For whatever reason I’m having trouble finding a straight answer for various reasons.
If we’re at 415 ppm, how many tons of CO2 must be released to get to 416 ppm? Is it around 20 billion tons of CO2?
1 part per million of atmospheric CO2 is equivalent to 2.13 Gigatonnes Carbon.
1 gigatonne of carbon equals 3.67 gigatonnes of carbon dioxide.
So around 7.8 Gigatonnes of CO2
It’s pretty simple to calculate an approximate answer.
Air pressure is just the weight of air above. 1 atmosphere is about 14.7 pounds per square inch, so there’s about 14.7 pounds of air above every square inch of earth.
The surface area of the earth is about 197 million square miles. 14.7 lb / inch^2 * 197 million mile^2 = 5.3e18 kg. That’s how much air the earth has.
When they talk about ppm concentration of CO2, they mean by volume. CO2 is heavier than air (44 g/mol vs. 29 g/mol). So 1 ppm of CO2 by volume is about 1.5 ppm by weight, i.e. 1.5 kg of CO2 in 1 million kg of air would be 1 ppm. So, 1ppm CO2 in 5.3e18 kg air would be 8.0e12 kg, or 8 trillion kg, or 8 billion tons.
I think I used too few significant digits, or made an approximation somewhere, but this is pretty close to the number given above.
By the way, when we burn fossil fuels, the carbon in the fuel combines with oxygen in the air to create CO2. So only 12/(16*2+12)=27% of the mass of the CO2 comes from fossil fuels (atomic mass of C is 12, O is 16). Which means we only need to burn 2.1 billion tons of carbon to create 8 billion tons of CO2.
And also, we humans burn about 100 million barrels of oil per day. A barrel contains about 0.12 ton of carbon, so that’s about 5 billion tons of carbon we burn just from oil.
Wikipedia also shows 8 billion tonnes of CO[sub]2[/sub] as 1 ppm.
However, this figure may be misleading. If you inject an extra 8 gigatonnes into the atmosphere, 2 gigatonnes will dissolve in the ocean, and another 2 gigatonnes may be absorbed on land, e.g. by plants. So perhaps OP should inject 16 gigatonnes if he needs to increase atmospheric CO[sub]2[/sub] by 1 ppm.