Answer:
The air in the room at 0.01% carbon monoxide at 43.8 min
Step-by-step explanation:
Let be the volume of CO in the room at time t, be v(t) and the total volume of the room be V. The volume percent of CO in the room at a given time is then given by:
[tex]p(t) = \frac{100\times v(t)}{V}[/tex]
Volume percent is the measure of concentration used in this problem. The "Amount" of CO in the room is then measured in terms of the volume of CO in the room.
Let the rate at which fresh air enters the room be f, which is the same as the rate at which air exits the room. We assume that the air in the room mixes instantaneously with the air entering the room, so that the concentration of CO is uniform throughout the room.
As you wrote, the rate at which the volume of CO in the room changes with time is given by
[tex]\frac{dvt}{dt} = 0 \times f -\frac{f}{v} \times v(t) = -\frac{f}{v} \times v(t)[/tex]
This is a simple first-order equation:
[tex]\frac{dv}{v} = -\frac{f}{v} dt[/tex]
[tex]ln(v) - ln(c) = -\frac{f}{v} \times t[/tex]
where ln(c) is the constant of integration.
ln [tex]\frac{v}{c} = -\frac{f}{v} \times t[/tex]
[tex]v(t) = c \times e^{(-f*\frac{t}{V})}[/tex]
In terms of volume percent,
[tex]p(t) = \frac{100*v(t)}{V}= (\frac{C}{V})*exp(\frac{-f \times t}{v})[/tex]
where C = 100*\frac{c}{V} is just another way of writing the constant.
Plugging in the values for the constants, we get:
[tex]p(t) = (\frac{C}{768 cu.ft.})* exp(\frac{-t}{7.68 min})[/tex]
Now use the initial condition (p(0) = 3% at t = 0) to solve for C:
3% = C
[tex]p(t) = (3\%)\times exp(\frac{t}{7.68 min})[/tex]
To find the time when the air in the room reaches a certain value, it is easier to rewrite this solution as:
[tex]\frac{p(t)}{3\%} = exp(\frac{-t}{7.68 min})[/tex]
[tex]t(p) = -(7.68 min)ln(\frac{p}{3\%} )[/tex]
[tex]= (7.68 min)*ln(\frac{3\%}{p})[/tex]
The question asks when p(t) = 0.01%. Plugging this into the above equation, we get:
[tex]t(0.01\%) = (7.68 min)*ln(\frac{3}{0.01}) = 43.8 min[/tex]
