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Not all atmospheric pollution is from industrial sources.

Volcanic eruptions can be a significant source of air pollution.

The Kilauea volcano in Hawaii emits 200-300 tons of SO2 per day.

If this gas is emitted at 800°C and 1.0 atm, what volume of gas is emitted?

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Answer: Minimum volume of gas emitted is 2.75 x 10^5 m³

Maximum volume of gas emitted is 4.126 x 10^5 m³

Explanation: If 200 to 300 tons are emitted per day, multiplying by 1000 give the amount in kilogrammes.

FOR MINIMUM VOLUME OF EMITTED GAS AT 200 TONS

1 ton = 1000kg

800°C = 800 + 273 = 1073K

Molar mass of SO2 = 32 + (16 x 2) = 64g = 0.064kg

Number of moles of SO2 is given by:

(200 x 1000kg) / 0.064 = 3.125 x 10^6 mol

V = nRT/p

V = [ (3.125 x 10^6 mol) x 0.08206dm³atm/mol/K x 1073K] / 1 atm

V = 2.75 x 10^8 dm³

V = 2.75 x 10^5 m³

FOR MINIMUM VOLUME OF EMITTED GAS AT 300 TONS

1 ton = 1000kg

800°C = 800 + 273 = 1073K

Molar mass of SO2 = 32 + (16 x 2) = 64g = 0.064kg

Number of moles of SO2 is given by:

(300 x 1000kg) / 0.064 = 4.687 x 10^6 mol

V = nRT/p

V = [ (4.687 x 10^6 mol) x 0.08206dm³atm/mol/K x 1073K] / 1 atm

V = 4.126 x 10^8 dm³

V = 4.126 x 10^5 m³

Therefore, Kilauea emits 2.75 x 10^5 m³ to 4.126 x 10^5 m³ daily

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