Deflategate

Despite everything else of note that is happening in the world, it seems that we can't shake our obsession with this somewhat trivial NFL scandal that has become known as "Deflategate." So, at the risk of prolonging this discussion even further, I have decided to use this space to explore some of the scientific questions raised by this issue. Is it possible that 11 of the 12 footballs used by the New England Patriots during the first half of their 45-7 win over the Indianapolis Colts lost 2 psi naturally or did someone tamper with these balls, willfully deflating them? Certainly, it is well known that a drop in temperature will result in a loss of gas pressure. If the balls were inflated at room temperature, could the drop to the outside playing temperature account for decrease from 12.5 psi, the league-mandated minimum, to 10. 5 psi?

If we assume ideal gas behavior (a reasonable assumption under these conditions), we can determine the likelihood of a temperature-induced pressure loss quite easily. From introductory chemistry, you, no doubt, have fond memories of the ideal gas law:

PV = nRT

Here, P is pressure, V is volume, n is moles (ah, moles), R is a constant, and T is temperature. If we assume conditions of constant volume (any change in football volume is likely to be minor), pressure and temperature, in units of kelvins, are directly proportional. This means we can compute the final pressure if we know the initial pressure, initial temperature, and final temperature:

P_final = P_initial(T_final/T_inital)              (1)

Now, the game-time temperature of the Pats-Colts game was 51 degrees Fahrenheit. Because determination of the physical properties of gases requires the use of the absolute temperature scale, we will need to convert Fahrenheit to Celsius first:

(51℉ - 32)(5/9) = 11℃

Converting this to kelvins yields

11℃ + 273 = 284 K

So, the playing temperature was 284 kelvins. Let's assume that the ball was inflated initially to 12.5 psi at 80℉ or 300 kelvins. The 12.5 psi measurement is likely differential pressure, meaning that the actual pressure inside the ball is 12.5 psi plus the external atmospheric pressure (roughly 14.7 psi). The initial absolute pressure would then be

12.5 psi + 14.7 psi = 27.2 psi

When taken outside, the ball temperature will drop from 80℉ to 51℉ (300 K to 284 K). According to equation (1) above, the resulting pressure would be

Remember this is the absolute pressure. To determine the final gauge pressure, we must subtract 14.7 psi:

25.8 psi - 14.7 psi = 11.1 psi

This corresponds to a 1.4 psi loss, less than the measured 2 psi loss. So, what would the inflation temperature have to be to account for a 2 psi drop to 10.5 psi. We can compute this by rearranging equation 1.

T_initial = T_final(P_inital/P_final)

Adjusting the 10.5 psi differential pressure to an absolute pressure of 25.2 psi (after adding 14.7 psi) yields the following initial temperature:

(284 K)(27.2 psi/25.2 psi) = 306 K

This corresponds to an initial temperature of 92 ℉, a hot room, indeed, but, certainly within the realm of possibility. Of course, the officials would have to measure the pressure soon after inflation before the temperature and pressure had a chance to drop significantly.

In the end, it is certainly possible that deflation could occur naturally, but the circumstances required would be quite unusual. During recent press conference, coach Bill Bellichick suggested that the pregame preparation of the footballs may have elevated the pressures to 12.5 psi at the time of measurement, so that the game time drop may have been greater. I'm not sure what this preparation would entail, but the Patriots claim that the rules allow it.

So, did the Patriots cheat? Could the pressure drop have occurred naturally without external intervention? I won't offer my opinion. Instead, I'll let you be the judge!