ & =& \text{290}\text{K}\end{array}$

$\begin{array}{rll}{\text{T}}_{\text{initial}}\left(\text{K}\right)& =& \left(\text{26}+273\right)\text{K}\\ & =& \text{299}\text{K}\end{array}$

The initial and final temperature of ${\text{SO}}_{\text{2}}$ are $\text{299}\text{K}$ and $\text{290}\text{K}$ respectively.

The relation between the initial and final pressure, volume and temperature of gas is shown below.

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}=\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$…(1)

Where,

• ${\text{P}}_{\text{1}}$ is the initial pressure.

• ${\text{V}}_{\text{1}}$ is the initial volume.

• ${\text{T}}_{\text{1}}$ is the initial temperature.

• ${\text{P}}_{\text{2}}$ is the final pressure.

• ${\text{V}}_{\text{2}}$ is the final volume.

• ${\text{T}}_{\text{2}}$ is the final temperature.

Substitute the values of initial and final temperature, pressure and volume into the equation (1).

$\begin{array}{rll}\frac{\text{741}\text{torr}\times \text{35}\text{.2}\text{L}}{\text{299}\text{K}}& =& \frac{\text{874}\text{torr}\times {\text{V}}_{\text{2}}}{\text{290}\text{K}}\\ {\text{V}}_{\text{2}}& =& \frac{\text{741}\text{torr}\times \text{35}\text{.2}\text{L}\times \text{290}\text{K}}{\text{299}\text{K}\times \text{874}\text{torr}}\\ & =& 29.9\text{L}\end{array}$

Therefore, the final volume of ${\text{SO}}_{\text{2}}$ gas is $29.9\text{L}$.

The volume of ${\text{O}}_{\text{2}}\left({\text{V}}_{{\text{O}}_{\text{2}}}\right)$ which will produce $29.9\text{L}{\text{SO}}_{\text{2}}$ is calculated as shown below.

$\begin{array}{rll}{\text{V}}_{{\text{O}}_{\text{2}}}& =& \frac{\text{1}\text{L}{\text{O}}_{\text{2}}}{\text{1}\text{L}{\text{SO}}_{\text{2}}}\times 29.9\text{L}{\text{SO}}_{\text{2}}\\ & =& 29.9\text{L}{\text{O}}_{\text{2}}\end{array}$

Therefore, the volume of oxygen gas needed is $29.9\text{L}$.

Conclusion

The volume of ${\text{O}}_{\text{2}}$ needed to form $\text{35}\text{.2}\text{L}$ of ${\text{SO}}_{\text{2}}$, when only the ${\text{SO}}_{\text{2}}$ is at $\text{741}\text{torr}$ and $\text{26}{}^{\text{o}}\text{C}$, but the ${\text{O}}_{\text{2}}$ is at $\text{17}{}^{\text{o}}\text{C}$ and $\text{847}\text{torr}$, is $29.9\text{L}$.