Chapter 5, Problem 5.131QP

(a)

Interpretation Introduction

Interpretation:

The rate of CO production and it take to build up a lethal concentration of CO have to be calculated.

Concept introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

  $\text{PV = nRT}$

Answer to Problem 5.131QP

The rate of $\text{CO}$ production is $\text{0}\text{.112}\text{mol}\text{CO/min}$

Explanation of Solution

To calculate the rate of CO production

  $\frac{\text{188}\text{g}\text{CO}}{\text{1}\text{h}}\text{×}\frac{\text{1}\text{mol}\text{CO}}{\text{28}\text{.01}\text{g}\text{CO}}\text{×}\frac{\text{1}\text{h}}{\text{60}\text{min}}\text{=}\text{0}\text{.112}\text{mol}\text{CO/min}$

The rate of $\text{CO}$ production is calculated by plugging in the values of the given weight of the $\text{CO}$, molecular weight of the $\text{CO}$ and time.  The rate of $\text{CO}$ production was found to be $\text{0}\text{.112}\text{mol}\text{CO/min}$.

(b)

Interpretation Introduction

Interpretation:

The rate of CO production and it take to build up a lethal concentration of CO have to be calculated.

Concept introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

  $\text{PV = nRT}$

Answer to Problem 5.131QP

The rate of $\text{CO}$ production per minute time is $\text{2}\text{.0}\text{×}{\text{10}}^{\text{1}}\text{min}$

Explanation of Solution

$\text{1000 ppm}$ means that there are $\text{1000 particles of gas per 1,000,000}$ particles of air. The pressure of a gas is directly proportional to the number of particles of gas. We can calculate the partial pressure of $\text{CO}$ in atmospheres, assuming that atmospheric pressure is 1 atm.

  $\frac{\text{1000}\text{particles}}{\text{1,000,000}\text{particles}}\text{×}\text{1}\text{atm}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-4}\text{3}}\text{atm}$

A partial pressure of $\text{1}{\text{.0×10}}^{\text{-3 }}\text{atm CO}$ is lethal.

The volume of the garage (in L) is:

$\text{(6}\text{.0m}\text{×}\text{4}\text{.0}\text{m}\text{×}\text{2}\text{.2}\text{m)×}\left(\frac{\text{1}\text{cm}}{\text{0}\text{.01}\text{m}}\right)\text{×}\frac{\text{1}\text{L}}{\text{1000}{\text{cm}}^{\text{3}}}\text{=}\text{5}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{L}$

From part (a), we know the rate of $\text{CO}$ production per minute. In one minute the partial pressure of $\text{CO}$ will be:

$\begin{array}{l}\text{PV = nRT}\\ \text{}{\text{P}}_{\text{CO}}\text{=}\frac{\text{nRT}}{\text{V}}\text{=}\frac{\text{(0}\text{.112}\text{mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(20+273)K}}{\text{5}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{L}}\text{=2}\text{.0}\text{×}{\text{10}}^{\text{1}}\text{min}\end{array}$

The rate of $\text{CO}$ production per minute time is calculated by plugging in the values of the given moles, volume and temperature.  The rate of $\text{CO}$ production per minute time was found to be $\text{2}\text{.0}\text{×}{\text{10}}^{\text{1}}\text{min}$

How many minutes will it take for the partial pressure of CO to reach the lethal level, $\text{1}\text{.0×10-3 atm?}$

$\text{?}\text{min}\text{=}\text{(1}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{atm}\text{CO)}\text{×}\frac{\text{1}\text{min}}{\text{5}\text{.1×}{\text{10}}^{\text{-5}}\text{atm}\text{CO}}\text{=}\text{2}\text{.0}\text{×}\text{10}{}^{\text{1}}\text{min}$

It takes for the partial pressure of $\text{CO}$ to reach the lethal level is calculated by plugging in the values of the given pressure of $\text{CO}$.  It takes for the partial pressure of $\text{CO}$ to reach the lethal levelwas found to be $\text{2}\text{.0}\text{×}{\text{10}}^{\text{1}}\text{min}$