This is an old revision of the document!

I thought I'd try going through high school chemistry equations to at least show that even if we haven't been causing global warming, we're certainly causing an increase in CO2 levels.

Gasoline is mostly octane with some other hydrocarbons or alcohols mixed in. “mostly” means about 90%, and those other hydrocarbons and alcohol burn almost the same way. When octane is burned, this happens:

<m>(1) </m><m> C_{8}H_{18} + 17O_{2} right 8CO_{2} + 9H_{2}O </m>

So 1 molecule octane produces 8 molecules carbon dioxide and 9 molecules water.

How many molecules in a litre of octane?

##### Constants
• Density of Octane = 0.703 g/mL
• Molecular Mass of Octane = 114.22852 g/mol

<m> 1L of Octane = 0.703kg </m>

<m> 1L of Octane = {0.703kg}/{0.1142kg/mol} </m>

<m>(2) </m><m> Octane=6.154 mol/L </m>

Note that “mol” is a molar, or <m>6.02*10^23</m> molecules.

So how many molars of <m>CO_2</m> are produced from one molar of octane?

<m>per(1), </m><m> 1 mol C_8 H_{18} right 8 mol CO_2 </m>

<m>per(2), </m><m> 1L C_8 H_{18} = 8(6.154) </m>

<m>(3) </m><m> 1L C_8 H_{18} = 49.232 mol CO_2 </m>

49.232 molars. But this is a gas, so it takes up a lot more volume than a liquid.

How much volume does 49.232 mol of <m>CO_2</m> occupy at ground level?

<m> N_A = 24.7 L/mol @STP </m>

<m>per(3), </m><m> 1L C_8 H_{18} = N_A * 49.232 mol(CO_2) </m>

<m> = 1215.03 L(CO_2) </m>

So 1L liquid octane burns into 1216.03L pure gaseous <m>CO_2</m> at standard temperature and pressure.

The atmosphere is currently at 387ppm <m>CO_2</m>. 407ppm is considered unprecedented (+20ppm).

<m> 1216.03 L(CO_2) * {1,000,000L}/{20L} </m>

<m>(5) </m><m> = 60,801,500L </m>

This means that it takes almost 61 million litres of atmosphere to dilute one litre of <m>CO_2</m> to unprecedented concentrations.

It sounds like a mistake, but it's not quite so insane if you think about it in volume. 1L is 10x10x10cm 1 cubic meter is 1000L. A cube 10m x 10m x 10m is 1 million litres. This means that if you burned a litre of gasoline in your neighbourhood, it would bring up the CO2 levels of a few people's lawns to high levels. Then it would float away and dissipate. Eventually the trees would reclaim it and everything is good right?

I use 1 tank of gas every 2 weeks in my car. That's 26 tanks per year.

My tank is 35L

I have been driving for 10 years.

How much atmosphere is required to absorb this?

<m>per(5), </m><m> 26 tanks * 10 years * 36L * 60,801,500L </m>

<m> =~ 5.53×10^11 L </m>

How big is that?

The atmosphere ends at 120km, but realistically, the density drops off to zero out there, so we can half that to get a reasonable number for a “dense” atmosphere. So it's 60km deep.

Assuming an atmosphere of uniform density and an altitude of 60km, what percentage of a square km is this?

<m> 60km*1km*1km </m>

<m> = 60*10^4dm * 1*10^4 * 1*10^4 </m> <m> = 6.0*10^13 L </m>

<m> { 5.53 * 10^11 polluted }/{6.0 * 10^13 volume} = 0.922% </m>

So my one car for 10 years has raised the CO2 levels to dangerous amounts for almost 1% of a square km of atmosphere.

• Multiply that by millions of cars over decades. 