14.7 PSI - 12 PSI at 1 mile altitude times square inches in 1 square mile.
WHY do you insist on trying to look intelligent by bringing up things that are totally irrelevant?
"STP". Look it up. Standard Temperature and Pressure.
I refer you back to the original question:
Now the question:
What is the mass of one cubic mile of air at sea level?
Assume STP, of course, and a perfect cube. In other words, imagine a strong steel box 1 mile by 1 mile by 1 mile.
If it were evacuated, sealed, and weighed at sea level, and then unsealed and air allowed to fill it, what is the increase over the tare weight?
The answer will astound you.
Since this question was first posted, there have been several inquiries on specifics and assumptions of the conditions. For example, is the pressure uniform? When you say to assume STP, then you are suggesting that pressure is to be considered uniform. Yet your most recent assertion that the pressure at the top of the cube would be less than at the bottom suggests that it is not. But if it is not, then is the median elevation at sea level? Or is it the top of the cube that is at sea level as opposed to the bottom?
Also, which version of STP are we to use? Scientific organizations define it differently. Is standard pressure 1 bar? Or is it 101,325 Pa? Is standard temperature at 0°C, or is it at 25°C? And then there's the composition of the air. Saturated air weighs less than unsaturated air. You would be better served by the members of this forum if you defined the question better.
There is also the matter of compressibility. Air is not an ideal gas. So compressibility would have to be factored in for the partial pressures of each component.
n = (PV) / ZRTIn the absence of said clarification, the following assumptions will be made:
Uniform pressure - 101.325 kPa
Standard Temp - 273.15 K (zero water vapor pressure - dry air)
Universal gravity constant - 9.807 m/s
2MW of dry air ≈ 0.028965 kg/mol
Ideal Gas constant ≈ 8.3145 (m
3 ⋅ Pa) / (K ⋅ mol)
Volume - 1 mi
3 ≈ 4168181825 m
3Z ≈ 0.9994
n = [(101,325 Pa) * (4168181825 m
3)] / [0.9994 * (8.3145 (m
3 ⋅ Pa) / (K ⋅ mol)) * 273.15 K]
n = 1.861 E11 moles
mass = (1.861 E11 moles) * (0.028965 kg/mol) = 5,389,644,414 kg
US mass = 5389644414 kg * (0.06852 slugs/kg) =
369,308,000 slugs
This is roughly 14% higher than the answer you provided in Nov 2012 which is likely the result of the assumption I made about universal pressure throughout the cube and your assumption that the top of the cube was 1 mile above the ocean's surface. If the cube sat along the dead sea, the top of the cube would be only ¾ miles above sea level, thus increasing the amount of air in the cube. And if the cube rested on top of the Arctic ocean at the North Pole during low tide, it would be closer to the center of the earth making the acceleration of gravity greater.
Again, there have been legitimate questions asked regarding this problem. Please extend the courtesy of explaining the rules. Thanks.