A bar opens at 6 and is quickly filled with customer,the majority of whom are cigarette smokers.The bar has ventilators which exchange the smoke-air mixture with for fresh air.Cigarette smoke contains 4% carbon monoxide anda prolonged exposure to a concentration of more then 0.012% can be fatal.The bar has a floor area of 20, by 15m and a height of 4m.It is estimated that smoke enters the room at a constant rate of 0.006m^3/min and that the ventilators remove the smoke-air mixture at 10 times the rate at which smoke is produced.(adsbygoogle = window.adsbygoogle || []).push({});

The question is when is it advisable to leave the bar?or in other words when does the concentration of carbon monoxide reach the lethal limit?

Formulate the differential equation for the changing concentration of carbon

monoxide in the bar over time.

(b) By solving the differential equation, establish the time at which the lethal

limit will be reached.

attempted solution

20*15*4=1200m^3=volume of room

smoke enter room at 0.006m^3/min

smoke leaves room at 0.06m^3/min

rate in = 0.04*0.006=0.00024

rate out = 0.06(y)/1200=0.00005(y)

dy/dt= 0.00024-0.00005y

dy/dt-1/500000(120-y)

dy/(120-y)=1/500000(dt)

ln(120-y)=1(t)/500000+c

120-y=Ae^(t/500000)

120-Ae^(t/500000)=y

second part 120-Ae^(t/500000)=0.012

119.088=e^(t/500000)

ln(119.088)(500000)=t

2.39*10^6=t

answer is slighty absurd so can anyone spot any mistakes?

thanks in advance

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# Differential equations problem rate in/rate out problem

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