Question

The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 m3/min and the mixed air flows out at the same rate. find the percentage p of carbon dioxide in the room as a function of time t (in minutes).

Answer 1

Answer:

$p(t)=\frac{1}{20}(1+2e^{-t/90})$

Explanation:

Let $p(t)$ be % of $CO_2$ at time $t$(time in minutes).

Amount of $CO_2$ in room=Volume of room $\times p$

Rate of change of $CO_2$ in room=Volume of room x $\frac{dp}{dt}$

$=180\times \frac{dp}{dt}$

Rate of inflow of $CO_2$=$(\% CO_2 \ in\ Fresh\ Air \times \frac{1}{100})\times(Rate \ of Inflow\ of \ air)$

$=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}$

Rate of $CO_2$ outflow

$(\% CO_2 \ in\ Fresh\ Air \times \frac{1}{100})\times(Rate \ of Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}$

Therefore rate of change of $CO_2$ is $\frac{1}{100}-\frac{p}{50}$

% rate of change is$100\times \ Rate of \ Change=\frac{1}{10}-2p$

Therefore we have:

$180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}$

Integrate both:

$\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}$

$\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C$

In$|20p-1|=-\frac{t}{90}+$In$K$

$+20In \ C=+In \ K$

Raise both sides to base $e$,

$20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})$

Initially, there's only 0.25% of $CO_2$,

Now substitute $p=0.25$ and $t=0$

$0.25=\frac{1}{20}(1+K)\\K=4$

$p=\frac{1}{20}(1+2e^{-t/90})$