In the absence of turbulent mixing, the partial pressure of eachconstituent of a

Question

In the absence of turbulent mixing, the partial pressure of eachconstituent of air would fall off height above sea level in theearth's atmosphere asPi=P0ie-Mz/RT wherePi is the partial pressure at the heightz,P0i is the partial pressure of component iat sea level ,g is the acceleration of gravity, R is the gasconstant, T is the absolute temperature and Mi is themolecular mass of the gas. As a result of turbulent mixing, thecomposition of the earth's atmosphere is constant below an altitudeof 100 Km, but the total pressure decreases with altitude asP=P0e-Mavegz/RT whereMave is the mean molecular weight of air. At sea level,xN2=0.78084 and xHe=0.00000524 andT=300K.              ...a)calculate the total pressure at 10 Km assuming a mean molecularmass of 28.9 g mol-1 and that T=300K throughout thisaltitude range........b)calculate the value thatxN2/xHe would have at 10 Km in the absence ofturbulent mixing. compare your answer with correct value.

Explanation / Answer

(a)
P=P0e-Mavegz/RT

P0 = 105Pa

R = 8.314472 J/K mol

M = 28.9*10-3 Kg/mol

z = 10Km =104 m


P =105*e(-28.9*10^-3*9.81*10^4/(8.314472*300))

P =32090.5 Pa

(b)PN2 =0.78084*32090.5

PN2 = 25057.546 Pa

PHE =0.00000524*32090.5

PHE =0.16815422Pa

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