A party is being held in a room that contains $1800$ cubic feet of air which is originally free of carbon monoxide. Beginning at time $t=0$ several people start smoking cigarettes. Smoke containing $6$ percent carbon monoxide is introduced into the room at the rate of $0.15$ cubic feet/minute and the well circulated mixture leaves at the same rate through a small open window. Extended exposure to CO concentration as low as $0.00018$ can be dangerous. When should a prudent individual leave the party?
How do we solve this problem by formulating a differential equation? I know that the answer is "no later than 36 minutes after the smoking starts".
I have done the following:
(1) Let $x$ be the amount of CO at any time $t$.
(2)$\displaystyle\frac{dx}{dt} = CO_{\text{in}} – CO_{\text{out}}$
(3) $CO_{\text{out}}=(x/1800)*(0.15)*(0.06)$.
(4) $CO_{\text{in}}$ I could not find
(5) initial CO=0 and final CO=0.00018.
How do I proceed?
Best Answer
You are on the right track with your answer, but you need to be more careful to distinguish between the total amount of carbon monoxide and the concentration of carbon monoxide.
Let $x$ be the total amount of carbon monoxide.
Then $$\frac{dx}{dt} = CO_{\text{in}} - CO_{\text{out}}$$ as you have written.
The amount coming in is given by: $CO_{\text{in}} = 6\% \times 0.15 \text{ cubic feet/minute}.$
The amount leaving is given by: $\displaystyle CO_{\text{out}} = \frac{x}{1800} \times 0.15 \text{ cubic feet/minute}.$
The initial amount is zero, so $x(0) = 0$. This information defines an initial value problem. Find the solution $x(t)$ and then solve for $t^{*}$ such that $$\frac{x(t^{*})}{1500} = 0.00018.$$