The air in a room with volume 180 m3

Question

The air in a room with volume 180 m³ contains 0.45% 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 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 → ∞ p(t) =  %

Answer 1

Michelle, I hope it is not too late for a solution to your problem. I just noticed it yesterday! In terms of concentration I get the following equation

C(t) = (C_o-C₁)exp(-Qt/V) + C₁ where C(t) is the concentration in the room at time t

C_o is the starting concentration in the room

C₁ is the concentration of the flow into the room

Q is the flow rate into the room

V is the volume of the room

Substituting your numbers into the equation yields (.45-.05)exp(-2t/180) + .05

or (.4)exp( -.01111t) + .05

As you can see, after a long time, the first term disappears and the final concentration is the room is .05 which is what you would expect.

And at time =0, the room concentration is .45-.05 +.05 = .45 which is what you know.