Smoking. Suppose people start smoking in a room of volume 60 m3, thereby int rod

Question

Smoking. Suppose people start smoking in a room of volume 60 m3, thereby int roducing air containing 5% carbon monoxide at a rate of 0.002 m3/min into the room. Assume that the smoky air mixes immediately and uniformly with the rest of the air. and that this mixture leaves the room at the same rate as the smoky air enters. Prove that the amount x(t) (in m3) of carbon monoxide in the room at time t minutes satisfies the differential equation: Solve this differential equation to obtain the amount of carbon monoxide in the room at time t, assuming that there is no carbon monoxide in the room initially. What happens to the concentration of carbon monoxide in the room in the long term? Sketch a curve of the concentration of carbon monoxide as a function of time. Medical tests warn that exposure to air containing 0.1% carbon monoxide for some time can lead to coma. How long does it take for the concentration of carbon monoxide in the room to reach this level?

Explanation / Answer

part a.

suppose

x(t)=the amount of carbon mono oxide in the room at time t.

dx/dt=change in the amount of carbon mono oxide over time

so change in carbon mono oxide is equal to carbon monooxide comes minus carbonmono oxide goes

so we can write this equation

dx/dt= rate in - rate out

dx/dt=1/10000-(2/1000)*(x/60)

dx/dt+x/30000=1/10000

part b.

simplify and variable separation

(1/x-3)dx=(-1/30000)dt

ln|x-3| = -t/30000+c

x-3=Ae-t/30000

apply initial condition

x=3-3e-t/30000

part c

from equation we can see that

when time tends to infinity e-t/30000   tends to zero

as a result

x will tend to 3

after long time

concentration of carbon monooxide will become stable and equal to 3.

part d.

carbon mono oxide =0.1%

so in the room carbon mono oxide should be (60*0.1)/100=0.06

put this in equation

0.06=3-3e-t/30000

3e-t/30000 = 2.94

e-t/30000 =2.94/3

e-t/30000 =0.98

-t/30000=ln(0.98)

-t/30000 = -0.02020270731

t=606 minute

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