This is a test of the math2 plugin for dokuwiki.  For the real article, check the one done using the math plugin
-----

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> =~ 5.53x10^11 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.