Question

The current mass of the atmosphere is approximately 5.3×10¹⁸ kg. How much fossil fuel, expressed as kilograms of carbon (C), must have been burned to raise the CO₂ concentration of the atmosphere from 210 ppm_v to 330 ppm_v? The bulk density of carbon is 2267 kg/m³. What is the volume in cubic kilometres of this much carbon?

Answer 1

Hi
Given the current mass of the atmosphere is 5.3*10^18 kg.
We may assume that atmosphere is made up of air and is at ambient conditions because all the air lies in the lower atmosphere.

Considering air as a mixture of nitrogen and oxygen we can say that atomic mass of air will be close to 29 and thus assuming the ideal gas we ca write.

Density =PM/RT
Here P is the pressure that we can take close to the atmospheric pressure and M is the molecule weight of the gas.
R is the universal gas conatnt and T is the temperature in K .

Using ambient condition we know the volume of one mole of the gas will be close to 24.4 L/mole.
So the total volume of the atmosphere =

Moles of air in atmosphere *volume per mole of atmosphere

Moles of air in atmosphere = Mass of atmosphere/molecular weight
5.3*10^21 /29 moles
1.82 *10^20 moles
Volume = 1.83*10^20 * 22.4 Litres
=4.4*10^18 m^3

Initial concentration of CO2 =210 ppmv
Final concentration of CO2 = 330 ppmv

Mass addition of CO2 = density *( Volume addition )
ppmv is nothing but 1 part volume in 10^6 part volume.
So the volume of CO2 initially =(4.4*10^12 )*210 m^3.
Volume of CO2 finally
= (4.4*10^12)*330 m^3.
Volume change of CO2.
=4.4*10^12*110 m^3
=4.84*10^14 m^3

Mass addition of CO2 = density*4.84*10^14 m^3
Density of CO2 = Density of ideal gas at the ambient conditions.

Density =PM/RT =101325*44*0.001/8.314*298
=1.79 kg/m^3
=8.66*10^14 kg
We know in the production of 44 g CO2 we need 12 gram of carbon.
So the equivalent mass of carbon required=

(8.66*12/44 )*10^14
=2.36*10^14 kg
Volume of carbon =Mass /density
=2.36*10^14/2267 m^3.
=1.04*10^11 m^3.
1km^3 = 10^9 m^3.
So the volume
=104 km^3.

Hope this helps

Thanks