Skeptically Speaking...(part II)

Here's part II of the problems I solved for Skeptically Speaking:

Problem 6:

Physicists say it's possible for all the air molecules in a room to spontaneously slide over into one corner, suffocating everyone in the room. What are the odds of that actually happening?

Solution:

Assuming an ideal gas^*, the probability of a molecule being on any side of the room is ½. The probability of n molecules all being on the same side is then (½)ⁿ. The density of air is about 1.2 kg/m³ at sea level and 20^° C. A reasonably large room might be 10 m by 10 m by 3 m giving a total volume of 300 m³. Multiplying the density of air by this volume gives 360 kg of air in total. Most of the air is made up of nitrogen with a molecular weight of 0.028 kg/mol. Dividing this into the total mass of air gives about 13,000 moles or n = 7.7x10²⁷ molecules.

Therefore, the probability of finding all the gas molecules on one side of the room is ½ raised to the 7.7x10²⁷th power. I’m not going to even attempt to write out this number. It is extraordinarily small. “How small?” you say. If you were to try to write out all the zeros, you would cover the surface of 10 million Earths.

^*A real gas is non-ideal. If you want to do this problem rigorously, you need to account for the fact that each molecule takes up a finite amount of space.

Problem 7:

If the Skeptically Speaking team got locked in the on-air booth, how long would it take for them to run out of air? We have 5 people in the booth every show, and the booth is about 10 ft by 6 ft.

Solution:

Technically, you’d never run out of air, but you would run out of breathable oxygen, which gets converted to CO₂. Assume you take a breath once every 3 seconds. According to Hypertextbook.com, human lungs have a volume of about 5000 cm³. Oxygen makes up about 20% of air by weight. A quick web search shows that about 25% of the oxygen we breath in gets converted to CO₂. Using the density of air from the previous problem, we can compute the amount of oxygen lost per second:

(5 people)

*(5000 cm³ per breath per person)

*(1/3 breaths per second)

*(1.2 kg of air per m³)

*(10^-6 m³ per cm³)

*(0.20 kg of O₂ per kg of air)

*(0.25 kg of O₂ lost per kg O₂ breathed in)

-----------------------------------------------------

5.0x10^-4 kg of O₂ lost per second

Assuming the room has10 ft ceilings, the total volume is then ~17m³. From this, the total starting mass of O₂ can also be calculated:

(17 m³)

*(1.2 kg of air per m³)

*(0.20 kg of O₂ per kg of air)

--------------------------------------

4.1 kg of O₂

Dividing the amount of O₂ lost per second into the total amount of O₂, we get ~8200 seconds or roughly 2.5 hours.

Problem 8:

How likely is it that we inhale a molecule breathed by Carl Sagan?

Solution:

This is a pretty hard problem for a number of reasons. For starters, air molecules are constantly reacting to form new molecules. Second, it's not clear how long it would take Carl Sagan's air molecules to mix evenly across the globe. Third, Carl Sagan may have breathed in the same air molecules more than once.

To simplify, let's assume that air molecules get evenly distributed around the world very quickly, that Carl Sagan never breathed the same molecule twice, and that air molecules don't react. Strictly speaking, only noble gases won't react, but this assumption might be approximately true for nitrogen since it's fairly inert. Dr.Sagan lived 62 years and there's roughly 32 million seconds in a year. From this and the numbers used in the previous problems, we can estimate the number of molecules breathed by Dr. Sagan:

(5000 cm³ per breath)

*(1/3 breaths per second)

*(3.2x10⁷ seconds per year)

*(62 years per lifetime)

*(1.2 kg of air per m³)

*(10^-6 m³ per cm³)

*(1 mole per 0.028 kg of air)

*(6.022x10²³ molecules per mole)

-----------------------------------------------

~8.5x10³¹ molecules breathed in a lifetime

According to Hypertextbook.com, the mass of the whole atmosphere is 5.3x10¹⁸ kg. If we assume that air molecules have an average molecular weight equivalent to nitrogen's 0.028 kg/mol, then there should be about 3.2x10⁴² molecules in the atmosphere. This means that the probability P of a molecule being "Saganized" is then,

P = (8.5x10³¹) / (3.2x10⁴²) ~2.7x10^-12

Here's where it gets a little tricky. The probability of a single molecule not being breathed by Sagan is (1-P). This is very close to 1, so you might be tempted to think it's very unlikely that to share any of your molecules with Sagan. However, the probability that none of the molecules you breath were ever breathed by Dr. Sagan is given by (1-P)ⁿ, where n ~ 8.5x10³¹ is extremely large. Since the exponent is so large, the probability of never breathing a Saganized molecule will be very close to zero. In short, you breathe a lot of molecules breathed by Carl Sagan!

Problem 9:

I met my girlfriend online, and she's totally the one, but she lives on the other side of the continent. I've been wondering about the odds: not just that we would meet, but that we'd both be born at this time in history when the Internet exists and allows us to interact over such a large distance.

Solution:

I did a similar problem in How Many Licks? In it, I estimated the probability of finding Mr./Mrs. Right is about 1 in 200,000, but this is assumed that special someone was born at the same time in history. What if the love of your life was born in a different time period, like in the Christopher Reeve/Jane Seymour movie Somewhere in Time?

To start, I looked up the world population as a function of time. If you assume an entire population dies off after about 70 years (1 lifespan) and is replaced by an entirely new population, you can trace back about 1000 different to 70,000 B.C. With this assumption, you can calculate the total number of humans who have ever lived on the planet and then calculate the fraction of people living today. Surprisingly, I computed about 30% of people who have ever lived are still living. I was a little skeptical of this, so I decided to look it up and found that other people have done this problem as well and they estimated 5.8%. Since these numbers are easily within an order of magnitude of each other, they're likely close to the actual result. As such, the chances of both lovers being born at the same time in history are much better than the chances that they are born in the same geographical location. This is very surprising, but it shows just how fast exponential growth is. In any event, the probability of finding that special someone is 1 in about 600,000.

That's all for now. Thanks again to Desiree, the Skeptically Speaking crew, and the audience for coming up with some great questions!