Xs, Os, and the Ideal Gas Law

Honk if you predicted that an AFC Championship game would lead to a debate over the proper application of the Ideal Gas Law.

Yesterday, I posted a calculation from WUSA, which explained the Ideal Gas Law and how it applies to the question of whether the weather — specifically, the temperature — could be largely responsible for the fact 11 of 12 footballs used by the New England Patriots in Sunday’s win over the Indianapolis Colts were deflated.

It said this:

We make the following assumptions, based on what we know about the procedure regarding regulation footballs in the NFL and about the Ideal Gas Law:

1) V, the volume of gas (air) in the ball should not change, since (according to procedure), no air is added to or subtracted from the ball after reaching the proper inflation,

2) n will not change for the same reason as above,

3) R does not change, since it is a universal constant.

Now, let’s just change the way the equation looks by moving all the letters to one side of the equation:

pV/nRT = 1
From here, we need to think of this as two different times: the pressure, temperature, etc. from when the balls were checked and the pressure, temperature, etc. out on the field. Let’s set those to be equal:

p1 V1 / n1 RT1 = p2 V2 / n2 RT2,
where the 1 represents the initial readings and 2 represents the readings on the field. Since the volume will not change (assuming no air is added or taken away from the ball), then V1 = V2, and those can be cancelled. For the same reason, n1 andn 2 can cancel. The R ‘s cancel, since R

is a constant. We are left with a simple equation:

p1 / T1 = p2 / T2
Now, we can start solving this puzzle quite easily!

Let’s assume that each ball was inflated to the minimum pressure required to meet the NFL rules regarding proper inflation: 12.5 psi. We convert psi (English) to pascals (Metric), which comes out to 86,184.5 Pa and assume room temperature (68ºF/20ºC) which converts to 293.15 K (Kelvin, the Metric equivalent). We now have,

86,184.5 Pa / 293.15 K = p2 / T2.
We’re down to two variables. But we also know the temperature on the field at the start of the game was reported as 51ºF/10.6ºC (283.15 K). Plug it in…

86,184.5 Pa / 293.15 K = p2 / 283.15 K
Neat! Look, we’re left with a solvable equation with one variable, p2, which is the pressure of the air inside the ball at game time! Let’s solve this riddle…

Isolate the lone variable:

(86,184.5 Pa / 293.15 K) * 283.15 K = p2
83,244.6 Pa = p2 —> 12.1 psi

In the mailbag today, we received this correction to the calculations from Martin Schmaltz, a professor of physics at Boston University, who had also been mentioned in yesterday afternoon’s post.

Dear Bob –

the calculation about the pressure change due to temperature change which you quote from WUSA is not quite right and it makes a difference.

When using the ideal gas law (which is where the formula p1/T1=p2/T2 comes from) you need to use absolute pressure. On the other hand, the
12.5 PSI that everyone quotes is “pressure above atmospheric pressure”.
But it’s easy to fix: you just add the value of atmospheric pressure
(14.5 PSI) to the 12.5 PSI to get absolute pressure. i.e. you do the same calculation that you quoted but use 27 PSI instead of 12.5 PSI. You also do not have to convert to Pascals, that unit conversion drops out in p1/T1=p2/T2.

So here is the corrected version of the calculation that you WUSA guy wanted to do (he wanted to calculate the pressure change for a temperature change from 68 to 51 degrees. I first solve the above equation for p2 and then plug in the numbers. Here goes:

p2=p1 T2/T1 = 27 PSI * ( 283 K / 293 K) = 26.1 PSI

so a drop in pressure by 0.9 PSI (not 0.4 PSI as your WUSA guy incorrectly found)

To summarize, a drop by about 1 PSI is to be expected from bringing the balls from inside to the outside. A 2 PSI pressure drop is a little large, that would require a strangely high indoors temperature, or a much colder outside temperature, or a slightly under-inflated ball to begin with.

Since this temperature induced effect is not small compared to the tolerances given in the rules, and given that it is easy to measure, it must be well-known to NFL experts and refs who inspect the balls.

Best wishes -Martin Schmaltz

Why didn’t any of you pick up that mistake?