The volume of air in a kitchen is 100 m^3. The stove in the kitchen has been lef

Question

The volume of air in a kitchen is 100 m^3. The stove in the kitchen has been left on, and is emitting CH_4 at a rate of 5 L/min. (1000 L = 1 m^3) This has been going on for some time, and the mole fraction of CH_4 in the kitchen has reached 4% by volume. Coming into the kitchen, we smell the gas and run to open two windows to let in fresh air. Fresh air comes into the room at a rate of 5000 L/min through one window, and out the other window at the same rate (5000 L/min). We run out of the house, without turning off the gas. How much time is needed until the concentration (volume fraction) of methane in the kitchen air drops down to 0.5%?

Explanation / Answer

Since the density does not vary much

Let x= outlet concentration of CH4

Nothing is added and CH4 is diluted

Rate of mass of CH4 in –Rate of mass of CH4 out= Rate of accumulation

0 -5000*x/100 = (d/dt(100*1000*x/100)

-5000x = d/dt(100*1000x)

=-5x = d/dt(100x)

Hence -dx/x = 100/5 dt

-lnx = 20*t+C, C is integration constant

At x=4, t=0

-ln4= C=-1.386

-lnx= 20*t-1.386

For x=0.5 % =0.5/100

-ln(5/100)= 20*t-1.386

Hence time t=4.38 min

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