Question

Although Earth's atmosphere is stratified, there is still exchange of air between, for example, troposphere and stratosphere. In the following, consider only the stratosphere and the troposphere. The lifetime of air in the stratosphere is 1.72 years. Determine the lifetime of air in the troposphere in years. Assume that the troposphere extends to 15.0 km in height and that the stratosphere extends from 15.0 km to 50.0 km. The molar mass of air is 28.96 g/mol. Assume a temperature of 15.0 °C.

Answer 1

Pressure = force/area

force = mass x acceleration due to gravity = m g

Acceleration due to gravity, g = 9.18 m/s²

Area of earth = 4r²

radius of earth, r = 6371 x 10³ m

As acceleration due to gravity and area is constant for troposphere and stratosphere.

So, Pressure is directly proportional to the mass

At steady state there will be no net flow of air between troposphere and stratosphere.

Troposphere ranges from 0-15 km and stratosphere ranges from 15 - 50 km

F(s --> t) = F (t--->s)

Mt/Ms = t/s

Pressure at an altitude can be calculated using formula : P at sea level e^{(-g M (h-h0)/RT}

g = 9.8 m/s² ,

h0 = 0 and h = 15000 m(15 km),

M = molar mass of air = 28.96 g/mol = 0.02896 kg/mol

and R - ideal gas constant = 8.314 Nm/mol K ,

P at sea level = 1 atm,

Temperature = 15 +273 = 288 K

P15km = 1 x e^{(-9.8 x 0.02896 X15000/8.314X288}) = e^{(-4257.12/2394.432)}= e^-1.7789 = 0.1688 atm

P50 Km = 1 x e^{(-9.8 x 0.02896 X50000/8.314X288}) = e^{(-14190.4/2394.432)}= e^-5.93 = 0.00266 atm

Mt/Ms = Pt/Ps = P(0 Km) -P(15 Km)/P(15 Km) - P(50 Km) = 1 atm - 0.1688 atm/0.1688 atm - 0.00266 atm = 0.8312/0.16614 = 5.003
Mt/Ms = t/s

The lifetime of air in the troposphere in years, t = Mt/Ms) x s = 5.003 x 1.72 = 8.60516 yrs = 8.61 years