[WUSA9] Let’s assume that each ball was inflated to the minimum pressure required to meet the NFL rules regarding proper inflation: 12.5 psi. We convert psi (English) to pascals (Metric), which comes out to 86,184.5 Pa and assume room temperature (68ºF/20ºC) which converts to 293.15 K (Kelvin, the Metric equivalent). We now have,

86,184.5 Pa / 293.15 K = p2 / T2.

We’re down to two variables. But we also know the temperature on the field at the start of the game was reported as 51ºF/10.6ºC (283.15 K). Plug it in…

86,184.5 Pa / 293.15 K = p2 / 283.15 K

Neat! Look, we’re left with a solvable equation with one variable, p2, which is the pressure of the air inside the ball at game time! Let’s solve this riddle…

Isolate the lone variable:

(86,184.5 Pa / 293.15 K) * 283.15 K = p2
83,244.6 Pa = p2 —> 12.1 psi

83,244.6 Pa is 12.1 psi, so, according to our calculations, the balls could have been under-inflated by 0.4 psi on the field. This makes sense given the very first equation, which shows that a decrease in temperature would force a decrease in pressure, assuming the same volume of air in the football.
Temperature drop accounts for .4 PSI. The balls were 2 PSI under inflated. And this assumes the refs did not wait, and measure the PSI immediately after the game out on the field.

http://www.sportsgrid.com/nfl/scienc...ith-our-magic/