# A calculation

I wanted to do a back-of-the-envelope calculation to find what volume of air would need to be processed every second in order to remove CO2 at the rate that it is currently being emitted by human activities. I don't consider CO2 emitted in other ways, and I don't consider CO2 being absorbed.

Current emissions = 40 billion tonnes per year = 4 x 10^13 kg / year ( = 1.27 x 10^6 kg/s = 1270 tonnes /s)

Current concentration of CO2 = 400 ppm so 1m3 air contains 0.0004 m3 CO2 (about the volume of a can of soda)

Gram molecular mass of CO2 = 12 + 2x16 = 44 g

Molar volume of any gas at STP = 22.4 L

Density of CO2 at STP = 44/22.4 = 1.96 g / L = 1.96 kg / m3

So mass of CO2 in 1 m3 air = mass of 0.0004 m3 CO2 = 1.96 x 0.0004 = 0.000786 kg

So (reciprocating) 1272 m3 air contains 1kg CO2

So volume of air containing 4 x 10^13 kg CO2 = 4 x 10^13 x 1272 = 5 x 10^16 m3 = 5 x 10^7 km3 air

To put that in perspective, what thickness of the Earth's atmosphere would that represent?

Surface area of Earth = 4 x pi x r² = 4 x pi x (6366 km)² = 5.09 x 10^8 km²

So depth of air that needs to be processed each year = 5 x 10^7 / 5 x 10^8 = 0.1 km = 100 m

So each year I would need to process a volume of air equal to a 100m thick layer covering the entire planet.

Put another way, what volume of air would need to be processed every second?

5 x 10^7 km3 / (365 x 24 x 3600) = 1.6 km3 / s = 1.6 x 10^9 m3/s

So if I build 10 000 CO2 capture plants, spread around the planet, how much air would each one have to process per second?

1.6 x 10^9 / 10^4 = 1.6 x 10^5 m3/s = 160 000 m3/s

The volume of Wembley stadium, one of the largest stadiums in the world, is 4 million m3, so each plant would need to process one stadium's worth of air every 25s since 4 x 10^6 / 1.6 x 10^5 = 25