Suppose that a room containing 35 m3 of air is free of carbo

Suppose that a room containing 35 m3 of air is free of carbon monoxide. At time t = 0 cigarette smoke containing 4% carbon monoxide is introduced at the rate of 2.8 x 10-3 m3/min, and the well-circulated mixture is vented from the room at the same rate. Find a formula for the percentage of carbon monoxide in the room at time t. Extended exposure to air containing 0.012% carbon monoxide is considered dangerous. How long will it take to reach this level?

Solution

a)since initially carbon monooxide was not present in the room after time t co present in the room will be 2.8*10^-3t. hence percent of co in the room will be 2.8*10^-3t*100/35 =8*10^-3t b)now8*10^-3t=0.012 implies t=1minute 30 seconds