# 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