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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....
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- Spring '07
- Calculus, #, gram molecular weight