Air Pressure Normal sea level pressure is expressed as 1013.2 mb (millibars) of

Question

Air Pressure Normal sea level pressure is expressed as 1013.2 mb (millibars) of mercury (a way of expressing force per square meter of surface area). At sea level standard atmospheric pressure is expressed in several ways: 14.7 lb in2 29.9213" of Hg (mercury) 1013.250 millibars 101.325 kilopascals (kilopascal = 10 millibars) . Some convenient conversions are helpful: . 1.0 in. (Hg)-33.87 mb 25.40 mm (Hg) 0.49 lb/in2 1.0 mb 0.0295 in. (Hg)-0.75 mm (Hg)-0.0145 Ib/in2 The standard atmosphere for pressure in millibars and altitude in kilometers is given in Table 7.1. This is used in the activity that follows the table. Table 7.1 Standard atmosphere for pressure and altitude Altitude (km) 10.00 12.00 14.00 16.00 18.00 20.00 25.00 30.00 35.00 40.00 50.00 60.00 70.00 Altitude (km) 0.00 Pressure (mb) Pressure (mb) 264.99 193.99 141.70 103.52 75.65 55.29 25.49 1013.25 954.61 898.76 1.00 1.50 2.00 2.50 3.00 4.00 795.01 746.91 701.21 616.60 540.48 472.17 411.05 5.75 2.87 0.79 7.00 8.00 9.00 0.06 1. Using the graph in Figure 7.1, plot the standard atmosphere of air pressure decrease with altitude presented in Table 7.1. After completing the plot, connect the data points with a line to complete the pressure profile of the atmosphere. The data points from 0 to 5 km are plotted for you

Explanation / Answer

3. a) 320mb is given. So in inches, 320×0.0295=9.44 in

b) Mount McKinley, 6124m air pressure= 460mb=460×0.0295 inches= 13.57 in

c) Mount Whitney, 4418m , air pressure= 580 mb= 580×0.0295 in= 17.11 in

d) at 2356 m, air pressure=760 mb=760×0.0295in=22.42 in

e) 452m, air pressure=960 mb=960×0.0295in =28.32 in

f) at 12000 m, air pressure from above= 193.99mb

While at the surface it is 1013.2 mb

So, the %of atmospheric pressure above the flight= (193.99/1013.2)×100=19.15%

And the% of air pressure below the flight= {( 1013.2-193.99)/1013.2}×100= 80.85%

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