Atmospheric pollution is a problem that has received much attention. not all pollution, however, is from industrial sources. volcanic eruptions can be a significant source of air pollution. the kilauea volcano in hawaii emits 200−300 t (1 t = 103 kg) of so2 each day. if this gas is emitted at 800 °c and 1.0 atm, what volume of gas is emitted?

For 200t per day, the volume of gas is 2.76×10⁵ m³

For 300t per day, the volume of gas is 4.13×10⁵ m³

Explanation:

It is stated that 1t = 10³ kg

Therefore,

For 200t per day:

The number of moles of SO₂ is calculated as follows,

n(SO₂)= m/(molar mass)

where

Thus,

n(SO₂)= (200×10³kg)/(64×10⁻³kg.mol⁻¹)

         = 3.13×10⁶ mol

Therefore, the volume is determined as follows,

V= nRT/P

where

Thus,

To convert from degrees celcius to Kelvin, we 273 to the given temperature.

V= ((3.13×10⁶ mol)(0.08206dm³.atm.mol⁻¹.K⁻¹)(800+273K))/1 atm

 = 2.76×10⁸ dm³

For the answer to be in m³, we divide by 10³. Thus,

V= 2.76×10⁵ m³

For 300t per day:

The number of moles of SO₂ is calculated as follows,

n(SO₂)= m/(molar mass)

where

Thus,

n(SO₂)= (300×10³kg)/(64×10⁻³kg.mol⁻¹)

         = 4.69×10⁶ mol

Therefore, the volume is determined as follows,

V= nRT/P

where

Thus,

To convert from degrees celcius to Kelvin, we 273 to the given temperature.

V= ((4.69×10⁶ mol)(0.08206dm³.atm.mol⁻¹.K⁻¹)(800+273K))/1 atm

 = 4.13×10⁸ dm³

For the answer to be in m³, we divide by 10³. Thus,

V= 4.13×10⁵ m³

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