Group Exercise Three

                            Solution


Part a:

1.8 mmol CO2/kJ x 10,000 kJ, x 4 persons x 365 days x 44.01 g
CO2/mol x 1 mol/1000 mmol  =  1.156 x 10^6 grams per year for four
persons.

Part b:

15,000 miles x 1 gal/20 miles x 4 qts/gal x 1 L/1.057 qt x 0.7025
g/mL  x 1000 mL/L x 1mol octane/114.23 g x 8 mol CO2 / 1mol octane 
x44.01 g CO2/molCO2  =  6.145 x 10^6 g CO2/year.

Part c:

1200 kwh/mo x 12 mo/yr x 3.6 x 10^6 J/kwh x 1/0.20 (the efficiency
factor,  needed in this problem) x 1 mol CH4/802 kj x 1 kJ/1000 J
x 1 mol CO2/mol CH4 x 44.01 g CO2/mol CO2 = 1.42 x 10^7 g CO2/year.

Part d:

Breathing:     1.156 x 10^6 g CO2 /4 person years
Automobile:    6.145 x 10^6 g CO2 /4 person years
Electricity:  14.2   x 10^6 g CO2 /4 person years

Total:        21.5   x 10^6 g CO2 /4 person years

From figure of 1.0 x 10^6 kJ consumed per person per day:

1.0x10^6 kJ/day x 4 persons x 365 days x 1mol CH4/802 kJ x 1 mol
CO2/mol CH4 x 44.01 g CO2 /mol = 8.01 x 10^7 g CO2/4 person years
(or 80.1 x 10^6)

Therefore, breathing , driving, and electricity account for about
27% of  our use of energy.

Part e:

8.01 x 10^7 g CO2 x 1 mol CO2/g CO2 x 10 yrs x 1/4 persons =
4.55x10^6  moles of CO2 per person in 10 years.

volume of atmosphere = 1.0 x 10^22 liters
1.0 x 10^22 L x .022 moles/L x 1 mol CO2/10^7 moles air (for 0.1
ppm)    = 2.2 x 10^13 moles.

2.2 x 10^13 moles/4.55 x 10^6 moles/person = 4,830,000 people in
ten
years

      (or 48 million people in one year)