6. National Football League regulations stipulate that regulation balls must be

Question

6. National Football League regulations stipulate that regulation balls must be inflated to a pressure between 12.5 psi and 13.5 psi. 1n the 2015 AFC game, in a scandal known as “deflategate”, the New England Patriots’ footballs were found to be underinflated by as much as 2 psi, prompting accusations of ball tampering. On average, the Patriots’ balls were measured at half-time at 11.1 psi. Some experts, including famous scientists such as Neil DeGrasse Tyson, claimed that such a large drop in pressure could not be accounted for with the ideal gas laws and plausible temperature drops. It later became clear that even experts confuse gauge pressures with absolute pressures. Don’t think it matters? Let’s see: (calculator permitted)

a. Assume that the balls were initially inflated to the lower allowed limit of 12.5 psi. Convert this to SI units. Is this gauge pressure or absolute pressure? How can you be sure?

b. The game was played in cold, wet conditions. The outside temperature was 50 °F. Use the ideal gas law [PV=nRT], assuming that the balls did not change volume, to estimate the indoor temperature at which the balls must have been inflated, to account for the outdoor (cold) reading of 11.1 psi. Do this both with the assumption that the readings were gauge pressure AND absolute pressure readings.

c. Comment on the plausibility that the deflation was simply caused by a temperature change. The Patriots were fined $1,000,000. Do you see why it really matters whether a reported pressure is a gauge pressure or an absolute pressure?

Explanation / Answer

(a) 12.5 psi - 86184.47 Pa since 1 psi - 6894.757 Pa. Now let's assume this is absolute pressure then it must be greater than atmospheric pressure which is 101325 Pa approximately because pressure inside the football should be more than outside to make it inflated. But that is not the case so it must be gauge pressure.

(b) Assuming 12.5 psi as absolute pressure -

For the outdoor case using ideal gas equation PV=nRT

So, the equation becomes 11.1V=nR50....(1)

For the indoor case 12.5V=nRT.....(2) now dividing equation (1) by (2) we get T= 56.31°F.

assuming 12.5 psi as gauge pressure-

then actual pressure becomes 12.5+atmospheric pressure (14.7) = 27.2psi

Now solving as the previous one we get T = 122.52°F

(C) From the above calculation we can see there is a drastic change in temperature as we change our assumption. So this really matters when you consider gauge pressure and

absolute pressure. And it is really plausible that deflation was caused by temperature change because change in temperature is causing change in pressure.

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