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Find the equivalent thickness of CO2 on Earth.

Earth’s atmospheric CO2 = 3.2 × 1015 kg

Earth’s radius = 6.4 × 106 m

Molar volume (@STP) = 22.414 L mol-1

User Jon Kyte
by
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1 Answer

1 vote

Answer:

h = 9483 m

Step-by-step explanation:

Assume Re as radius of earth and h as equivalent thickness of co_2

total volume occupied by
co_2 = (4)/(3) \pi Re^3 +(4)/(3) \pi (Re+ h)^3

mass of
co_2 = 3.2 * 10^(15) kg

1 mole of co_2 has 44 g mass

1 g has = 1/44 mole


3.2 * 10^(15) kg  has = 1/44 * 3.2* 10^(18) = 7.27 * 10^(16) mole

total volume by
co_2 =  7.27 * 10^(16) * 22.414 L = 163.01* 10^(16) L


163.01* 10^(16) L = frac{4}{3} \pi Re^3 +(4)/(3) \pi (Re+ h)^3

SOLVING RIGHT SIDE WE GET


h ( Re^2 +h^2 + Reh)  = (3* 163.01* 10^(16))/(4\pi)


h ( Re^2 +h^2 + Reh)  =  (3* 163.01* 10^(16))/(4\pi)

solving for h we get

h = 9483 m

User Nochkin
by
8.1k points
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