Answer:
h = 9483 m
Explanation:
Assume Re as radius of earth and h as equivalent thickness of co_2
total volume occupied by [tex]co_2 = \frac{4}{3} \pi Re^3 +\frac{4}{3} \pi (Re+ h)^3[/tex]
mass of [tex]co_2 = 3.2 \times 10^{15} kg[/tex]
1 mole of co_2 has 44 g mass
1 g has = 1/44 mole
[tex]3.2 \times 10^{15} kg has = 1/44 \times 3.2\times 10^{18} = 7.27 \times 10^{16} mole[/tex]
total volume by [tex]co_2 = 7.27 \times 10^{16} \times 22.414 L = 163.01\times 10^{16} L[/tex]
[tex]163.01\times 10^{16} L = frac{4}{3} \pi Re^3 +\frac{4}{3} \pi (Re+ h)^3[/tex]
SOLVING RIGHT SIDE WE GET
[tex]h ( Re^2 +h^2 + Reh) = \frac{3\times 163.01\times 10^{16}}{4\pi}[/tex]
[tex]h ( Re^2 +h^2 + Reh) = \frac{3\times 163.01\times 10^{16}}{4\pi}[/tex]
solving for h we get
h = 9483 m
