[MMaschin]

Not sure what you are trying to say. If you go to the link I provided you will see that the chart is relative to normal air - that is NTP normal temperature and pressure. So yes, it is relational to normal atmospheric pressure.

[lepton]

The volume of the air in the ball would have dropped 2% for each 10 degree F drop.

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Using that equation, (for which the same constant would work for V or P if the other is constant):

.96 (12.5 + 14.7) = 26.1

26.1-27.2 = -1.1, not 0.5.

.5 is the solution only if you don’t include the 1 atm base as changing too, which only would if your 12.5 was the absolute pressure, which only would be so if the area outside the ball were in a vacuum.