# If you were to empty a full tank 16 gallon tank of fuel, how

1. Oct 20, 2015

### tyvus

I am kind of wondering how much for example CO2 you would put out or other greenhouse gases would be emitted everytime i use a full tank of gas. It would help me if you could explain it in volume, like how much it would take to contain the emissions on its own, and how much it would take to contain the emissions if it were to be compressed. I want to be able to imagine a sort of smoke bubble around my car representing how much gases i emitted into the atmosphere.

I think one of the real questions is how would I find out how the liquid would convert to a gas and how that would change the proportional container volume.

2. Oct 21, 2015

### Simon Bridge

You get w=19.64lbs CO2 per gallon and it has a density d of 0.12 lbs/cf (at room temp) and there are k=0.133 cf/gal ...
So X gal of gasolene turns into V=Xw/(kd) gallons of CO2.

3. Oct 21, 2015

### bigfooted

All of your gas will be converted to CO2 and water*. They are both greenhouse gases, so your entire tank will be converted to greenhouse gases.

To compute how much volume an amount of gas occupies, use the ideal gas law: https://en.wikipedia.org/wiki/Molar_volume
To compute how much CO2 and H2O you get from burning gasoline, you can use the combustion reaction of octane:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
So 1 mole of octane produces 8 moles of CO2

so: convert mass of fuel into moles of fuel, multiply by 8 to get moles of CO2, then multiply by molar volume to find volume of gaseous CO2.

*There will be small amounts of CO, NO and other stuff, but it is negligible compared to CO2 and H2O

4. Oct 21, 2015

### Simon Bridge

Ideal gas law is only good for monatomic gases. It can be adjusted for di-atomic gasses via equipartition of energy but when you get complicated bendy molecules like H2O and CO2 it gets trickier - best to just look up the densities in a table.

5. Oct 22, 2015

### cjl

For these purposes, at reasonably low density and pressure, ideal gas works fairly well, even for complicated molecules.

6. Oct 22, 2015

### Simon Bridge

Be that as it may, we need to hear back from OP.