How do you calculate the volume of carbon dioxide produced ( in liters at 102.0 kPa and 25 degrees celsius) by driving a car for 15000 miles asuming gasoline is octane?

Answer 1

Assume the car gets 30 miles per gallon of octane burned.

Voct = 15000 mi / 30 mi/gal = 500 gal

moct = Voct / doct = ( 500 gal ) / ( 6.0 lbm / gal ) = 83.33 lbm octane

noct = moct / Moct = ( 83.33 lbm ) / ( 114.29 lbm/lbmol ) = 0.7291 lbmol

nCO2 = ( 0.7291 lbmol C8H18 ) ( 8 mol CO2 / mol C8H18 ) = 5.833 lbmol CO2

nCO2 = ( 5.833 lbmol CO2 ) ( 453.6 gmol CO2 / lbmol CO2 ) = 2646 gmol CO2

VCO2 = ( nCO2 ) ( R ) ( T ) / ( P )

VCO2 = ( 2646 mol ) ( 0.08206 atm - L / gmol - K ) ( 298.2 K ) / ( 1.01 atm )

VCO2 = 64100 L at 25 C and 102 kPa <---------------

Answer 2

Since gasoline is not pure octane, let's throw that out. And we have to assume your car gets enough air that it can transform all the carbon in the fuel to CO2 with no CO production at all.

The US government claims 1 gallon of gasoline will produce 9.08 kg of carbon dioxide. A little web Surfing shows me that at 1 atmosphere of pressure and temperature of 25 degrees C, one cubic meter of CO2 weighs 1.7845 kg.

Therefore:

First, determine the number of gallons of gasoline you need to drive your car 15,000 miles.

Multiply that by 9.08 to get the right number of kilos of CO2.

Next, divide by 1.7845 to find the number of cubic meters of CO2 produced.

Finally, convert cubic meters to liters and you're done. (Your answer should be in the millions of liters.)

Answer 3

the car gets 25mpg