I was trying to solve this problem, very simply it is:

what is the weight of water over lake superior?

I did some research and deciced top solve the problem in atmosphere levels
But I had no idea of how to use units of volume, mass, and mainly pascals(air preassure)

this is what my plan was :
Thinking of Lake Superior as a square I will use its surface area value as the values for its length and width. I will calculate the space between each layer of atmosphere and use it as a height value. Using these values I will calculate the volume of each layer of atmosphere.

I multiply each volume accordingly to its respective air density to solve for the weight of air in each layer.

I will sum the values to have a net weight in kg of air over Lake Superior

Sounds fine, as long as you have data for the density of air as a function of altitude above the lake.

A much easier apporach is to simply multiply the area of the lake by the air pressure at the surface. A nominal value for air pressure is 14.7 psi. Keep your units straight - given the air pressure in pounds per square inch, multiply that by the surface area of the lake in square inches to find the total weight of air in pounds. You'll end up with a mighty big number!

Hi,
But at that altitude, can we consider the atmospheric pressure as 14.7 psi?
Ravi

Quote:

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Originally Posted by Ravi42

Hi,
But at that altitude, can we consider the atmospheric pressure as 14.7 psi?
Ravi

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Indeed Lake Superior is about 600 ft above sea level, so yes - on average the atmospheric pressure on the surface of Lake Superior is only about 95% of sea level pressure.