Answer:
the final temperature would have been 2.81 °C if the pressure drop was 2 psi
Step-by-step 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