3. A caveman discovered kerosene and is using a kerosene lamp to light his cave.

Question

3. A caveman discovered kerosene and is using a kerosene lamp to light his cave. Sadly, he does not know about air pollution, and thus is exposing himself to SO2. The lamp burns 0.115 g of kerosene each minute and contains 1.3% sulfur by weight. a) (12 pts) What is the emission rate of SO2 in g/s? b) (15 pts) There is a breeze through the cave (which has a volume of 50 m3) that leads to an air flow of 1 m/min through the cave. After a very long time (i.e. once steady state is reached), what is the concentration of SO2 in the cave (in g/m3) if SO2 decays with a first order rate constant of 6.39 x 10-5 s?

Explanation / Answer

Answer 1

Amount of kerosene burned per minute = 0.115 g

Amount of sulphur per minute = 1.3*0.115/100 = 1.495 x 10^-3 g

Moles of sulphur = 1.495/(32*1000) = 4.672 x 10^-5 moles per minute = 7.7864 x10^-7 moles per second

Thus SO2 emission = 7.7864 x10^-7 * 64 = 49.83 micro gram per second

Answer 2

Once the steady state is reached we will have rate of emission of SO2 equal to rate of destruction of SO2 and will have no change in concentration of SO2

Thus concentration of SO2 in steady state = 49.83/50 = 0.9966 micro gram per cubic metre

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