The air in a room with volume 180m^3

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

Your initial conditions are: Volume of the Room, V_r; Initial concentration c_1.

Fresh air comes in with: concentration c_2, at a rate of R_in.

We know that R_in=R_out.

We define the volume of concentration = V_c. The concentration as a function of t will be V_c/V_r. Then just move the decimal two places to the left to find the percentage as a function of time of the final answer.

The key to solving this is dimensional analysis.

V_c = V_r * c_1 + c_2 * R_in * t - R_out (c_1 + c_2) * t.