Is there enough Oxygen in the atmosphere?
Have you ever wondered if there is enough Oxygen in our relatively tiny atmosphere to run all our cars, planes, and so on?
Oxygen is burned (and therefore taken out of our atmosphere) in these cars, planes, trains, fires in prodigious amounts. Will there be a concentrated enough amount of Oxygen for us, other animals and plants to breath comfortably for long?
I've wondered about this.
What about all the Carbon Dioxide that results from this burning? Only about 0.03% of the atmosphere consists of Carbon Dioxide -- a Greenhouse gas. How much is being added to the atmosphere each year?
What about burning Natural Gas for heating, generating electricity?
So the following are some calculations to check on this.
The following calculations are only for the world's Gasoline (Petrol in England) consumption.
We are NOT considering the burning of Natural Gas, Coal, Diesel, Wood or any other fuel.
(1) What is the total mass of the atmosphere?
(a) Calculate the surface area of the Earth:
Area of a sphere = 4 x Pi x Radius2, Radius2 means Radius Squared.
The Radius of the Earth is 6.37 x 106 m
Area of the Earth = 4 x Pi x ( 6.37 x 106m)2
= 5.099 x 1014 m2 (5.1 x 1014 m2)
(This agrees with E. Kossina's calculations.)
Area of the Oceans is 3.61 x 1014 m2
Area of the Continents is 1.49 x 1014 m2
Mean Height of the Continents is 875 m
Fraction of Earth's Surface that consists of Continents is 1.49/3.61 = 0.292
Mean Height above Sea Level of both Continents and Oceans is 0.292 x 875m = 256 m
Atmospheric Pressure at Sea Level is 101.325 KPa (kiloPascal) or 101,325 N/m2
At 256 m elevation the atmospheric pressure is less, 0.97039 times the pressure at sea level.
0.97039 x 101,325N/m2
= 98,325 N/m2
Total Force of the atmosphere on Earth's surface is 98,325N/m2
x 5.10 x1014m2
= 5.02 x 1019 N
Equivalent mass of atmosphere: F=mg ---> m=F/g ---> m = 5.02 x 1019N / 9.8N/kg
m = 5.12 x 1018 kg
(This agrees closely with 5.14 x 1021grams or 5.14 x 1018kg given in the Handbook of Chemistry and Physics)
(2) How much Volume and Mass of Oxygen is in the Atmosphere?
|Gas||% of Atmospheric Molecules||Molecular Mass (amu)||% times Molecular Mass|
|Carbon Dioxide (CO2)||0.03||44||1.32|
Fraction of atmosphere consisting of Oxygen by mass: 670.4 / 2895.44
= 0.23153 or 0.23
The volume of 1 mole of any gas is 22.4 L at STP.
Mass of Oxygen (O2) in the atmosphere is: 0.23 x (5.12 x 1018kg) = 1.1776 x 1018 kg (1.2 x 1018 kg)
This is in the right "ball park". The estimate in the Handbook is 1.5 x 1018 kg.
1 mole of O2 has a mass of 0.032 kg
Number of moles of O2 in the atmosphere is: 1.2 x 1018kg / 0.032 kg/mole = 3.75 x 1019 mole
Number of Litre of O2 in the atmosphere is: 3.75 x 1019mole x 22.4L/mole = 8.4 x 1020 L.
(3) How many Litre of O2 are used when 1 Litre of gasoline is burned (assuming 100% efficiency)?
Gasoline consists of C8H18, that is each molecule
of Gasoline consists of 8 Carbon and 18 Hydrogen atoms.
Every 2 molecules of gasoline needs 25 molecules of Oxygen and gives off 16 molecules of Carbon Dioxide and 18 molecules of water:
2C8H18 + 25O2 --> 16CO2 + 18H2O
C8H18 + 12.5O2 --> 8CO2 + 9H2O
So, 1 mole of gasoline needs 12.5 mole of Oxygen.
1 mole of gasoline (C8H18) has a mass of 8 x 12 + 18 x
1 = 114 gram
An aside: 12.5 mole of O2 has a mass of 12.5 x 32 = 400 gram
Density of gasoline is 0.75 g/mL (gram/milliLitre)
Using Density = Mass/Volume, therefore Volume = Mass/ Density, the Volume of 1 mole of gasoline is: 114g / 0.75g/mL = 152 mL
So, the number of mole in 1 Litre of liquid gasoline is: 1L/0.152L/mole = 6.6 mole
1 mole (152 mL) of liquid gasoline uses 12.5mole of O2 gas. Therefore 1L of gasoline uses 12.5L/mole of O2 x 6.6mole = 82.5 mole of O2.
The Volume of 82.5mole of O2 is: 82.5mole x 22.4L/mole = 1848 L.
An Alternate method for the above calculation:
The Volume of 12.5 mole of Oxygen is 12.5mole x 22.4 L/mole is: 280 L of Oxygen.
1L of liquid gasoline uses 6.6 times as
much Oxygen as 1 mole of gasoline would use.
Therefore, 6.6 x 280 L of Oxygen is: 1848 L of Oxygen.
Each Litre of gasoline uses 1848 L of Oxygen.
(4) How much gasoline is used by the World in one year?
Stats from earthtrends.wri.org: For 1997 the
population of the Earth was: 5,821,127,100. For 2004 it is: 6,364,315,000,
an increase of 9.3%.
The amount of gasoline used in the world per capita in 1997 was: 161.4 L/y.
In comparison North America 1997 per capita consumption of gasoline was: 1519.8 L/y.
Therefore the total gasoline consumption in
the world for 1997 was: 161.4L/person/year x 5,821,127,100person = 9.395 x 1011
Total amount of O2 used in burning this amount of gasoline is: 1848L of O2/L of gasoline x 9.395 x 1011L of gasoline/y
= 1.7 x 1015 L of O2 per year for the whole world.
(5) How many years at this rate of burning will all the O2 in the atmosphere be used?
To use all the O2 would require:
8.4 x 1020L/1.7 x 1015L/y = 494,118 years or rounded off to about
There would be no Oxygen left, none even to breath. BUT, this is a long, long time!
How long would it take to decrease the Oxygen
from 20.95% of the atmosphere to 19.95% of the atmosphere?
1% x 494,118y/20.95% = 23,585 y or about 24,000 years.
In terms of a human's lifetime this seems like a long, long
Remember we are NOT considering the burning of any other fossil fuel or wood or anything that needs Oxygen.
(6) What about Carbon Dioxide, a greenhouse
How much does it increase with the burning of gasoline alone?
1 mole of gasoline produces 8 mole of Carbon
There are 6.6 mole of gasoline in 1 L.
So, for every Litre of gasoline burned produces 6.6 mole of gasoline x 8mole of CO2/mole of gasoline = 52.8 mole of CO2.
Volume of CO2 gas given off when burning 1 L of gasoline is: 52.8mole x 22.4L/mole = 1164.8 L
The fraction of the atmosphere by mass that is
CO2 is: 1.32/2895.44 = 0.00046 or 4.6 x 10-4
Mass of CO2 in the atmosphere: 4.6 x 10-4 x 5.12 x 1018kg = 2.355 x 1015 kg (or 2.4 x 1015kg)
1 mole of CO2 has a mass of 0.044 kg.
Number of moles of CO2 in the atmosphere is: 2.4 x 1015kg/0.044kg/mole = 5.5 x 1016 mole
Number of Litre of CO2 in atmosphere is: 5.5 x 1016mole x 22.4L/mole = 1.23 x 1018 L.
From earlier, the volume of gasoline used
worldwide (1997) in one year was 9.395 x 1011 L/y.
Total Amount of of CO2 produced is: 9.395 x 1011L of gasoline/y x 1164.8 L of CO2/L of gasoline = 1.1 x 1015 L of CO2/y.
% increase in CO2 per year = 1.1 x
1015/1.23 x 1018 x 100% = 0.089% or 8.9 x 10-4
Total % increase in CO2 in one year is: 8.9 x 10-4 x 0.03 = 2.7 x 10-5
So, in one year the Carbon Dioxide content of the atmosphere would change from 0.03% to 0.030027%.
To increase CO2 by a third: 1.23 x 1018/3
= 4.1 x 1017.
would take: 4.1 x 1017/1.1 x 1015 = 372 years.
Remember this is only due to burning gasoline, NOT Natural gas, coal, diesel etc.
(7) What about Natural Gas?
Natural Gas is 93% methane (CH4).
When it burns, 1 Litre of methane uses 2 Litre of Oxygen and produces 1 Litre of Carbon Dioxide. and 2 Litre of water:
CH4 + 2O2 --> CO2 + 2H2O
An e-mail from Brother John:
O. Hooge, B.C. Canada
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