Natural sources add sulfur dioxide (SO_2) to the atmosphere at a rate of about 1

Question

Natural sources add sulfur dioxide (SO_2) to the atmosphere at a rate of about 10^8 metric tons of S per year. The background concentration of atmospheric SO_2, measured in remote areas where anthropogenic sources are not likely to have much influence, is about 0.2 parts per billion, by volume [ppb(v)]. What is the residence time, in days, of atmospheric SO_2 in remote regions? Useful info: molar mass of air = 28.96 g/mol, mass of Earth's atmosphere is 5.14 Times 10^21 g, and ppb(v) is equivalent to mole fraction. 2. A stable and highly soluble pollutant is dumped into a lake at the rate of 0.16 metric tons per day. The lake volume is 4 Times 10^7 m^3 and the average water flow-through rate is 8 Times 10^4 m^3/day. Ignore evaporation from the lake surface and assume the pollutant is uniformly mixed in the lake. What eventual steady-state concentration (in ppm) will the pollutant reach?

Explanation / Answer

2)

The amount of water in the lake, Mw = 4 x 107 m3

The average water flow through the lake, Fw = 8 x104 m3/d

Then the residence time of the water in the lake is:

Tw = Mw/Fw

Tw = 4 x 107 m3/ (8 x 104 m3/d) = 0.5 x 103 day

Tw = 500 days

Because the pollutant is uniformly mixed in the lake, the residence time of the pollutant will equal the residence time for the lake water.

Tw = Tp = 500 days

Then it follows that the steady state stock of pollutant is simply the input rate times the residence time:

Mp = FpTp = (0.16 tonnes/days) x 500 days

Mp = 80 tonnes

To get the concentration we need to know how much water there is in the lake in tonnes. If we multiply the volume of water by the density of water (1 g/cm3 = 1 kg/1000 cm3 = 1000 kg/m3, we find that 1 cubic meter has a mass exactly 1 metric ton (1000 kg) which is equal to 1 tonne.

The steady state concentration of the pollutant, Cp = 80 tonnes/(4 x 107 tonnes) = 2 x 10-6

Cp = 2 parts out of every million (by mass)

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