[sten]

NFL balls are to be inflated to 12.5 - 13.5 psi.

therefore, a 2 psi loss could be 11.5 psi when measured after the game while assuming the highest possible number prior to game (but without having measured it)

playing within the rules, they could fill their balls to 12.5 psi in a 90F room and then take them out to the 40F game. what would be the change?

P1 = 12.5 psi
T1 = 90F == 305K
T2 = 40F == 277K

therefore:

P2 = P1 * T2 / T1
P2 = (12.5) * (277) / (305)
P2 = 11.35 psi

[TexasGator]

P1 = 12.5 psi
T1 = 90F == 305K
T2 = 40F == 277K

therefore:

P2 = P1 * T2 / T1
P2 = (12.5) * (277) / (305)
P2 = 11.35 psi

---

P1 = 12.5 psig + 14.7
T1 = 90F == 305K
T2 = 40F == 277K

therefore:

P2 = P1 * T2 / T1
P2 = (12.5 + 14.7) * (277) / (305) = 24.7 psia
P2(gauge) = 24.7 psia - 14.7 = 10 psig

[lepton]

Your 12.5 should be 27.2 (14.7+12.5= 27.2)

...otherwise your starting condition is less than one atmosphere of pressure...which does other odd things. ;)