Chapter 5, Problem 5.124P

Interpretation Introduction

Interpretation:

The number of metric tons of each gas are emitted per year is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.124P

The number of metric tons of ${\text{CO}}_{\text{2}}$ emitted per year is $4.83\times {10}^2\text{ t/year}$.

The number of metric tons of $\text{CO}$ emitted per year is $9.61\text{ t/year}$.

The number of metric tons of ${\text{H}}_{\text{2}}\text{O}$ emitted per year is $1.5\times {10}^2\text{ t/year}$.

The number of metric tons of ${\text{SO}}_{\text{2}}$ emitted per year is $1.70\times {10}^2\text{ t/year}$.

The number of metric tons of ${\text{S}}_{\text{2}}$ emitted per year is $4\times {10}^{-1}\text{ t/year}$.

The number of metric tons of ${\text{H}}_{\text{2}}$ emitted per year is $2.1\times {10}^{-1}\text{ t/year}$.

The number of metric tons of $\text{HCl}$ emitted per year is $6\times {10}^{-1}\text{ t/year}$.

The number of metric tons of ${\text{H}}_{\text{2}}\text{S}$ emitted per year is $2\times {10}^{-1}\text{ t/year}$.

Explanation of Solution

The expression to calculate the $\text{moles of gas}/\text{day}$ is as follows,

$PV=nRT$        (1)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole ofgas and $R$ is the gas constant.

Rearrange the equation (1) to calculate $n$ as follows,

$n=\frac{PV}{RT}$        (2)

Substitute the value $\text{1}\text{.0 atm}$ for $P$, $298\text{ K}$ for $T$, $\text{1}\text{.5}\times {\text{10}}^{\text{3}}{\text{ m}}^{\text{3}}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (2).

$\begin{array}{c}n=\frac{\left(\text{1}\text{.0 atm}\right)\left(\text{1}\text{.5}\times {\text{10}}^{\text{3}}{\text{ m}}^{\text{3}}\right)\left(\frac{1000\text{ L}}{{\text{1 m}}^{\text{3}}}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{298 K}\right)}\\ =6.13101\times {10}^5\text{ mol/day}\end{array}$

The expression to calculate the $\text{moles of gas/year}$ is as follows,

$\text{Moles of gas/year}=\left[\begin{array}{l}\left(\text{Moles of gas/day}\right)\\ \left(\text{Total number of days in year}\right)\end{array}\right]$        (3)

Substitute the value $6.13101\times {10}^5\text{ mol/day}$ the $\text{moles of gas}/\text{year}$ and $\text{365}\text{.25 day/year}$ for total number of days in year in the equation (3).

$\begin{array}{c}\text{Moles of gas/year}=\left[\left(6.13101\times {10}^5\text{ mol/day}\right)\left(\text{365}\text{.25 day/year}\right)\right]\\ =2.23935\times {10}^7\text{mol}/\text{year}\end{array}$

The expression to calculate the mass of ${\text{CO}}_{\text{2}}$ is as follows:

${\text{Mass of CO}}_{\text{2}}=\left[\begin{array}{l}\left({\text{Mole fraction of CO}}_{\text{2}}\right)\left(\text{Moles of gas/year}\right)\\ \left({\text{Molar mass of CO}}_{\text{2}}\right)\end{array}\right]$        (4)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.4896$ for mole fraction of ${\text{CO}}_{\text{2}}$ and $44.01\text{ g/mol}$ for molar mass of ${\text{CO}}_{\text{2}}$ in the equation (4)

$\begin{array}{c}{\text{Mass of CO}}_{\text{2}}=\left(0.4896\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(44.01\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =482.519\text{ t/year}\end{array}$

The expression to calculate the mass of $\text{CO}$ is as follows:

$\text{Mass of CO}=\left[\begin{array}{l}\left(\text{Mole fraction of CO}\right)\left(\text{Moles of gas/year}\right)\\ \left(\text{Molar mass of CO}\right)\end{array}\right]$        (5)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.0146$ for mole fraction of $\text{CO}$ and $28.01\text{ g/mol}$ for molar mass of $\text{CO}$ in the equation (5)

$\begin{array}{c}\text{Mass of CO}=\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(28.01\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =9.15773\text{ t/year}\\ \approx 9.16\text{ t/year}\end{array}$

The expression to calculate the mass of ${\text{H}}_{\text{2}}\text{O}$ is as follows:

${\text{Mass of H}}_{\text{2}}\text{O}=\left[\begin{array}{l}\left(\text{Moles of gas/year}\right)\\ \left({\text{mole fraction of H}}_{\text{2}}\text{O}\right)\left({\text{Molar mass of H}}_{\text{2}}\text{O}\right)\end{array}\right]$        (6)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.3710$ for mole fraction of ${\text{H}}_{\text{2}}\text{O}$ and $18.02\text{ g/mol}$ for molar mass of ${\text{H}}_{\text{2}}\text{O}$ in the equation (6)

$\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{O}=\left(0.3710\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(18.01\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =149.70995\text{ t/year}\\ \approx 1.5\times {10}^2\text{ t/year}\end{array}$

The expression to calculate the mass of ${\text{SO}}_{\text{2}}$ is as follows:

${\text{Mass of SO}}_{\text{2}}=\left[\begin{array}{l}\left(\text{Moles of gas/year}\right)\\ \left({\text{mole fraction of SO}}_{\text{2}}\right)\left({\text{Molar mass of H}}_{\text{2}}\text{O}\right)\end{array}\right]$        (7)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.1185$ for mole fraction of ${\text{SO}}_{\text{2}}$ and $64.06\text{ g/mol}$ for molar mass of ${\text{SO}}_{\text{2}}$ in the equation (7)

$\begin{array}{c}{\text{Mass of SO}}_{\text{2}}=\left(0.1185\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(64.06\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =169.992\text{ t/year}\\ \approx 1.7\times {10}^2\text{ t/year}\end{array}$

The expression to calculate the mass of ${\text{S}}_{\text{2}}$ is as follows:

${\text{Mass of S}}_{\text{2}}=\left[\begin{array}{l}\left(\text{Moles of gas/year}\right)\\ \left({\text{mole fraction of S}}_{\text{2}}\right)\left({\text{Molar mass of S}}_{\text{2}}\right)\end{array}\right]$        (8)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.0003$ for mole fraction of ${\text{S}}_{\text{2}}$ and $64.12\text{ g/mol}$ for molar mass of ${\text{S}}_{\text{2}}$ in the equation (8)

$\begin{array}{c}{\text{Mass of S}}_{\text{2}}=\left(0.0003\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(64.12\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =0.4307614\text{ t/year}\\ \approx 4.3\times {10}^{-1}\text{ t/year}\end{array}$

The expression to calculate the mass of ${\text{H}}_{\text{2}}$ is as follows:

${\text{Mass of H}}_{\text{2}}=\left[\begin{array}{l}\left(\text{Moles of gas/year}\right)\\ \left({\text{mole fraction of H}}_{\text{2}}\right)\left({\text{Molar mass of H}}_{\text{2}}\right)\end{array}\right]$        (9)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.0047$ for mole fraction of ${\text{H}}_{\text{2}}$ and $2.016\text{ g/mol}$ for molar mass of ${\text{H}}_{\text{2}}$ in the equation (9)

$\begin{array}{c}{\text{Mass of H}}_{\text{2}}=\left(0.0047\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(2.016\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =0.21218\text{ t/year}\\ \approx 2.1\times {10}^{-1}\text{ t/year}\end{array}$

The expression to calculate the mass of $\text{HCl}$ is as follows:

$\text{Mass of HCl}=\left[\begin{array}{l}\left(\text{Moles of gas/year}\right)\\ \left(\text{mole fraction of HCl}\right)\left(\text{Molar mass of HCl}\right)\end{array}\right]$        (10)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.0008$ for mole fraction of $\text{HCl}$ and $36.46\text{ g/mol}$ for molar mass of $\text{HCl}$ in the equation (10)

$\begin{array}{c}\text{Mass of HCl}=\left(0.0008\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(36.46\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =0.6531736\text{ t/year}\\ \approx 6\times {10}^{-1}\text{ t/year}\end{array}$

The expression to calculate the mass of ${\text{H}}_{\text{2}}\text{S}$ is as follows:

${\text{Mass of H}}_{\text{2}}\text{S}=\left[\begin{array}{l}\left(\text{Moles of gas/year}\right)\\ \left({\text{mole fraction of H}}_{\text{2}}\text{S}\right)\left({\text{Molar mass of H}}_{\text{2}}\text{S}\right)\end{array}\right]$        (11)

Substitute the value $2.23935\times {10}^7\text{mol}/\text{year}$ for moles of $\text{gas}/\text{year}$, $0.0003$ for mole fraction of ${\text{H}}_{\text{2}}\text{S}$ and $34.08\text{ g/mol}$ for molar mass of ${\text{H}}_{\text{2}}\text{S}$ in the equation (11)

$\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{S}=\left(0.0003\right)\left(2.23935\times {10}^7\text{mol}/\text{year}\right)\left(34.08\text{ g/mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{1\text{ t}}{1000\text{ kg}}\right)\\ =0.228951\text{ t/year}\\ \approx 2\times {10}^{-1}\text{ t/year}\end{array}$

Conclusion

The number of metric tons of ${\text{CO}}_{\text{2}}$ emitted per year is $4.83\times {10}^2\text{ t/year}$.

The number of metric tons of $\text{CO}$ emitted per year is $9.61\text{ t/year}$.

The number of metric tons of ${\text{H}}_{\text{2}}\text{O}$ emitted per year is $1.5\times {10}^2\text{ t/year}$.

The number of metric tons of ${\text{SO}}_{\text{2}}$ emitted per year is $1.70\times {10}^2\text{ t/year}$.

The number of metric tons of ${\text{S}}_{\text{2}}$ emitted per year is $4\times {10}^{-1}\text{ t/year}$.

The number of metric tons of ${\text{H}}_{\text{2}}$ emitted per year is $2.1\times {10}^{-1}\text{ t/year}$.

The number of metric tons of $\text{HCl}$ emitted per year is $6\times {10}^{-1}\text{ t/year}$.

The number of metric tons of ${\text{H}}_{\text{2}}\text{S}$ emitted per year is $2\times {10}^{-1}\text{ t/year}$.