Mass of atmospheric carbon dioxide

What is the mass of CO2 in the Earth's atmosphere?

The diameter of the Earth at the equator is 12756.32 km, so let's call the Earth's radius $6 \times 10^{6}$ m.

The atmospheric pressure at the Earth's surface is approximately 1 bar, or $10^{5}$ Pa.

The atmosphere is a mass of various gases gravitationally bound to the Earth's surface. The pressure is the force exerted by this mass of gas per unit area of surface. Consequently the total force exerted by the atmosphere can be obtained by multiplying the atmospheric pressure by the Earth's surface area ($4 \pi r^{2}$, where $r$ is the Earth's radius). That is,

(1)
\begin{split} F &= P \times A \\ & = 4 \pi r^{2} P \\ & = 4 \pi \times \left( 6 \times 10^{6} m\right)^{2} \times 10^{5} Pa. \end{split}

Incidentally, since we are using SI units, we don't really need to think too much about what we are doing. Our answer will automatically also be in SI units. More on this later.

The key to solving this problem comes from Isaac Newton, in particular via the second law $F = m a$, or here in the particular case of gravity $F = m g$, where $g$ is the acceleration due to gravity. At the Earth's surface, $g$ has a value of 9.81 m s-2, which for our purposes is close enough to 10 m s-2.

We can rearrange our equations for $F$ and solve for $m$, which in this case is the total mass of atmospheric gases pressing down on the Earth's surface. (Incidentally, it's reasonable to assume that the entire atmosphere is located at the Earth's surface because the thickness of the atmosphere - about 50 km - is less that 1 % of the radius of the Earth.)

(2)
\begin{split} m \times g &= P \times A \\ & = 4 \pi r^{2} P \end{split}

and so

(3)
\begin{split} m & = \frac{4 \pi r^{2} P}{g} \\ &= \frac{4 \pi \times \left( 6 \times 10^{6} m\right)^{2} \times 10^{5} Pa}{10 m s^{-2}} \\ & \simeq 4.5 \times 10^{18} kg. \end{split}

The fraction of CO2 in the atmosphere is about 400 ppm, or 0.04 %, or about 4 x 10-4. However, this fraction is calculated per unit volume, and because the molar mass of CO2 (0.044 kg) is about 1.5 x the average molar mass of air (about 0.8 x 0.028 kg + 0.2 x 0.032 kg, or about 0.0295 kg) we can estimate the fraction of CO2 by mass to be more like 6 x 10-4.

Multiply 4.5 x 1018 kg by 6 x 10-4 and you get (about) 3 x 1015 kg. This is the correct value, to one significant figure. We could improve it by (for example) not truncating everything to one significant figure. Try it!

PS About those SI units, if you don't trust me, try dimensional analysis. Units of more complex quantities (such as force or pressure or energy) can always be broken down into the units of the more fundamental quantities from which they are constructed (such as mass, length, and time). All you have to do is remember one equation that gives you (say) an energy in terms of these quantities, and you have the SI unit of energy. For example, work = force x distance and force = mass x acceleration (and acceleration = distance/time2). Therefore any other kind of energy has the same units as work, namely, mass x distance2/time2, and the SI unit for energy (the joule, named after James Prescott Joule) is equivalent to 1 kg m2 s-2.

Returning to our calculation above, and neglecting constants, we have distance2 x pressure (i.e. mass x acceleration/distance2) divided by acceleration, in other words our final quantity is indeed a mass. And with everything else in SI units, our final mass is in kg.

page revision: 13, last edited: 06 Aug 2009 16:01