Chapter 20, Problem 20.65QP

A concentration of 8.00 × 10² ppm by volume of CO is considered lethal to humans. Calculate the minimum mass of CO in grams that would become a lethal concentration in a closed room 17.6 m long, 8.80 m wide, and 2.64 m high. The temperature and pressure are 20.0°C and 756 mmHg, respectively.

Interpretation Introduction

Interpretation:

The minimum mass of carbon monoxide would become poisonous to humans in a closed room has to be calculated.

Concept Introduction:

Partial pressure:

The product of total pressure of gases and number of mole of individual gas is the partial pressure of a gas.

$\begin{array}{l}{\text{P}}_{\text{i}}{\text{=P}}_{\text{total}}{\text{X}}_{\text{i}}\\ \\ {\text{X}}_{\text{i}}\text{is}\text{mole}\text{fraction}\end{array}$

${\text{P}}_{\text{i}}$ is individual pressure of a gas

${\text{P}}_{\text{Total}}$ is total pressure of all gases.

Mole fraction:

Mole fraction is the ratio number of moles of a species to that of total number of moles of solution.

$\text{X}\text{=}\frac{\text{Moles}\text{of}\text{a}\text{species}}{\text{Total}\text{moles}\text{of}\text{solution}}$

Ideal gas equation:

The ideal gas equation is given as

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

Where,

P is the pressure

V is the volume

n is number of moles

R is the gas constant

T is the temperature

The volume of the closed room is calculated as,

$\text{Volume}\text{=}\text{Length}\text{×}\text{Wide}\text{×}\text{height}$

Answer to Problem 20.65QP

The minimum mass of carbon monoxide would become poisonous to humans in a closed room is $378gCO$

Explanation of Solution

Given,

Temperature $\text{=}\text{20}\text{.0}{}^{\text{o}}\text{C}$

Total pressure $\text{=}\text{756mmHg}$

Length of the closed room $\text{=}\text{17}\text{.6m}$

Wide of the closed room $\text{=}\text{8}\text{.80m}$

Height of the closed room $\text{=}\text{2}\text{.64m}$

$\text{8}\text{.00}\text{×}{\text{10}}^{\text{2}}\text{ppm}$ of carbon monoxide is poisonous to humans.

The volume of the closed room is calculated as,

$\begin{array}{l}\text{Volume}\text{=}\text{Length}\text{×}\text{Wide}\text{×}\text{height}\\ \\ \text{=}\text{17}\text{.6m}\text{×}\text{8}\text{.80 m}\text{×}\text{2}\text{.64 m}\\ \\ \text{=}\text{4}\text{.09}\text{×}{\text{10}}^{\text{2}}{\text{m}}^{\text{3}}\\ \\ {\text{1m}}^{\text{3}}\text{=}\text{1}\text{×}{\text{10}}^{\text{3}}\text{L}\\ \\ \text{Volume}\text{=}\text{4}\text{.09}\text{×}{\text{10}}^{\text{5}}\text{L}\end{array}$

The mole fraction of $\text{8}\text{.00}\text{×}{\text{10}}^{\text{2}}\text{ppm}$ by volume of carbon monoxide is calculated as,

$\frac{8.00\times 10^2}{1\times 10^6}=8.00\times 10^{-4}$

From the total pressure and mole fraction of carbon monoxide, the partial pressure is calculated as,

 $\begin{array}{l}{P}_{CO}=X_{{CO}_{}}{P}_{Total}\\ \\ =(8.00\times 10^{-4})(756mmHg)\times \frac{1atm}{760mmHg}\\ \\ =7.96\times 10^{-4}atm\end{array}$

The number of moles of carbon monoxide is calculated as,

$\begin{array}{l}{n}_{CO}=\frac{PV}{RT}\\ \\ =\frac{(7.96\times 10^{-4}atm)(4.09\times 10^{5}L)}{(0.0821L.atm/K.mol)(293K)}\\ \\ =13.5mol\end{array}$

The mass of carbon monoxide is calculated using molar mass of carbon monoxide as,

$\begin{array}{l}Mass=13.5mol\times \frac{28.01gCO}{1molCO}\\ \\ =378gCO\end{array}$