Calc II: Stirred tank problem (differential equations)

buttz123

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CO2 is leaking into the interior of a car, whose volume is 3000 liters, at the rate of 0.01 mg per minute. Air enters and exits from the car at the rate of 3 liters per minute.

Assume the air in the car contains no CO2. Write the differential equation whose solution will be the mg of CO2 in the car as a function of time.

I know I need to use the equation m'(t)= (concentration in the inflow)(rate of flow into tank) - m(t)(rate of flow/volume of tank). The issue I'm running into is the fact that there are two flows: the flow of CO2 and the flow of air. Usually when you plug in numbers to the equation, their units cancel, but both the flow of CO2 and the air are in per minute forms. I'm not sure how to correctly set it up because of this. /: pls help
 
CO2 is leaking into the interior of a car, whose volume is 3000 liters, at the rate of 0.01 mg per minute. Air enters and exits from the car at the rate of 3 liters per minute.

Assume the air in the car contains no CO2. Write the differential equation whose solution will be the mg of CO2 in the car as a function of time.

I know I need to use the equation m'(t)= (concentration in the inflow)(rate of flow into tank) - m(t)(rate of flow/volume of tank). The issue I'm running into is the fact that there are two flows: the flow of CO2 and the flow of air. Usually when you plug in numbers to the equation, their units cancel, but both the flow of CO2 and the air are in per minute forms. I'm not sure how to correctly set it up because of this. /: pls help
I would say that there are three flows! There is the CO2 coming in, there is air coming in, and there is the combination of air and CO2 going out.
The CO2 is coming into the car at 0.01 mg per minute. The air coming in we can ignore because it contains no CO2. Now, the air going out contains CO2. To determine how much CO2 is going out, we need to determine how much CO2 there is in every liter of air in the car. Let the volume of CO2, in liters, in the car at time t, which is what we are trying to find, be "V(t)". Then the concentration of CO2, at time t is \(\displaystyle \frac{V(t)}{3000}\) so that the 3 liters of air going out carries with it \(\displaystyle \frac{3V(t)}{3000}= \frac{V(t)}{1000}\) liters of CO2. The CO2 is coming in at 0.01 mg per minute and going out at \(\displaystyle \frac{V(t)}{1000}\) liters per minute.

There are, actually, two complications that must be dealt with. First, the problem says that CO2 is coming in at 0.01 mg per minute while the air is coming in and going out at the same rate! That should mean that the volume of gas, air and CO2, in the car is increasing, NOT the constant "3000" liters given! But because:
1) that would make the problem more complicated
2) 0.01 mg is so small compared with 3000 liters
3) The problem says that the volume in the car (not just the "initial volume") is 3000 liters
I think we are intended to ignore that. Take the volume in the car to be the constant 3000 liters as I did above.

The other complication is that while everything else is given in liters, the inflow or CO2 is given in "mg per minute". Was that what you were talking about when you complained about the units? (But you only mentioned "per minute" which is pretty standard.) But that is easy to handle, we just need to convert form mg to liters- although, strictly speaking, a "mg" is a measure of mass not volume, for a gas like CO2, a mg mass takes up a milliliter of volume and a milliliter is 1/1000 of a liter. "0.01 mg per minute" is 0.01/1000= 0.00001 liters per minute.
So you have have CO2 coming in at 0.00001 liters per minute and going out at \(\displaystyle \frac{V(t)}{1000}\) liters per minute: \(\displaystyle \frac{dV}{dt}= 0.00001- \frac{V}{1000}\) with V(0)= 0.
V(t) is measured in liters and t in minutes.
 
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