rate math problem?

Question

the air in a room with volume 180m3 contains 0.15% carbon dioxide itially. 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. What happens in the long run?

Answer 1

Assume that the air in the room got perfectly mixed, that means that the concentration is the same everywhere in the room. Then the leaving air has the same concentration as the mixed air i n the room
The change of carbon dioxide content in the room is given by
V dc/dt = F & middot (cin - c)
where
V - volume of the room
F - flowrate
c - carbon dioxide concentration
cin - inlet concentration of carbon dioxide

You can solve the diffrential equation by separation of varibales
∫ 1/(cin-c) dc = F/V dt
=>
cin - c = C·exp{-F/V · t}
=>
c = cin - C·exp{-F/V · t}
The constant of Integration C can be evaluted from the initial condition:
for t = 0 c = c₀
=>
c₀= cin - C·exp{-0}
c = cin + (c₀-cin)·exp{-F/V · t}
For the given values
c = 0.05% + 0.1% · exp{-1/60 min^-1 · t} = 0.05% + 0.1% · exp{-1h^-1 · t}

For long term operation the exponential expressions decreases to zero and the concentration in the room approaches the inlet concentration:
lim t→ ∞ (c) = cin + (c₀-cin)·0 = cin

Answer 2

There are 2/180 = 0.011 air changes per minute. Initially, exiting air will be 0.15% CO2 and will continue to be so for a time depending on the amount of mixing in the room.

After 1 minute, the concentration is:
(0.15*178 + 0.05 *2)/180 = 14.9%

Generally:
K = (0.15(180-2t) + 0.05*2t)/180
K = 0.15 - 0.0011t

Eventually, the room will reach a steady state approaching 0.05% CO2. By the equation