Month: October 2017

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?

Math that matters (Part I–Missing Women)

If we want kids/adults to learn math, we might as well make it relevant. Here are a few relevant calculations (that employ nothing more than algebra) which I find very relevant to our future. Imagine these calculations being taught to an 8th grade algebra class! Here is the first installment:

Missing women

Most people are not aware that females were systematically removed from the population during the 20th Century and it is a practice that continues today. How do we know? Well, as Nobel Prize winning economist, Amartya Sen, noted back in the 1990s, if we look at sex ratios of nations, we find several that have ratios that are far from 1:1. Pakistan and China have ratios of 0.94:1 and India has a ratio of 0.93:1 (in 2016) (these numbers are pretty much the same as they were in 1990, though Pakistan has improved slightly from 0.91:1). Given that women live longer than men, nations should have sex ratios above 1–most European nations are above 1.03:1. Given these “small” differences among nations, one might just dismiss the low ratios as “normal” variation. Unfortunately, this would be a huge mistake. Here is the math to determine what a ratio of 0.93:1 means, in comparison to a 1.03:1.

First, let’s define the variables needed:
F = number of females in a population
M= number of males in a population
T = total population = F + M
R = sex ratio = F/M

So the above two equations have 4 variables (F, M, T, & R)…if you know two (and you do, T and R, from Internet sources), you should be able to use simple algebra to compute the other two, F and M.

Again, the equations are: (1) T = F + M and (2) R = F/M

Here is how you solve these two equations:
Solving (2) for M yields (3) M = F/R, substituting (3) into (1) yields, F + F/R = T; this can be rewritten as: F(1+(1/R)) = T
which can be rewritten as
(4) F = T/(1 + 1/R)

So, you can determine how many females are in a population using this equation. This can be considered the Actual Females (Fact).

So, with a population of 1 billion (1,000,000,000; which is smaller than both India’s and China’s current population) and a sex ratio of R=0.94, we use equation (4) to solve for Fact as such:

Fact = 1,000,000,000/(1 + 1/0.94) = 485 million
So, Mact = 1 billion – 485 million = 515 million

Now to determine the Expected Females (Fexp) in a “healthy” society, with F/M = 1.03, we use equation (4) again with this new R value.
Fexp = 1,000,000,000/(1 + 1/1.03)) = 507 million
So, Mexp = 493 million

Now you can determine the “missing females” (Fmiss) using this simple formula:
Missing Females = Fmiss = Expected Females – Actual Females = Fexp – Fact

In our example above (the hypothetical nation of 1 billion people), we find:

Fmiss = 507 million – 485 million = 22 million

Is this a large number? Well, when one considers that between 50-60 million people died in World War II, I’d say it is! Also, this is only for one country (say China or India). If you were to add up all the nations in the world with “missing women,” it comes to close to 100 million! Now that is an abominable figure, isn’t it? Yet, how many of you have heard of this figure before? If you are wondering why women are missing, do some research. It isn’t a pretty story. (I wrote about this issue over 10 years ago and got it published in a local paper’s front page. Sadly, as I recall, it hardly drew any attention.)

Just to put these numbers in perspective it is sometimes valuable to imagine what a sex ratio looks like when you bring it down to a scale that we can see. Let’s say, if you had a party of 100 people and a sex ratio of 0.94, you would have 52 men and 48 women. This would hardly be noticeable, would it? Hence, now we see why we need to do the large-scale calculations to expose something very sinister.