AtmosphericPressure

AtmosphericPressure - A $ p h h A p h h" A p h" A h...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Atmospheric Pressure Decreases Exponentially Suppose that p ( h ) represents the atmospheric pressure at altitude h above the surface of the earth. We will show that this function has the form p ( h ) = p (0) e ! Mg RT h where M is the gram molecular weight of the atmosphere, R is the universal gas constant, T is the absolute temperature, and g is the gravitational acceleration. Of course, these must all have values using consistent units. Let ! V be a small unit of volume which we visualize below. In the figure A is the cross- sectional area of the small volume and ! h is the height. The force supporting this unit volume is the difference between the force due to the pressure on the bottom of the volume, given by p ( h ) ! A and the force on the top of the volume given by p ( h + ! h ) " A . This vertical force should just balance the weight of the gas in the volume which is given by ! W = " ( h ) # ! h # A # g where ! ( h ) is the density of the atmosphere at altitude h . After some manipulation and taking a limit, we have the following differential equation. ! W = ( h ) # A # ! h # g = p ( h )
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: # A $ p ( h + ! h ) # A p ( h + ! h ) " A # p ( h ) " A ! h = # $ ( h ) " A " g dp dh = ! ( h ) # g Using the ideal gas law, we can now related the pressure, p ( h ) , with the density, ( h ) , of the atmosphere at altitude h . the ideal gas law is given below. p ! V = n ! R ! T ! h A ! V " # $ p ( h ) ! A p ( h + ! h ) " A Here p is the pressure, V is the volume of gas under consideration, n is the number of moles of the gas, and R and T are as stated above. If M is the gram molecular weight of the gas, then n ! M is the weight. This leads to the equation p = n ! M V " # $ % & ' R ! T M = R ! T M ! ( Thus we have dp dh = ! M " g R " T " p ( h ) The solution of is differential equation is just p ( h ) = p (0) e ! Mg RT h For standard units assuming that the gram molecular weight of the atmosphere is 28.8 g and standard temperature T Mg RT = 1.24426778 ! 10 " 4 per meter So, we get p ( h ) = p (0) e ! 1.24426778 " 10 ! 4 h when we measure h in meters....
View Full Document

This note was uploaded on 07/14/2011 for the course MAC 3472 taught by Professor Jury during the Spring '07 term at University of Florida.

Page1 / 2

AtmosphericPressure - A $ p h h A p h h" A p h" A h...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online