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
Temperature at game time was 50F, which is 282K
187.54/297.04 = P2/272.04
P2=178.07kPa
171.76kPa= 25.82psi
subtract atmospheric pressure and
25.82-14.7= 11.15psi
187.54/297.04 = P2/272.04
P2=178.07kPa
171.76kPa= 25.82psi
subtract atmospheric pressure and
25.82-14.7= 11.15psi
4 comments :
The end-game temperature was in the upper 40s, so the numbers would have changed again.
http://www.accuweather.com/en/weather-news/deflategate-patriots-football/40975001
Not sure by how much, but i've heard rumors of 47/48F at agme end.
Glad to see that you corrected for absolute pressure. We do not know a lot of things about the method and conditions of the measurements. Sticking to just the temperature issue for a moment, the rain falling through very cold air higher in the atmosphere would be much colder than ground air conditions.Not below freezing but possibly close. If you assume 10F colder rain (38F) then you would get an expected 2 Deg F drop.
Minor correction to the above post. You would get "an expected 2 psi drop"
Also worth mentioning that leather expands when wet, increasing volume and thus decreasing pressure even further.
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