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 $\mathbf{100}\mathit{f}{\mathit{t}}^{\mathbf{3}}\mathbf{/}\mathit{m}\mathit{i}\mathit{n}$. 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?

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)

The dimensions of the room are12 ft by 8 ft by 8 ft. So, the volume of the room is$\left(12ft\right)·\left(8ft\right)·\left(8ft\right)=768f{t}^3$

We are given that att = 0fresh 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,

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

We are given that theair 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,

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

The air initially contained3% carbon monoxide. So,

$$\begin{gathered} \frac{x\left(0\right)}{768}=3\% \\ \frac{x\left(0\right)}{768}=\frac{3}{100} \\ x\left(0\right)=\frac{\left(768\right)\left(3\right)}{100} \\ x\left(0\right)=23.04 \end{gathered}$$

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

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 $e^{\frac{25}{192}t}$,

$$\begin{gathered} e^{\frac{25}{192}t}·\frac{dx}{dt}+e^{\frac{25}{192}t}·\frac{25·x\left(t\right)}{192}=0 \\ \frac{d}{dt}\left(x{e}^{\frac{25}{192}t}\right)=0 \end{gathered}$$

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

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" $$\begin{gathered} 0.0768=\left(23.04\right)e^{-\frac{25}{192}t} \\ {e}^{\frac{25}{192}t}=\frac{23.04}{0.0768} \\ {e}^{\frac{25}{192}t}=300 \\ \frac{25}{192}t=ln300 \\ t=\frac{\left(192\right)ln300}{25} \\ t=43.8min \end{gathered}$$

Hence, the amount of carbon monoxide in the air will reach0.01% after 43.8 min.