volume of room ...
let r = percentage of room air changed in 1 minute ...
let p = percentage volume of CO in the room at any time t
The air in a small 12 ft by 8 ft by 8 ft room is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon dioxide is blown into the room at a rate of 100 cubic feet per minute. If air in the room flows out through a vent at the same rate, when will be the air in the room be 0.01% carbon monoxide?
Here's another way using the integrating factor thing. More roundabout than the skeets though.
The rate in of the air is
The equation is then:
The integrating factor is
Use the IC to solve for C. The amount of air at t=0 is
Using y(0)=744.96, we find that C=-23.04
.01% CO means 99.99% air. .9999(768)=767.9232
Solving or t we find
You can also do this in terms of the CO instead of air. Either way should give the same solution.