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Ever since the AFC championship game, and into the run-up to the Super Bowl, there's been a lot of talk about deflated footballs. And in particular, to what degree can the pressure of a football go down-- to what degree can it get deflated-- based on temperature differentials from where it was inflated to where it is actually played. Especially if it's inflated in a warmer environment, say indoors, and then it's played outdoors, in a colder environment. And so to think about this, we'll take out one of my favorite laws or equations, the ideal gas law. And it tells us that pressure times volume is equal to the number of molecules of our ideal gas we're dealing with-- it's measured in terms of moles, which is the measure of the number of molecules-- times our ideal gas constant, times our temperature. Or if we just want to rearrange it-- and I do want to rearrange it, because I just want to solve for P here-- we can divide both sides by V, and we would be left with P is equal to nRT over V. Now let's think about the exact case of what happened with the New England football-- what the argument is for why they might have deflated naturally-- why there might not be foul play. The argument is maybe they were inflated indoors where the temperature was room temperature-- so roughly 75 degrees Fahrenheit, which is roughly 24 degrees Celsius, and in order to apply the ideal gas law, we'd want to convert it to Kelvin, so this is approximately 297 Kelvin. That's the indoor temperature. So if we're infated here, indoors, and then we play outdoors, where the temperature is roughly 50 degrees Fahrenheit-- and that, from what I've read, was actually the temperature on the field during the New England game. 50 degrees Fahrenheit, which is approximately 10 degrees Celsius, which is approximately 283 Kelvin. So the argument being made is, look, you have this temperature drop, could that temperature drop, if this temperature drops by a certain percent, could that account for the pressure drop that was actually measured? And the pressure drop that was actually measured, gauge pressure-- and this is an important point, because it was glossed over in some of the initial analyses-- the gauge pressure indoors, where the ball was inflated, was 12 point 5 pounds per square inch, and the gauge pressure outdoors, where it was felt the balls were deflated somehow, or where they had less pressure, was 10 point 6 pounds per square inch. So, to see if these numbers are consistent with no foul play, let's simplify our ideal gas law a little bit-- make it a little bit particular for this circumstance. So in this circumstance, we're going to assume no foul play. If we assume no foul play, that means no air is let into or out of. So the number of molecules aren't going to change. The ideal gas constant obviously shouldn't change, it's a universal gas constant, so this should be constant. And then our volume-- let's just assume the football is made out of leather, it's relatively rigid-- its volume might have changed a little bit, but for the sake of our analysis, we'll say that it pretty much held up its shape. So its volume stayed constant as we went from indoors to outdoors. So to simplify this equation, let's just call all of this stuff in green K. If you take your n, multiply it times R, divide it by V, that's just equal to some constant called K. So we can simplify all of this to P, pressure, in this case, should be some constant times temperature. Because once again, we're assuming volume doesn't change, we're assuming no air is let in or out. So when you look at it this way, any percent change on temperature should have the same percent change in pressure. So let's see if the percent change in temperature is consistent with the percent change in pressure that was actually observed. And so when we look at the temperature, we should be doing it in Kelvin, so we should look at this change here, going from 297 Kelvin to 283 Kelvin. So what is the drop here? Let's see. It's a drop of 14 Kelvin, let me get my calculator out. It's a drop of 14 Kelvin. And we started at 297, so divided by 297, let's just say if we round to the nearest percentage, gets us zero point zero four seven, that's rougly a five percent drop. So this is a five percent drop. And I'll write approximately 5 percent drop. Now let's see what the drop in pressure is. So this is going from 12 point 5 psi to 10 point 6 psi. That's 1.9 psi drop. So that's 1.9 psi drop divided by our start, which is 12 point 5, which looks like a 15 point 2 percent drop, or roughly a 15 percent drop. So things are starting to look shady. This looks like a 15 percent drop. And this is actually what some of the initial analyses did, they said, look, something clearly shady happened, because temperature by itself should only account for a 5 percent drop in pressure, but there was a 15 percent drop in pressure, so maybe some air was let out somehow. But there was actually a mistake in that initial analysis. Gauge pressure, the 12 point 5 psi, that's actually not the absolute pressure. What gauge pressure is a measure of is how much more pressure you have inside than outside. Outside the ball, you do have pressure, standard atmospheric pressure. And it might change depending on the weather, et cetera. But there is pressure. The pressure is caused by the weight of the atmosphere. So, in order to figure absolute pressure, and to figure out the correct percentage change, you need to add how much more pressure there is inside the ball to what the outside pressure is. And I don't know the exact readings for that day in New England, but standard atmospheric pressure is 14 point 7 psi. So plus 14 point 7 psi gets us to, 27 point 2 psi. And if we add the 14 point 7 over here, again, to get the true absolute pressure-- once again, the gauge reading is just how much more pressure inside the football than outside. So if you want the true pressure, the absolute pressure, you add these. And this is going to be 25 point 2 psi. Now we can calculate the actual percentage drop, and compare it to this 5 percent drop. Minor mistake, this is 25 point 3 psi. So we still have the same drop, we have a 1 point 9 psi drop, to go from 27 point 2 to 25 point 3. But it's over a larger base. So it should be a lower percentage. Let's calculate what that is. We have 1 point 9 psi drop, divided by 27 point 2 psi. Is equal to a little under a seven percent drop. So approximately a seven percent drop. Now these two things are not exact. But they are a lot closer. And this degree, you're like, okay, there might have been other factors at play. Maybe some of the measurements weren't done exactly right. Maybe actually the ambient pressure was somehow different inside and outside. There's also some possibility, especially when you're outside, that some water, some precipitation is on the ball. As that precipitation evaporates, it might cool down the ball further, which would drop the pressure even more. So this isn't definitive, but at least the numbers here, when we apply the ideal gas law properly, they pretty much account for most of the pressure drop. So at least here, I wouldn't get too caught up in all of the conspiracy theories.