Per Hamas: (You know, a MENSA certified genius)
Quote:
PV=nrT
The ideal gas constant is .08206
T is the temperature in Kelvin
V is the volume of the football
P is the pressure in pascals
n is the number of moles of gas.
Average PSI of the 11 footballs was 11.1. A PSI=6900 Pascals (basically)
Now, NFL footballs must be inflated to 12.5 PSI
12.5*6950=86250 Pa
11.1*6950=77145 Pa
Now, if you are assuming that there was no air removed, then the number of moles of gas will be identical for both, which means that we can ascertain the pressure of the football by using the temperature in Kelvins
86250=nrT
86250*1.0*10^-6*4237=n*.08206*293 (room temp)
77145*1.0*10^-6*4237=n*.08206*283 (game temp)
(Volume differences will be negligible here
Inside the locker room n= 15.1 moles of gas
So, if there are 15.1 moles of gas in a football that is 4237*10^-6 cubic meters at a gauge pressure of 12.5 psi and room temperature, then the pressure of the footballs outside, which was 283 K would be:
P*4237*10^-6=15.1*.08206*283
Pressure would be .967 of what it was inside.
.967*86250=83463 Pascals, which is 12.1 PSI.
Thus, the weather deflated them, at most, by 0.4 PSI. Where did the other pound go, Patriots fans?
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Sorry Dane. Once again you're arguing on the wrong side. But hey, you're DaneMcCloud, and you make music!