Author’s Note: What follows is not normal TAH fare. But I’ve seen so much scientifically illiterate speculation and bogus commentary on this particular issue that I could damn near hurl. And all our regular readers know how I am about running numbers to ground. (smile)<\/i><\/p>\n
That illiteracy includes a whopper of a “rookie” mistake that appeared in the original version of this article (forgetting to change from gauge pressure to absolute pressure – hey, it’s been approaching 40 years since I took thermodynamics, so I plead the proverbial “senior moment” here).\u00a0 It’s corrected below.<\/em><\/p>\n . . .<\/b><\/p>\n We\u2019ve all heard about \u201cdeflate-Gate\u201d recently.<\/p>\n A quick refresher: 11 of the 12 game balls used by the NE Patriots in their conference championship game were found to be substantially underinflated. The Patriots claim that the balls were inflated to the lower end of the NFL\u2019s specification (between 12.5 and 13.5 PSI – gauge pressure – inclusive) prior to inspection by the game’s referee. What I’ve seen indicates the underinflated balls were reportedly later all found to be approx 2 PSI below the lower limit.<\/p>\n Each team provides their own game balls for use on offense during the game. The balls are inspected prior to the game, and marked as compliant by the game\u2019s referee. They are then in the possession of that team throughout the game.<\/p>\n Various theories have been offered. Chief among them is, \u201cThe cold temperature during the game caused the pressure to drop.\u201d The NE Patriots claim no wrongdoing, and to have followed the NFL\u2019s rules \u201cto the letter\u201d.<\/p>\n Well, let\u2019s look at this.<\/p>\n A Bit of Science Background Provided its chemical composition remains unchanged, gas behavior in a closed system is governed by the ideal gas law<\/p>\n PV=nRT<\/p>\n where<\/p>\n P is pressure – absolute, not gauge This equation is equivalent algebraically to<\/p>\n P = nRT\/V<\/p>\n Since R is a constant, if there\u2019s no change in chemical composition of the gas only 3 things can change the pressure of a gas in a closed system. Those are (1) a change in temperature, (2) a change in volume, (3) removal of some of the gas molecules, or (4) a combination of the three. It really is that simple.<\/p>\n Change in Temperature<\/span><\/p>\n OK, now let\u2019s assume the change was due solely to a change in temperature.<\/p>\n 10.5 and 12.5 PSI (gauge) equate to 25.2 and 27.2 PSI (absolute), respectively. The balls in question were found to be inflated to (25.2 PSI) \/ (27.2 PSI) = 0.9265 of the minimum allowable pressure. However, they were inflated in (presumably) environmentally controlled conditions \u2013 let\u2019s say 70F. The game conditions were in the high 40sF \u2013 let\u2019s say 47F. Could the game temperature being 47 F have caused the discrepancy?<\/p>\n In a word: no.<\/p>\n Unlike a balloon, a football\u2019s volume doesn\u2019t change much if any when the internal pressure rises from 25.2 PSI to 27.2 PSI. Rather, the football\u2019s volume remains essentially constant. And if no gas was removed, that means the number of gas molecules is constant between the two pressure measurements.<\/p>\n Let\u2019s let P1 and T1 be the temperature when inspected; P2 and T2 are similarly the later pressure and temperature.<\/p>\n That means we have<\/p>\n 0.9265 = P2\/P1 = (nRVT2) \/ (nRVT1) = T2\/T1<\/p>\n A common mistake here is to fail to convert to degrees Kelvin (yeah, it does indeed matter here). 70F is 294.26 K. Doing the math, we find that<\/p>\n T2 = 0.9265 x T1 = 0.84 x 294.26 K = 272.62 K<\/p>\n 272.68 K is about 31 F. Since the game was played in conditions in the high 40\u2019s F, um, no that isn\u2019t the case here. This was a cool and rainy – but not exceptionally cold – winter day in New England, but the rain wasn’t freezing.<\/p>\n The game temperature (47 F) only explains a pressure drop of about 1.2 PSI, or somewhat more than half of the observed discrepancy. What I’ve seen indicates the 11 underinflated balls were all reportedly around 2 PSI below the lower limit \u2013 or 0.8 PSI lower than can be explained by the game temperature alone.<\/p>\n Change in Volume<\/span><\/p>\n Well, NFL teams reportedly \u201cprepare\u201d the surface of game balls after inspection to ensure the surface is to their quarterback\u2019s liking after they\u2019ve been inspected by the referee. If this \u201cpreparation\u201d process somehow added enough volume to the ball to account for the measured discrepancy, that might explain the incident.<\/p>\n However, we now know the actual pressure drop due to the temperature change from 70 F to 47 F \u2013 1.2 PSI. That makes P1\u2019 \u2013 the expected pressure at 47 F in the absence of tampering – 26 PSI (absolute), or about 11.3 PSI (gauge). So let\u2019s see how much additional volume at 47 F would be required to account for an additional 0.8 PSI pressure drop. Here, T1=T2=47 F. As above,<\/p>\n V1\/V2 = 25.2\/26.0 = 0.969 = P2\/P1\u2019 = (nRT2V1) \/ (nRT1V2)<\/p>\n which means that<\/p>\n V2 = V1\/.969 = 1.032<\/p>\n The observed pressure of 10.5 PSI (gauge) would require an increase in volume of 3.2% at 47 F. Thus, if the \u201cpreparation\u201d process causes a gain in volume of the football by about 3.2%, the observed low-pressure conditions in the 11 of 12 game balls would be explained.<\/p>\n Let\u2019s see if that\u2019s plausible.<\/p>\n
\n<\/span><\/p>\n
\nV is volume
\nn is the number of gas molecules present
\nR is the universal gas constant
\nand T is temperature – Kelvin, not Celsius or Fahrenheit<\/p>\n