Question

The air in a room with volume 180 m3 contains 0.4% 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).
p(t) = ___________

What happens with the percentage of carbon dioxide in the room in the long run?

lim t → [infinity] p(t) =________ %

Answer 1

Answer:

(A) The percentage p(t) of carbon dioxide in the room as a function of time t (in minutes) is $p(t)=0.55\%\left(0.63e^{-\frac{t}{90}}+0.09\right)$.

(B) The percentage of carbon dioxide will approach 0.0495%.

Step-by-step explanation:

Mixing problems are an application of separable differential equations. They’re word problems that require us to create a separable differential equation based on the concentration of a substance in a vessel.

Let y(t) be the total volume of carbon dioxide in the room.

The main equation that we’ll be using to model this situation is

Rate of change y(t) = Rate at which y(t) enters the room - Rate at which y(t) exits the room

where

Rate at which y(t) enters the room: (flow rate of gas entering) x (concentration of substance in gas entering)

Rate at which y(t) exits the room: (flow rate of gas exiting) x

(concentration of substance in gas exiting)

Fresh air enters the room at rate $2 \frac{m^3}{min}$, and 0.05% of this air is carbon dioxide, so the rate at which carbon dioxide enters the room is $2 \cdot 0.0005 = 0.001 \frac{m^3}{min}$.

The air mixture leaves the room at rate $2 \frac{m^3}{min}$, and the concentration of carbon dioxide in this air is $\frac{y}{180}$, so the rate at which carbon dioxide leaves the room is $2 \cdot \frac{y}{180} = \frac{y}{90} \frac{m^3}{min}$.

Thus, the differential equation that governs the system is

$\frac{dy}{dt}=0.001-\frac{y}{90}$

Rearrange the above equation

$\frac{dy}{0.001-\frac{y}{90} }=dt$

Integrate both sides

$\int \frac{dy}{0.001-\frac{y}{90} }=\int dt\\\\-90\ln \left|0.001-\frac{y}{90} \right|=t+C$

Solve for y

$-90\ln \left|0.001-\frac{y}{90} \right|=t+C\\\\\ln \left|0.001-\frac{y}{90} \right|=-\frac{t}{90} +\frac{C}{90}\\\\\left|0.001-\frac{y}{90} \right|=e^{-\frac{t}{90} +\frac{C}{90}}\\\\0.001-\frac{y}{90}=\pm e^{-\frac{t}{90} +\frac{C}{90}}\\\\\frac{y}{90}=0.001-(\pm e^{-\frac{t}{90} +\frac{C}{90}})\\\\y=0.09+Ce^{-t/90}$

Now we need to use the initial condition: the room contains 0.4% carbon dioxide at time t = 0, so $y(0) = 180 \cdot 0.004 = 0.72 \:m^3$. Then

$0.72=0.09+Ce^{-0/90}\\C=0.63$

The particular solution that describes this system is

$y(t)=0.09+0.63e^{-t/90}$

where y(t) is the volume of carbon dioxide.

(A) The percentage of carbon dioxide is:

$p(t)=100\%\cdot \frac{y(t)}{180}\\\\p(t)=100\%\cdot \frac{0.09+0.63e^{-t/90}}{180}\\\\p(t)=0.55\%\left(0.63e^{-\frac{t}{90}}+0.09\right)$

(B) To find out what happens in the long run, we can take the limit as $t\rightarrow \infty$

$\lim _{t\to \infty }\left(0.55\left(0.63e^{-\frac{t}{90}}+0.09\right)\right)\\\\0.55\cdot \lim _{t\to \infty \:}\left(0.63e^{-\frac{t}{90}}+0.09\right)\\\\0.55\left(\lim _{t\to \infty \:}\left(0.63e^{-\frac{t}{90}}\right)+\lim _{t\to \infty \:}\left(0.09\right)\right)\\\\0.55\left(0+0.09\right)\\\\0.0495$

The percentage of carbon dioxide will approach 0.0495%