Question

Problem 5 Deflategate (14 pts) In 2014, a controversy erupted in American football known as Deflategate. In this controversy, it was alleged that the New England Patriots tampered with their footballs on the day of a championship game by reducing their pressure, so that they would be easier to grip. On that day, the temperature in the locker room was between 67 – 71 OF. On the field, the temperature was 48 OF. a) (4 pts) If the footballs were inflated to the minimum legal gauge pressure of 12.5 psi (pounds per square inch) in the locker room, calculate the range of pressures that the footballs would have on the field. (Gauge pressure is the difference between the absolute pressure and atmospheric pressure, and 1atm = 14.7 psi). Below is a graph of the gauge pressures of the New England Patriots’ 11 footballs on the day of the championship game, measured at half-time. The arrow points to the gauge pressure that the Patriots claimed to have filled the balls to before the game. b) (3 pts) Is the observed pressure change of the footballs consistent with your analysis using the ideal gas law? Or was there likely foul play? Briefly explain. Does your conclusion depend on which gauge you use? c) (3 pts) List three assumptions you had to make to answer this question. d) (2 pts) Choose one of these assumptions and determine the effect of the assumption. Namely, if the assumption is not valid, does that weaken your conclusion or reinforce it? e) (2 pts) The footballs were brought back inside the locker room for the half-time measurement. What further info would you want to know about the situation before making a judgment?

Answer 1

1. Gauge pressure is 12.5 psi, lets put it in atm

12.5 / 14.7 = 0.85034 atm

P gauge = P abs -P atm

P absolute = 0.85 + 1 = 1.85 atm for total pressure

now we are said we have a locker temperature of 67 - 71 F, lets convert 67 F

K = (67 + 459.67 )/1.8 = 292.549

71 F will be 294.817 K

Temperature on the field was 48 F which is kelvin is 282 K

Now let´s apply gay-lussac´s law

P / T is constant

P₁ / T₁ = P₂ / T₂

P2 = P₁ T₂ / T₁

P₂ = 1.85 * 282 /   292.549 (for 67 in locker room) = 1.78 atm

for 71 F

P2 = 1.85 * 282 /   294.817 = 1.7695 atm

The absolute pressures in the ball wil go from 1.77 to 1.78 atm for gauge pressure just substract 1 atm so

gauge pressure are between 0.77 to 0.78 atm

just multiply these values by 14.7 psi

0.77 * 14.7 = 11.31 psi

0.78 * 14.7 = 11.466 psi

this result makes sense, if the temperature raises for a constant volume the pressure will raise, if the temperature decreases the pressure will decrease

B) The ideal gas law says that if you decrease the temperature the pressure will drop too, so this analysis is consistent, we can also say that since the pressure is below 12.5 psi, there is a clear evidence that there was a foul play.

c) assumptions: The ball was inflated with air.

Air behaves like an ideal gas

The ball will get the same temperature of the environment, this is that if temperature is 50F the temperature of the air in the ball must be 50F which may not be necessarily true since air is not a good heat conductor

d) Let´s assume that the air in the ball is not 48 F , temperature from the field, let´s assume that the temperature is between 48 and the locker room of 71F

48 + 71 = 119

119 / 2 = 59.5 F

This is 288.42 Kelvin, lets retake equation from part b

P2 = 1.85 * 288.42 /   294.817 = 1.809 atm

P gauge = 1.809 - 1.= 0.809 atm

0.89 * 14.7 = 11.8923 psi,

temperature is still below the minimum pressure, so even if the temperature of the air in the ball does not equal the temperature of the environment the pressure will drop below the legal rule.

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