Balls lose pressure in the cold

Regarding deflate-gate

Amounton’s Law says

“The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas’s absolute temperature.”

So P/T = Constant

For comparing the same substance under two different sets of conditions, the law can be written as:

P1/T1 = P2/T2

I’m also going to do this in metric to avoid other potential errors. And I apologize that I am 30 years from doing this sort of work

12.5 psi = 86.18 kPa ( kiloPascals)
75F=23.89C = 297.04K
30F=-1.11C = 272.04K

So

86.18/297.04 = P2/272.04
0.29= P2/272.04
P2=78.93kPa
78.93kPa= 11.45PSI

Temperature change from 75F to 30F degrees would result in a football ball losing ~1 pound of pressure per square inch.

Now this does not account for ALL of the claimed 2lbs PSI that was measued as having been lost — but we don’t know that the ball began at 12.5 — what if they began at 12? What if they were inflated in a 78 degree room?

I think early calls dismissing temperature as a force are hasty.

http://en.wikipedia.org/wiki/Gay-Lussac%27s_law#Pressure-temperature_law

UPDATE
Someone pointed out that

Pressure gauge= Pressure absolute – Pressure atmosphere
which means I need to do my calculations with absolute pressure to use the Ideal Gas Law

Atmospheric pressure is 14.7 psi

12.5 + 14.7 = 27.2psi = 187.54 kPA

187.54/297.04 = P2/272.04
0.29= P2/272.04
P2=171.76kPa
171.76kPa= 24.91psi
subtract atmospheric pressure and
24.91-14.7= 10.21psi

!!!!

A ball inflated at 75 degrees to 12.5 psi will be 10.2 psi at 30 degrees!!!  This matches the readings by the NFL Officials!!

UPDATE REDUX