How do I solve this (preferably using integrals)? Air in a room with volume 180m^3 initially contains...?

Question

Air in a room with volume 180m^3 initially contains 0.20% carbon dioxide. Air with only 0.07% carbon dioxide flows into the room at 2m^3 per minute, and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What is the percentage of the carbon dioxide in the room after half an hour?

Answer 1

The total rate of change of the CO2 volume is the sum of two rates of change of the CO2 volume, namely:

+0.07k going in, (where k = 2m^3/min)

and

-(VCO2/VT)k going out, (where the volume of CO2 over the total volume is the concentration of CO2. This times the rate of total volume going out, k, gives the rate of CO2 volume going out.)

Then:

d(VCO2)/dt = 0.07k - (VCO2)/(VT)*k

Now that we know what we're talking about, let us change variables for the volume of CO2 and the total volume:

VCO2 ---> v for simplicity, and VT ---> V, then

dv/dt = 0.07k - vk/V

We now must solve this differential equation . . . there's a classic trick. Multiply the equation by e^(kt/V) and rearrange:

e^(kt/V)(dv/dt) + e^(kt/V)(vk/V) = e^(kt/V)(0.07k)

Note that this can be equally expressed as:

d/dt(v(e^(kt/V))) = e^(kt/V)(0.07k)

Integrate both sides from tinitial, I take to be zero, and tfinal, which I'll just write as t (bad notation, deal with it.) Don't forget the fundamental theorem of calculus . . .

v(t)e^(kt/V) - v(t=0) = e^(kt/V)(0.07V) - 0.07V

So:

v(t) = 0.07V + (v(t=0) - 0.07V)e^(-kt/V)

v(t) = V[ 0.07 + 0.13e^(-kt/V)]

The percentage, or concentration, C, of CO2 is v/V:

C(t) = 0.07 + 0.13e^(-kt/V)

After half an hour this equals

0.07 + 0.13e(-1/3) = 0.16 or 16%

This solution does what it should, it starts off at the initial concentration of CO2 (C(t=0) = .2) and exponentially decays to the point where the concentration of CO2 (v) is the concentration of CO2 of the gas coming in, 0.07.

Answer 2

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