Answer
$ρ(z) = 0.60kg/m3$
$M = 5.092 x 10^(18) kg$
Work Step by Step
Assume that the atmospheric gas behaves as an Ideal gas
Given, thickness of atmosphere = 25km.
Radius of Earth = 6377km
A)
This question contains the data, variation of density of atmospheric air with elevation,in tabular form. So to solve the problem we need equation solving softwares (function fitting softwares) like EES . Plot graph using the data and obtain a polynomial equation.
Here let us use EES,
Define a function ρ(z) = a+z in equation window, where is the density and z is the vertical distance from earth at sea level
Choose new parametric tables from Tables and input the data given.
Plot the graph and obtain the 2nd order polynomial equation.
We get
$ ρ(z) = a + bz +cz2 = 1.20252 - 0.10167z + 0.00223747z2 kg/m3$
……………………………..(1)
$= (1.20252 - 0.10167z + 0.00223747z2) x 109 kg/km3$
At z = 7km(given), substituting in eq(1)
$ρ(z) = 1.20252 −0.10167×(7) + 0.00223747×(72) kg/m3 =0.60kg/m3$
$ρ(z) = 0.60kg/m3$
B)
Mass of the atmosphere can be evaluated as follows
Density = mass/volume
So mass m = Density x Volume =$ (ρdv)$, dv is the differential volume
Mass of Atmosphere = M
M = $Integral(ρdv)$, = $ Integral from z=0 to
h(((a+bz+c^(2))(r0 +h)^2) dz $
r0 is the radius of Earth = 6377
On integrating and substituting with h= 25km , a = 1.20252, b = 0.10167, c = 0.00223747 and multiplying by 109
$M = 5.092 x 10^(18) kg$
