Question

Carbon monoxide pollution An executive conference room of a corporation contains 4500 ft³ of air initially free of carbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the room at the rate of 0.3 ft³/min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft³/min. Find the time when the concentration of carbon monoxide in the room reaches 0.01%.

Accepted Answer

Define the variable $$Q(t) as $$the volume of carbon monoxide (in cubic feet) in the room at time $$t$$ minutes. Initially, since the room is free of carbon monoxide, we have $$Q(0) = 0. $$Set up the rate of change of carbon monoxide in the room. The rate in is the concentration of carbon monoxide in the incoming air multiplied by the inflow rate: $$0.04 	imes 0.3 ft$$³/min. The rate out is the concentration of carbon monoxide in the room multiplied by the outflow rate: $$\frac{Q(t)}{4500} 	imes 0.3 ft$$³/min. Write the differential equation expressing the rate of change of $$Q(t) as $$the difference between the inflow and outflow rates:

$$\displaystyle \frac{dQ}{dt} = 0.04 	imes 0.3 - \frac{Q(t)}{4500} 	imes 0.3. $$Simplify the differential equation and solve it as a first-order linear ordinary differential equation. Use an integrating factor or recognize it as a separable equation to find the general solution for $$Q(t). $$Use the condition that the concentration reaches 0.01%, which means $$\frac{Q(t)}{4500} = 0.0001$$, to solve for the time $$t$$ when this occurs.

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