I will admit that I am a little confused. If a football is inflated and the PSI is measured, then later the PSI is measured again, the difference between the two is a loss/gain in guage pressure. Now comes the confusion part, if you are doing an experiment on paper, you have to use the absolute pressure to account for atmospheric pressure?
Here you go. (If anyone sees any errors please let me know.)
The speak about Ideal Gas Law in the Wells report which is: PV=nRT
P = Pressure
V= Volume (how big is the container (football in this case)
R = Ideal gas constant (How much energy)
n = moles of gas (air) (How much)
T = Temperature
Since V, R and n are constant in that those values will not change in solving for a football, they used a simplified version of the Law called the Gay-Lussac's Law which is:
P1/T1 = P2/T2 If you want to solve for P2 you would convert that to P1 x (T2/T1)= P2
So in the Patriots case we would use these values to solve for P2:
P1 = 12.5 (gauge reading) + 14.7 (atmospheric pressure)
Atmospheric pressure changes with altitude. Why it is important to add to the gauge reading is that in order for the football to start to inflate it would have to overcome the atm pressure to do so. So in order for you to reach 12.5 psi you would have to exert 27.2 psi of pressure inside the football. The gauge only measures the relative pressure of 12.5. So if you only used 12.5 for P1 you are not including all of the pressure that exists.
T1 = 70 deg F = 294.26 Kelvin
Fahrenheit is not an absolute temperature either. 0 degree F is not the lowest temperature in the universe. 0 Kelvin is the coldest temperature. So you can use K as an absolute temperature. It's kind of like using a measuring cup with 0 cups marked at half and negative 1 marked at bottom. If you wanted 1 cup of water you could simply think I'll just fill it up to 0 and that will be 1 cup in total but in a mathematical equation the results wouldn't work out so good.
So after knowing that you can solve for any pressure or temp.
(12.5 + 14.7) x 282.04/294.26 = 26.07 - 14.7 = 11.37 psi
Note: Once you have solved the absolute pressure of P2 you will have to subtract atmospheric pressure to get back a gauge pressure reading.
Note: 282.04 K = 48 deg F
Mazz once stated that in order for a ball to deflate 2 psi the temperature would have to be -11 psi. Here is how that happened.
12.5 x 249.26/ 294.26 = 10.6
Note: notice that atmospheric pressure was not added. Also 249.26 K = -11 deg F.
I kept the formula in the same format for demonstration purposes.
