Gay-Lussac's Law & the Patriots

Question

This may be a question better suited for xkcd what if? section...but here goes: with all the hoopla around the patriots / colts game and allegations of cheating by deflating footballs, I've read/heard only superficial reference to the notion that inflating the balls at a different temperature than the ambient field temperature could result in a swing in pressure, always by patriots apologists and never by anyone who seems to actually know the science. High school and college me is kicking current me for not being able to do a back of the envelope calculation on this, but I'm in the midst of a site launch with a project at work, so I turn to the stack exchange community:

Assume that the footballs were inflated at ~70 deg F and using the weather underground archive data with a mean of 34 deg F, min of 15 deg F, and max of 53 deg F. Is it at all feasible that the footballs were in fact not purposefully deflated, but instead lost pressure due to the temperature difference where they were inflated versus the field?

Answer 1

The pressure argument

This would indeed be a straight-forward application of Gay-Lussac's law. In the case of the Patriots, we are told that the pressure was 2 psi less than the league minimum of 12.5 psi (select the "Ball" option in the first drop down menu, you'll see it clearly in Section 1 of Rule 2 on the first page).

Gay-Lussac's law says that $$ P_2=P_1\frac{T_2}{T_1} $$ If it was pumped to 12.5 psi (86184.466 N/m$^2$) at 70$^\circ$ F (294.26 K), then it's pressure at 34$^\circ$ F (274.26 K) would be, $$ P_2=86184\frac{294}{274}=80326.779\,\rm\frac{N}{m^2}=11.65\,psi $$ So while it could account for some of the deflation, it cannot account for all of it. But note that the 34$^\circ$ F is the mean temperature for the day, recorded temperatures say it was around 51$^\circ$ F at game time, this leads to $$ P_2=12.05\text{ psi} $$ which, while below the required minimum, is still far from the discovered 10.5 psi. This strongly suggests tampering.

Note also that there has been nothing said in the news about the Colt's balls being deflated by any amount; assuming their balls were also inflated to the league minimum, theirs too should be around 12.05 psi at Gilette stadium (at the league maximum of 13.5, this would drop to a still-legal 13.0 psi). This discrepancy between the two sets of balls suggests tampering of some kind.

Thus, this "temperature" argument is effectively bogus.

Outcome of the game

The Patriots led the game 17-7 at halftime. From the NFL's own investigation,

While the evidence thus far supports the conclusion that footballs that were under-inflated were used by the Patriots in the first half, the footballs were properly inflated for the second half and confirmed at the conclusion of the game to have remained properly inflated.

By the end of regulation, the Patriots scored 4 more times while the Colts refused to score. Former NFL quarterback Matt Leinart has stated that the decreased pressure would actually decrease the velocity and distance of the throw. New England's offensive statistics by half are:

• First Half
  • 95 yards on 11 completions from 21 pass attempts (52% completion, 3 completed passes more than 10 yards)
  • 93 rush yards on 20 rushes
• Second half
  • 131 yards on 12 completions from 14 pass attempts (86% completion, 4 completed passes more than 10 yards)
  • 84 rush yards on 20 rushes

Which suggests that Leinart's assessment is at least plausible and that the Indianapolis Colts were actually at the advantage with the under-inflated balls. But with a sample of 1, we cannot really draw any conclusions from this.

I presume here that the balls were re-inflated on the field so that the temperature (and thus pressure) remained constant, but there is no such information at the moment that states one way or the other.

Answer 2

Gay-Lussac's law requires absolute temperatures and pressures. Absolute pressure is gauge pressure plus atmospheric pressure (about 15 psi), so the initial pressure is 27.5 to 28.5 psi (making lots of assumptions about how the balls were inspected before the game). A drop by 2 psi is a drop of about 7% (maybe anywhere in the range of 5% to 10% depending on precision of the numbers quoted).

If the balls were pumped up in a warm locker room at a temperature of a little below 300 K, then a 7% drop would be about 21°C, or about 38°F. So, if they were inspected accurately and found to be in regulation at 72°F, then by 34°F they would have lost 2 psi.

I had heard the game temperature was not quite that cold. If game temp was 45°F, that corresponds to a drop of 27°F or 15°C, which is still 5%, or about 1.4 psi.

Another contributing factor could be changes in atmospheric pressure. It would not be unheard of for a high-pressure front to come through between the inspection and the end of the game, increasing atmospheric pressure by 1 psi or more, which would drop the gauge pressure by the same amount. So even if the temperature swing were not sufficient to produce the drop, a combination of a smaller temperature drop with a rise in barometric pressure could certainly account for it.

EDIT: I looked up the weather conditions on weatherforyou.com (National Weather Service data). Game temperature in Foxborough was 52°F. The atmospheric pressure actually dropped by about 0.05 psi from the time the balls would have been inspected until half time, so this would have had a negligible effect on gauge pressure. That means the weather-induced pressure drop would have been about 1 psi. Whether that means the Patriots tampered depends on the precision and accuracy of the measurements, but any additional pressure changes they may have caused would have been pretty small.

P.S. Full disclosure: I live in Seattle and am rooting for the Seahawks!

Answer 3

The Real Gas Law states

$$P=\frac{ZnRT}{V}$$

Let's assume the initial volume $V_i \approx V_f$ (this may be part of the error seen in final answer below). The universal gas constant, $R$, and number of moles of gas inside the ball, $n$, remains the same.

Air is mostly nitrogen so I found the compressibility factors for nitrogen at 70 $^\circ $F (294.26 K) and 34 $^\circ$F (274.26 K) assuming the same atmospheric pressure (14.7 psia) using this calculator.(again, this may be part of the error seen in final answer below)

$Z_i=0.99999608122222$ and $Z_f=0.99999336822222$

The ratio of initial and final pressure

$$\frac{P_i}{P_f}=\frac{Z_i n R T_i}{V} \cdot \frac{V}{Z_f n R T_f}= \frac{Z_i T_i}{Z_f T_f} \approx 1.072926$$

and the final pressure from this ratio is (assuming initial pressure, $P_i=12.5$ psig)

$$P_f = \frac{P_i}{1.072926} =\frac{12.5}{1.072926} \approx 11.65 \ \text{psig}$$

Repeating the calculation with the same assumptions, except using the 15 $^\circ$F you cited, the final pressure would be $\approx 11.128$ psig.

Answer 4

The temperature during the 1st half was around 50F. I recall this being shown on my TV screen. This is a drop of only 20degrees (from 70f, or absolute ~530R). By simple gas physics, the pressure could only have dropped about 4%, or about 1/2 PSI. (20/530 = ~3.7%). As a second order effect, the ball would tend to want to shrink a bit too. This would actually tend to reduce the amount of pressure drop. Conclusion: intentional tampering.

Answer 5

@Sofia, can you please make a correction to the edited equation for adiabatic pressurization process? The pressure ratio (P2/P1) is the group [(gamma-1)/gamma]. In other words, if gamma = 1.4, then the exponent = ~0.2857...

This adiabatic pressurization concept was added to argue that If filled to 12.5 psig (27 psia) from just 25% below the target total pressure starting at 70F, then an internal temperature of 85F would result at the time the 12.5 psi was first achieved.

I disagree with Kyle that the temperature argument is bogus.

Finally, I offer this humble calculation based on ideal gas law (in particular, Gay-Lussac's observation for cooling a gas in a closed system):

T1 = 85F (the internal temperature of the ball when intial measurement was taken) T2 = 45F (ambient temperature at which the second measurement was taken)

Then P2 = P1 (T2/T1) (in absolute terms)

P2 = (27 psia)[45F + 460R]/[85F + 460] - 14.5 psia = 10.5 psig (or 2 psi below 12.5)

Therefore, It is quite plausible. This says nothing about intent. It could accidental, or intentional.

Answer 6

Dear Sirs and Madams: A quick pressurization can be approximated as an adiabatic-reversible process.

The increase in temperature in adiabatic compression is estimated as follows:

$T_2/T_1 = (P_2/P_1)^n$, $n=(\gamma-1)/\gamma$); where $\gamma$ is the specific heat ratio, $C_p/C_v = 1.4$ for air.

For example, an 8% increase in absolute temperature could occur with a 31% increase in absolute pressure, given a rapid pressurization. This would be enough to explain the issue.

Its possible that when the pressure was measured at 12.5 psia, the internal temperature was at an internal temperature higher than the local ambient. When the balls later cooled to ambient, an effective degrade in pressure could occur.

To clarify, this adiabatic pressurization concept was added to argue that If filled to 12.5 psig (27 psia) from just 25% below the target total pressure starting at 70F, then an internal temperature of 85F would result at the time the 12.5 psi was first achieved.

I disagree with Kyle that the temperature argument is bogus.

Finally, I offer this humble calculation based on ideal gas law (in particular, Gay-Lussac's observation for cooling a gas in a closed system):

$T_1 = 85^\circ F$ (the internal temperature of the ball when intial measurement was taken) $T_2 = 45^\circ F$ (ambient temperature at which the second measurement was taken)

Then $P_2 = P_1 \frac{T_2}{T_1}$ (in absolute terms)

$P_2 = 27 \mbox{psia}\frac{45F + 460R}{85F + 460} - 14.5 \mbox{psia} = 10.5 \mbox{psig}$ (or 2 psi below 12.5)

Therefore, It is quite plausible. This says nothing about intent. It could accidental, or intentional