To put in perspective the CO2 emissions by human transportation, plant photosynt
ID: 888003 • Letter: T
Question
To put in perspective the CO2 emissions by human transportation, plant photosynthesis and ethanol production, do the following calculation.
a) What are the annual CO2 emissions based on your vehicle use, vehicle's fuel consumption and a justified general combustion equation? Please use 10,000 miles as a reasonable estimate for annual vehicle use and use a published gas mileage for any specific car.
b) Let's assume a green area produces 330 g m-2 year-1 of plant material (50% is carbon) through photosynthesis. Based on this information, what surface or green area is needed to fix all your CO2 emissions (calculated in part a) into biomass?
c) What energy do you use annually for your transportation needs, assuming that all energy comes from the complete combustion of the gasoline (cf part 1a)? If you would run on E85, (assuming you need 85% of your energy supply from ethanol), how much ethanol would you use? Assuming all ethanol comes from corn: In this
case you could get 2.5 gallons of ethanol per bushel (~25kg) of corn. Or a typical corn yield is 125 bushels per acre. How large of a corn plantation do you need to supply for your annual gas needs? If you think that the US has more than 200 million cars, could this be a reasonable alternative?
Explanation / Answer
a) What are the annual CO2 emissions based on your vehicle use, vehicle's fuel consumption and a justified general combustion equation? Please use 10,000 miles as a reasonable estimate for annual vehicle use and use a published gas mileage for any specific car.
The rate, 22 mpg = 22 miles per gallon
Calculate total fuel consumption = 10000 mi / (22 mi / gal) = 454.54 gallons of fuel required (1 year )
assume this si mainly octane
0.703 g/mL = 0.703 kg/L
454.54 gal = 1720.62107 liters
total mass = D*V = 0.703 kg/L * 1720.62107 L = 1210 kg = 1.21*10^3 kg = 1.21*10^6 grams of octane
recall that
1 mol of octane = 8 mol of CO2
114.23 g of octane = 8*44 = 352 g of CO2
(1.21*10^6 g of octane ) ( 352 g of CO2 / 114.23 g of Octane ) = 3728617.70113 g of CO2 = 3.73*10^6 g of CO2
b) Let's assume a green area produces 330 g m-2 year-1 of plant material (50% is carbon) through photosynthesis. Based on this information, what surface or green area is needed to fix all your CO2 emissions (calculated in part a) into biomass?
3.73*10^6 g of CO2
1 mol of CO2 = 1 mol of C
44 g of CO2 = 12 g of C
(3.73*10^6 g of CO2) ( 12 /44) = 1017272.72727 g of C = 1.02*10^6 g of C
1 year basis:
Total Area = Mass / Density = (1.02*10^6 g of C) / ( 330 g of C /m2 ) = 3090 m2 required
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