Differential Equation for calculating concentration of mixture

Question

I am self studying linear algebra at the moment and I have been stuck at this section of the chapter.

The questions says:

A 10,000 cu-ft room contains 20% carbon dioxide. Pure oxygen will be pumped into the room at a rate of 5 cu-ft/min. The well-mixed air escapes from the room at a rate of 5 cu-ft/min.

How long will it take to reduce the carbon dioxide level to 5%?

I have tried to set up my dA/dt equation like this

dA/dt = 5 - (5A/10000)

And

A(t) = 10000 - (1/e^t)^(1/2000) + C

A(0) is 8000 so...

C = -1999

And that gives me

A(t) = 8001-(1/e^t)^(1/2000)

But when i try to find when the carbon is at 5%, which means the oxygen is at 10000(0.95) = 9500. But whatever number i try, it either gives me 8000 or an error on my calculator.

I can't seem to find my mistake on this particular question. So if stack exchange could lend me a little help, that'd be awesome!

Answer 1

You should define that $A$ is the number of cubic feet of oxygen in the room. Your solution for $A(t)$ is not correct. The $C$ should multiply the $e^{-t/2000}$ term. Then you get $$A(t)=10,000+Ce^{-t/2000}\\C=-2000\\A(t)=10,000-2000e^{-t/2000}$$ and now solving for $A(t)=9500$ should be no problem.

Answer 2

Ross Millikan has already explained the errors in your solution in another answer. I think, however, that you’re better off working directly with the carbon dioxide volume $V_{\text{CO}_2}$ and concentration $\phi_{\text{CO}_2}$. There is no new carbon dioxide entering the system, so the resulting equations should have fewer terms. Why make things harder for yourself than they need to be? The magic gases mix instantly, so the instantaneous rate at which carbon dioxide volume changes is simply $$\dot V_{\text{CO}_2} = - 5 \phi_{\text{CO}_2}.$$ To get the change in concentration, divide by the constant total gas volume: $${\dot\phi}_{\text{CO}_2} = -\frac1{2000}\phi_{\text{CO}_2}.$$ You can take it from here, I’m sure.