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Solution

You know density (1.3 kg/m3), you know gravity (9.8 m/s2), and you know the change in height (100 m). It's just a matter of algebra to solve for the change in pressure. Don't worry if you get strange-looking units.

Solution:

change in pressure = - change in height * density * gravity

change in pressure = - 100 m * 1.3 kg/m3 * 9.8 m/s2

change in pressure = about -1300 kg/m/s2
Strange units, eh? You were expecting, maybe, inches of mercury? Well, the appropriate scientific units of pressure are force per unit area, or kg * m / s2 / m2, or kg/m/s2, so this is right. One kg/m/s2 is called a Pascal, or Pa for short. For historical reasons, meteorologists usually work with a slightly different unit, a millibar (mb), defined as 100 Pa. Converting our answer to millibars, we find that over 100 m, the air pressure changes by 13 mb.

How big is 13 mb? A typical sea level pressure is about 1000 mb, so 13 mb represents 1.3% of the total atmospheric pressure.

Another simple question for the ambitious or physically inclined: if the atmosphere were everywhere the same density, 1.3 kg/m3, how tall would it be? Try to solve this one now if you want to, or come back to it if you finish the module early. (Hint: at the top of the atmosphere, the air pressure would be zero.)

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Questions or Comments

Technical: E-mail John Fulton < jdfult@nimbus.met.tamu.edu >
Scientific: E-mail Dr. John Nielsen-Gammon. < nielsen@ariel.met.tamu.edu >

Copyright © 1996-2003 Texas A&M University, Texas A&M Atmospheric Sciences Department and Dr. John Nielsen-Gammon. All rights reserved.