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Q3.2-6E

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Found in: Page 100

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# The air in a small room 12 ft by 8 ft by 8 ft is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon monoxide is blown into the room at a rate of ${\mathbf{100}}{\mathbit{f}}{{\mathbit{t}}}^{{\mathbf{3}}}{\mathbf{/}}{\mathbit{m}}{\mathbit{i}}{\mathbit{n}}$. If air in the room flows out through a vent at the same rate, when will the air in the room be 0.01% carbon monoxide?

The amount of carbon monoxide in the air will reach 0.01% after 43.8 min.

See the step by step solution

## Step 1: Analyzing the given statement

It can view the room as a compartment containing air. If we let $x\left(t\right)$ denote the amount of carbon monoxide in the room at a time t, we can determine the concentration of carbon monoxide in the room by dividing $x\left(t\right)$by the volume of the room at a time t. one uses the mathematical model described by the following equation to solve for x(t),

$\frac{dx}{dt}=$Input rate – Output rate ......................(1)

## Step 2: To determine the volume of air in the room

The dimensions of the room are 12 ft by 8 ft by 8 ft. So, the volume of the room is $\left(12ft\right)·\left(8ft\right)·\left(8ft\right)=768f{t}^{3}$

## Step 3: To determine the input rate of carbon monoxide in the room

We are given that at t = 0 fresh air containing no carbon monoxide is blown into the room at a rate of $100f{t}^{3}/min$. So, one concludes that the input rate of carbon monoxide in the room is,

$\left(100f{t}^{3}/min\right)·\left(0\right)=0$

## Step 4: To determine the output rate of carbon monoxide from the room

We are given that the air in the room flows out through a vent at the same rate as it is blown into the room i.e., $100f{t}^{3}/min$. So, one concludes that the output rate of carbon monoxide from the room is,

$\left(100\right)·\frac{x\left(t\right)}{768}=\frac{25·x\left(t\right)}{192}$

## Step 5: Determining the initial value problem

The air initially contained 3% carbon monoxide. So,

$\frac{x\left(0\right)}{768}=3%\phantom{\rule{0ex}{0ex}}\frac{x\left(0\right)}{768}=\frac{3}{100}\phantom{\rule{0ex}{0ex}}x\left(0\right)=\frac{\left(768\right)\left(3\right)}{100}\phantom{\rule{0ex}{0ex}}x\left(0\right)=23.04\phantom{\rule{0ex}{0ex}}$

So, one set the initial value $x\left(0\right)=23.04$.

Substituting the input and output rates from step 3 and step 4 into the equation (1), one will get the following initial value problem as a mathematical model for the mixing problem,

i.e., $\frac{dx}{dt}=0-\frac{25x\left(t\right)}{192},x\left(0\right)=23.04$

i.e; $\frac{dx}{dt}+\frac{25x\left(t\right)}{192}=0,x\left(0\right)=23.04$

## Step 6: To find the solution to the initial value problem obtained in step 4 to find the amount of carbon monoxide in the air at the time t min

The differential equation obtained in step 4 is

$\frac{dx}{dt}+\frac{25x\left(t\right)}{192}=0$ …… (2)

Integrating factor, I.F.= ${e}^{\int \frac{25}{192}dt}={e}^{\frac{25}{192}t}$.

Multiplying both sides of equation (2) by ${e}^{\frac{25}{192}t}$,

${e}^{\frac{25}{192}t}·\frac{dx}{dt}+{e}^{\frac{25}{192}t}·\frac{25·x\left(t\right)}{192}=0\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(x{e}^{\frac{25}{192}t}\right)=0\phantom{\rule{0ex}{0ex}}$

Now, integrating both sides,

$x·{e}^{\frac{25}{192}t}=C$

where C is an arbitrary constant.

$x=C{e}^{-\frac{25}{192}t}$ …… (3)

At t=0, x=23.04

Therefore, from equation (3),

$C=23.04$

Substituting this value of C in equation (3),

$x=\left(23.04\right){e}^{-\frac{25}{192}t}$

So, the amount of carbon monoxide in the room after the time, t min is $x\left(t\right)=\left(23.04\right){e}^{-\frac{25}{192}t}$

## Step 6: To determine the time after which the amount of carbon monoxide in the air will reach 0.01%

Firstly, we will find the amount of 0.01% of carbon monoxide by multiplying 0.01% by the volume of the room,

$\left(\frac{0.01}{100}\right)\left(768\right)=0.0768$

Substituting this value of $x\left(t\right)=0.0768$in equation (4),

role="math" localid="1664260534765" $0.0768=\left(23.04\right){e}^{-\frac{25}{192}t}\phantom{\rule{0ex}{0ex}}{e}^{\frac{25}{192}t}=\frac{23.04}{0.0768}\phantom{\rule{0ex}{0ex}}{e}^{\frac{25}{192}t}=300\phantom{\rule{0ex}{0ex}}\frac{25}{192}t=ln300\phantom{\rule{0ex}{0ex}}t=\frac{\left(192\right)ln300}{25}\phantom{\rule{0ex}{0ex}}t=43.8min\phantom{\rule{0ex}{0ex}}$

Hence, the amount of carbon monoxide in the air will reach 0.01% after 43.8 min.