After the 2015 AFC Championship football game between the New England Patriots and the Indianapolis Colts, it was alleged that the New England Patriots had under inflated eleven of their twelve footballs. NFL rules require that the footballs be inflated between 12.5 and 13.5 psi and the Pat's footballs were under by ~2 psi. Claims were made that the lower pressure was due to the change in temperature from the locker room were the footballs were initially inspected and the cold field where the game was played. Assuming that the footballs were properly inflated to the minimum pressure, 12.5 psi, in the locker room where it was 76.5°F (24.7°C), what would the temperature, in °C, have been on the field to explain a loss of 2 psi in pressure.

You will need to assume no change in the volume of the football and no leaks which may or may not be true.

Question

Kaiya Rodriguez

Engineering

27 October, 00:06

After the 2015 AFC Championship football game between the New England Patriots and the Indianapolis Colts, it was alleged that the New England Patriots had under inflated eleven of their twelve footballs. NFL rules require that the footballs be inflated between 12.5 and 13.5 psi and the Pat's footballs were under by ~2 psi. Claims were made that the lower pressure was due to the change in temperature from the locker room were the footballs were initially inspected and the cold field where the game was played. Assuming that the footballs were properly inflated to the minimum pressure, 12.5 psi, in the locker room where it was 76.5°F (24.7°C), what would the temperature, in °C, have been on the field to explain a loss of 2 psi in pressure.

You will need to assume no change in the volume of the football and no leaks which may or may not be true.

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Answers (1)

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Elias Hart · Answer 1

the final temperature would have been 2.81 °C if the pressure drop was 2 psi

Explanation:

if we assume that is no change in volume, there are not leaks present and also that the gas inside the football behaves as an ideal gas, we have:

initial state) P1 V = n R T1

final state) P2 V = n R T2

where P = absolute pressure, V = volume occupied by the gas, n = number of moles of gas, R = ideal gas constant T = absolute temperature

therefore since V = constant (constant volume) and n = constant (no leaks), if we divide both equations

P2/P1 = T2/T1

therefore

T2 = T1 * (P2/P1)

since P1 absolute = P1 relative + P atmospheric (14.7 psi) = 12.5 psi + 14.7 psi = 27.2 psia

also P2 = P1 - 2 psi = 25.2 psia

T1 = 24.7°C + 273 °C = 297.7 K

therefore

T2 = T1 * (P2/P1) = 297.7 K (25.2 psia/27.2 psia) = 275.81 K

thus T2 = 275.81 K = 2.81 °C