Carbon monoxide pollution An executive conference room of a co...

Question

Carbon monoxide pollution An executive conference room of a corporation contains 4500 ft³ of air initially free of carbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the room at the rate of 0.3 ft³/min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft³/min. Find the time when the concentration of carbon monoxide in the room reaches 0.01%.

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A well-mixed tank holds 2000 L of solution and initially contains 1.5 kg of a dissolved chemical. Beginning at T equals 0, a feed solution containing 0.02 kg per liter of the chemical flows in at 10 L per minute. The mixed solution leaves the tank at the same rate, 10 L per minute. Find the time T in units of minutes when the concentration in the tank reaches 0.01 kg per liter. OK. So it appears for this particular prompt, we're ultimately asked to solve for the time T, so we're solving for what does T equal and units of minutes, using all the information that's provided to us. What is the time T when the concentration in the tank reaches 0.01 kg per liter. So now that we know that we're ultimately trying to solve for this time value T in units of minutes. Our first step that we need to take is we need to define what our unknown variables are, so let us define the unknown, and let us also state that A of T will be in units of kilograms, and this will be the mass of the chemical in the tank at time T. And the concentration in the tank will be C T is equal to AT divided by 2000 kg per liter. Fantastic. So once again, A of T in units of kilograms once again is the mass of the chemical in our tank at time T, and we know what CFT, which is the concentration of the tank is. So our second step now is we need to consider our rates in and out. Meaning we need to note that inflow brings chemicals into our tank at a rate, at a specific rate, meaning we have our inflow, which I'll call I subscript F, so our inflow. Is equal to parentheses of 0.02 kg per liter multiplied by parentheses of 10 L per minute. Which will be equal to 0.2 kg per minute, because as you can see, our units of liters will cancel out leaving us with units of kilograms per minute. And we need to note that our outflow removes the chemical at a rate of O subscript FOF is equal to parentheses of. 10 L per minute, so 10 L per minute multiplied by A of T divided by 2000. Which will be equal to. A of T divided by 2000. Or actually, since we know we got all reduced because we're multiplying it by 10, so it'll be equal to AFT divided by 200, and once again, we need to note that our units will be kilograms per minute. Fantastic. Moving on to our third step, we now need to form the ODE and we need to also note that the mass balance will yield DA. By DT. Will be equal to 0.2 minus a divided by 200. And A of 0 will be equal to 1.5. So we need to solve now, so this is step 4, we're solving for our linear ODE. So we need to write it first in standard form, so DA by DT plus 1 divided by 200 multiplied by A will be equal to 0.2. And we need to integrate the factors or integrating factor mu of T will be equal to e to the power of the integral of 1 divided by 200. DT Which will be equal to e to the power oft divided by 200, and we now need to multiply through and integrate. So we will find that D by DT. Of A multiplied by e to the power of T divided by 200. Will be equal to 0.2 multiplied by e to the power of T divided by 200. Which will mean that a multiplied by e to the power of T divided by 200. Will be equal to 0.2 multiplied by 200 multiplied by e to the power oft divided by 200. Plus C. Which will be all equal to 40 multiplied by e to the power of T divided by 200 plus C. Where C denotes some constant variable. So, thus, we can go ahead and write that A of T is equal to 40 + C multiplied by E to the power of net divided by 200. Fantastic. So we need to use A of 0 is equal to 1.5, and this will help us find our C is equal to 1.5 minus 40. Which will be equal to, when we plug that into our calculator, -38.5. Therefore, we could safely write that a oft will be equal to 40 minus 38.5 multiplied by e to the power of net divided by 200. Fantastic. So now, for our fifth step, so this is step five, we need to now impose the target concentration. So let us note that our target concentration is 0.01 kg per liter, and this corresponds to our target mass. So our A target, which I'll call a subscript T. Will be equal to 2000 multiplied by 0.01, which will be equal to 20 kg, and we now need to solve A of T is equal to 20. So we'll be able to write 20 is equal to 40 minus 38.5 multiplied by e to the power of net divided by 200. And we need to note that we are rearranging this expression ultimately to isolate and solve fort, and when we begin to do that, we will have e to the power of net divided by 200 is equal to 40 minus 20 divided by 38.5, which will be equal to 40 divided by 77. Which will mean that T is equal to -200 multiplied by the natural log of 40 divided by 77. And don't forget that our units are in minutes. So our final answer without a doubt will be equal to 200 multiplied by the natural log of 77 divided by 40, and our units are once again in minutes and that's it. That is our final answer fort. Hooray, we did it. So looking at the prom itself, we could safely say once again that our final answer for the time t without a doubt will be. 200 multiplied by the natural log of 77 divided by 40 minutes. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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