Quote:
Originally Posted by DementedLogic
The temperature change is negligible unless the patriots were inflating the balls in boiling water.
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Not true at all. Let's say they inflated them at 80 degrees then it dropped to 45 degrees outside.. AND it was wet so the transfer of heat will go faster AND the pressure dropped outside as well (I don't have that figure so I'll ignore that aspect and pretend it is 14.7 and didn't change)
80F = 299.8K
45F = 280.4K
Ok so you have 12.5 as what the inflated to... but this is a relative reading so we need to add 14.7 to get a pressure of 27.2.
Now lets look at the formula, PV=nRT We can assume that V, n and R don't change (or not enough to matter in this case). So that means we have P1V=nRT1 and P2V=nRT2
27.2*V =nR*299.8 and P2*V =nR*280.4
Use V to match it all up..
(nR*299.8)/27.2=(nr*280.4)/P2
1/P2=299.8/(27.2*280.4)
P2=27.2*280.4/299.8=25.44
25.44-14.7=10.74
12.5-10.74 =1.76 PSI drop
SWEET! I pulled those numbers for temps out of my ass and it came out pretty close!
So, it's more than reasonable to explain it this way.