Math that matters (Part II–Atmospheric Changes)

Everyone now knows (or denies) that CO2 has been increasing in the atmosphere for the past couple hundred years and this increase is largely responsible for the climate changes that have occurred over that time. And 400 ppm (parts per million) is a number that more and more people are familiar with as well; it represents the average concentration of CO2 in the atmosphere (up from 280 ppm when the Industrial Revolution commenced, ~1750). Well, there are a few other very interesting/compelling numbers that we can determine based on this information alone, including:

(1) What is the average annual growth rate of the CO2 over this period (1750-2017)?
(2) What is 400 ppm in percentage terms?
(3) How many molecules of CO2 are there in a regular sized balloon?
(4) Considering CO2 changes alone, by what percentage has the composition of the atmosphere changed since Industrialization?

Let’s do some calculations:

(1) What is the average annual growth rate of the CO2 over this period (1750-2017)?

Atmospheric CO2 concentration has risen exponentially so we use an exponential growth equation to model it:

(a) Cf = Ci x e^(rt),
where, Cf = final concentration, Ci = initial concentration, e = 2.718 (Euler’s number), r = annual rate of growth, and t = duration of time (^ is used as an exponential symbol)

The variable we want is r (as we have all the other numbers). Solving for r requires using a little “Algebra II”:

First move Ci over by dividing both sides of (a) by Ci and flipping the equation around; this leaves:
e^rt = Cf/Ci

Now, if we take the log of both sides (here the natural log, ln) and use the fact that ln(e) = 1, we get:
ln(e^rt) = ln(Cf/Ci)  rt x ln(e) = ln(Cf/Ci)  rt = ln(Cf/Ci)

Now, if we divide both sides by t, we get an equation solved for r, the desired variable:
r = ln(Cf/Ci)/t

Inputting the values of the three known variables (Cf = 400 ppm, Ci = 280 ppm, and t = 2017-1750 = 267 years) yields the annual growth rate:
r = ln(400/280)/267 = 0.00134 (or 0.134% per year)

If we just look at the growth of CO2 concentration since 1950 (when industrialization became global; CO2’s concentration in 1950 was ~310 ppm), we can once again use the exponential equation above to determine the growth rate in more recent times:
r = ln(400/310)/67 = 0.0038 (or 0.38% per year)

In either case, notice that humans are increasing CO2’s concentration only very slightly each year. (Over the past 267 years, CO2’s concentration has gone up 43%.) However, what may appear “slight” is definitely not slight in terms of its current and future impacts.

(2) What is 400 ppm in percentage terms?

Answering this merely requires us to understand what ppm (parts per million) means. Unlike some words in English, ppm means literally what it says. If the atmospheric concentration of CO2 is 400 ppm then for every million parts (or molecules) of atmospheric gas, CO2 will be 400 of them. With this understanding we can compute a percentage by expressing the amount of CO2 (relative to everything in the atmosphere) as a ratio:
% of CO2 in atmosphere = ppm of CO2/1,000,000 parts of atm = 400/1,000,000 = 0.0004 (or 0.04%)

Again, this is a very, very small amount but as we know this small amount of CO2 is responsible for a significant amount (~30-40%) of the Natural Greenhouse Effect (NGE) and the bulk (~50-60%) of the Anthropogenic Greenhouse Effect (AGE), associated with “global warming” and “climate change.”

(3) How many molecules of CO2 are there in a regular sized balloon?

While (2) tells us what proportion of gases in the atmosphere are CO2 it doesn’t really tell us how much CO2 is a particular volume. If we take a typical balloon (1 foot in diameter; let’s assume it is a sphere), we can find the volume of the balloon using this formula:
V = 4/3 x pi x r^3,
where V = volume, pi = 3.14159, and r is the radius of the balloon. Plugging in the numbers leads to:

V = 4/3 x 3.14159 x (0.5)^3 = 0.52 cubic feet

Since 1 liter = 0.0353147 cubic feet, the volume of the balloon is:
V = 0.52 cubic feet x (1 liter/0.0353147 cubic feet) = 14.7 liters

This seems quite high when one considers a two liter bottle of soda but rest assured this is the correct value.

Now that we have the volume, we’ll need to determine how much CO2 is in such a volume. Here we have to use some chemistry laws; here the most well-known of them all, the Ideal Gas Law:

(c) PV = nRT, which can be rewritten as, n = PV/RT,
where P = pressure, V = volume, n = number of moles of gas, R = the Ideal Gas Constant = 0.082 L-atm/mole-K) and T = temperature in Kelvin degrees.

Assuming that we have a standard atmospheric pressure of 1 atmosphere and we are at room temperature of 27 deg Celsius (equivalent to 81 degrees Fahrenheit or 300 K), we can plug the numbers into equation (c):
n = (1 atm)(14.7 L)/((0.082 L-atm/mole-k)*(300 K)) = 0.60 moles

So what is a mole? Chemists define one mole as 6.02 x 10^23 molecules (or 602,000,000,000,000,000,000,000 molecules).

Notice that our use of the Ideal Gas Law was for all the molecules of gas, not just CO2.

Thus, if the 14.7 L (of a balloon) contains 0.60 moles of gas, we calculate the number of molecules of gas in the balloon as such:

# of molecules of gas (in balloon) = 0.60 moles x (6.02 x 10^23 molecules/mole) = 3.61×10^23 molecules

Now to determine how many of these molecules are CO2 we use the information determined in the previous question thusly:

# molecules of CO2 in a balloon = 3.61 x 10^23 molecules of gas x (0.0004) = 1.4 x 10^20 molecules of CO2

There we have it. There are 140,000,000,000,000,000,000 molecules of CO2 in every balloon (assuming it isn’t filled with helium 😊).

Now, while this balloon’s gas is very light, it certainly is filled with a lot of molecules of CO2. This might lead one to wonder how much does all the CO2 in the atmosphere weigh. Well, this calculation is a bit trickier. According to web references, the entire atmosphere weighs approximately 1.1 x 10^19 pounds (or 5.5 quadrillion tons). We might want to just multiply this by the percentage of CO2 in the atmosphere in order to determine the total weight of CO2 in the atmosphere but we would be wrong to do so; though it wouldn’t be a terribly bad approximation (if we did so we get 4×10^15 pounds). The different molecules of gas have different weights. Thus, we would have to do a calculation that would include all of these different gases and their weights. This is a bit more complicated than I choose to get right now, so suffice it to say that all the CO2 in the atmosphere weighs approximately six thousand times more than all the humans on the Earth combined (which is approximately 7×10^11 pounds).

(4) Considering CO2 changes alone, by what percentage has the composition of the atmosphere changed since Industrialization?

Calculation (1) clarified how small CO2’s concentration is the larger scheme of things, while (2) and (3) suggested how much CO2 there is. Notice, it all depends on how we look at things, as do most things.

The question asked here aims to provide a sense of how much change has occurred in our atmosphere since industrialization.
We know that the atmosphere’s CO2 concentration has gone from 280 ppm to 400 ppm (a 43% increase). And we calculated that CO2 comprises 0.04% of the atmosphere. Thus, to determine how big a change in the atmospheric composition has been caused by Industrialization, we need only to multiply these two percentages together, as such:

Change in Atmospheric Composition since industrialization = % Change in CO2 concentration x % of atm. that is CO2.

Putting numbers in yields,

Change in Atmospheric Composition since industrialization = 43% x 0.04% = 0.017%

Since most don’t work with percentages this small, what does this amount of change mean. Well, if you had 5,814 molecules of atmospheric gas–represented by 5,812 red polka dots (each representing a molecule of gas other than CO2) and 2 blue polka dots (each representing a molecule of CO2)—over the 267 years of Industrialization, one more molecule of CO2 has been added to the atmosphere (or one more blue polka dot). Another way to look at it: If you had 5,814 fans sitting in a gymnasium, 267 years later, one more fan would join the fray. Would you notice a change of this amount? Most definitely not (unless of course the fan was decked out in the rival’s colors 😊). But we can thank modern science for building equipment that can measure such small (yet powerful) changes in the atmosphere. Perhaps the most salient question is: have we built a political/social apparatus to respond to the future/predictions that this information clarifies?

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