Question

(a) A commercial 737 jet transporting 143 passengers and 5 crew members from Kansas City to Baltimore burned 11,800 lbs. (about 1700 gallons) of Jet A fuel en route. Jet A fuel is kerosine based, consisting primarily of C_nH_2n+2 hydrocarbons, with n = 6 to 16, so the carbon: hydrogen ratio is close to 1:2. During this flight, how much CO₂ was released into the atmosphere? Assume the combustion of the fuel was complete, so all the fuel was burned to form CO₂ and H₂O. Give both the mass of CO₂ produced (in kg and lb) and the volume it would occupy at 298 K, 1 atm.

(b) How much CO₂ would be released into the atmosphere is those passengers and crew made the trip instead, in pairs, in hybrid cars at 40 miles per gallon. Assume the density of the gasoline is 0.75 kg/L and that carbon and hydrogen dominate the composition in a ration of 1:2. The road trip is 1082 miles.

Note: This is in the complete problem as written in the textbook, in a chapter about gas laws (both ideal and nonideal).

Answer 1

a)

n = 143 passng + 5 crew members

m = 11800 lbs of fuel (CnH2n+2) with n = 6-16, ratio is 1:2

find CO2 fouind in the atms... assume combustion is complete

CH2 + 3/2O2 --> CO2 + H2O

then..

mass of fuel = 11800 lb = 11800*454 = 5357200 g of fuel

assume:

MW of CH2 = 12+2 = 14 g/mol

so

mol of fuel = mass/MW = 5357200/14 = 382657.142 mol of CH2

so ratio is 1:1 with respect to fuel + CO2

so

382657.142 mol of CO2 produced

mass of CO2 = mol*MW = 382657.142*44 = 16836914.2 g of CO2 = 16836.91 kg of CO2

lb = 16836.91 /0.454 = 37085.70 lb of CO2

Volume:

PV = nRT

V= nRT/P = (382657.142)(0.082)(298)/1 = 9350609.92 liter of CO2

b)

in pairs = 2 persons

find gallons required:

1082 miles * gal / 40 mile = 27.05 gallons required

to liter =27.05 gal = 102.39539 liters

mass = D*V = 0.75 kg / L * 102.39539 L = 76.796 kg of fuel = 76796 g of fuel

mol = mass/MW = 76796 /14 = 5485.428 mol of CH2

so

mol of CO2 = 5485.428 mol of CO2 per car

no of cars = 143 passenger --> 143/2 = 71.5 roundup = 72 cars requred

mol of CO2 total = 5485.428 *72 = 394950.816 mol of CO2

so..

mass of CO2 = 394950.816*44 = 17377835.904 g of CO2 =17377.84 kg of CO2