Chapter 10, Problem 141AP

Interpretation Introduction

Interpretation:

The rate of $\mathrm{CO}$ production and the time taken to build up a concentration of $1000\text{ ppmv}$ of $\mathrm{CO}$ are to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown as

$PV=nRT$

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

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Answer to Problem 141AP

Solution:

(a) $0.112\text{ mol}/\text{min}$

(b) $\text{20 minutes}$

Explanation of Solution

a)The rate of CO production in mol/min

The molar mass of carbon monoxide is $28\text{ g}/\text{mol}$.

The rate of production of mass $m_{\mathrm{CO}}=188\text{ g}/\text{hr}$.

Calculate the number of moles of carbon monoxide as follows:

$n_{\mathrm{CO}}=\frac{m_{\mathrm{CO}}}{M_{\mathrm{CO}}}$

Substitute $188\text{ g}/\text{hr}$ for $m_{\mathrm{CO}}$ and $28\text{ g}/\text{mol}$ for $M_{\mathrm{CO}}$ in the above equation as follows:

$\begin{array}{c}{n}_{\mathrm{CO}}=\frac{188\cancel{\text{g}}/\text{hr}}{28\cancel{\text{g}}/\text{mol}}\\ =6.71\text{ mol}/\text{hr}\end{array}$ $\begin{array}{c}{n}_{\mathrm{CO}}=\frac{188\text{ g}/\text{hr}}{28\text{ g}/\text{mol}}\\ =6.71\text{ mol}/\text{hr}\end{array}$

In 1hour, there are 60minutes.

Convert hour to minute as follows:

$\begin{array}{c}6.71\text{ mol}/\text{hr}=\left(6.71\frac{\text{mol}}{\cancel{\text{hr}}}\right)\left(\frac{\text{1 }\cancel{\text{hr}}}{60\text{ min}}\right)\\ =0.112\text{ mol}/\text{min}\end{array}$

Hence, the number of moles of $\mathrm{CO}$ produced is $0.112\text{ mol}/\text{min}$.

b)Time to build up a lethal concentration of CO of $1000\text{ ppmv}$.

Length $l=6\text{ m}$

Breadth $b=4\text{ m}$

Height $h=2.2\text{ m}$

Temperature $T=20°\text{C}$

The concentration is $1000\text{ ppmv}$, which means $1000$ particles of the gas for air particles of count $1,000,000$. The atmospheric pressure is $1\text{ atm}$. So, the partial pressure of $\mathrm{CO}$ is

$\begin{array}{c}{P}_{\text{CO}}=\frac{1000}{1,000,000}\times 1\text{ atm}\\ =0.001\text{ atm}\\ =1\times {10}^{-3}\text{ atm}\end{array}$

The volume of the garage is

$V=\left(l\right)\left(b\right)\left(h\right)$

Substitute $6\text{ m}$ for $l$, $4\text{ m}$ for $b$, and $2.2\text{ m}$ for $h$ in the above equation

$\begin{array}{c}V=6\text{ m}\times 4\text{ m}\times 2.2\text{ m}\\ =52.8{\text{ m}}^3\end{array}$

In 1cubic meter, there are 1000liters.

$\begin{array}{c}52.8{\text{ m}}^3=\left(52.8\cancel{{\text{m}}^3}\right)\left(\frac{1000\text{ L}}{1\cancel{{\text{m}}^3}}\right)\\ =5.3\times {10}^4\text{L}\end{array}$

The temperature is $20°\text{C}$.

The conversion of temperature from degree Celsius to Kelvin is done by using the formula given below:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(20+273.15\right)\\ =293.15\text{ K}\end{array}$

The equation for anideal gas is as follows:

$PV=n_{{\text{CO}}_2}RT$

Rearrange the above equation for pressure of carbon monoxide as follows:

$P=\frac{n_{{\text{CO}}_{}}RT}{V}$

Substitute $5.3\times {10}^4\text{ L}$ for $V$, $293.15\text{ K}$ for $T$, $0.112\text{ mol}$ for $n$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation

$\begin{array}{c}P=\frac{\left(0.112\cancel{\text{mol}}\right)\left(0.08206\cancel{\text{L}}\text{.atm}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(293.15\cancel{\text{K}}\right)}{\left(5.3\times {10}^4\cancel{\text{L}}\right)}\\ =\frac{2.69}{5.3\times {10}^{-4}}\text{ atm}\\ =0.5083\times {10}^{-4}\text{ atm}\\ \approx 5.1\times {10}^{-5}\text{ atm}\end{array}$ $\begin{array}{c}P=\frac{\left(0.112\text{ mol}\right)\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(293.15\text{ K}\right)}{\left(5.3\times {10}^4\text{ L}\right)}\\ =\frac{2.69}{5.3\times {10}^{-4}}\text{ atm}\\ =0.5083\times {10}^{-4}\text{ atm}\\ \approx 5.1\times {10}^{-5}\text{ atm}\end{array}$

Now, the time taken to reach the lethal level is as follows:

$\begin{array}{c}\frac{1\times {10}^{-3}\cancel{\text{atm}}}{5.1\times {10}^{-5}\cancel{\text{atm}}}=0.196\times {10}^2\text{ minutes}\\ \approx \text{20 minutes}\end{array}$ $\begin{array}{c}\frac{1\times {10}^{-3}\text{ atm}}{5.1\times {10}^{-5}\text{ atm}}=0.196\times {10}^2\text{ minutes}\\ \approx \text{20 minutes}\end{array}$

Hence, the time taken to build $1000\text{ ppmv}$ of $\mathrm{CO}$ is $\text{20 minutes}$.