Question

The atmosphere is a giant sedimentation equilibrium experiment. Nitrogen and oxygen are pulled down by the gravitational potential, and diffuse back upwards due to entropy. Assuming the earth’s gravitational acceleration is uniform and 9.802 m s–2, and the average temperature of the atmosphere is –20°C, and the pressure of N2 and O2 at sea-level are 0.79 and 0.20 bar respectively, calculate the pressure of nitrogen and oxygen at the elevation of a transatlantic airplane flight (10 km).

Answer 1

Ans.

pressure of nitrogen at 10 km  

P = Po [ 1 – (Lh/To) ]^(gM/RL)

where

P is pressure above level = ?
Po of N2 at sea level = 0.79 bar
h altitude in meters = 10 km = 10000 m
L temperature lapse rate 0.0065 K/m
To Sea level standard temperature = -20oC = -20 + 273 K = 253 K
g = Earth-surface gravitational acceleration = 9.802 m/s²
M molar mass of gas = 0.028 kg/mol

R universal gas constant = 8.314 J/(mol•K)

P = Po [ 1 – (Lh/To)]^(gM/RL)

    = (0.79) [ 1- (0.0065x 10000 /253)]^(9.802 x 0.028) /( 8.314 x 0.0065)

= 0.175 bar

P= 0.175 bar

Therefore, pressure of nitrogen at 10 km = 0.175 bar

pressure of oxygen at 10 km  

P = Po [ 1 – (Lh/To) ]^(gM/RL)

where

P is pressure above level = ?
Po of O2 at sea level = 0.2 bar
h altitude in meters = 10 km = 10000 m
L temperature lapse rate 0.0065 K/m
To Sea level standard temperature = -20oC = -20 + 273 K = 253 K
g = Earth-surface gravitational acceleration = 9.802 m/s²
M molar mass of gas = 0.032 kg/mol

R universal gas constant = 8.314 J/(mol•K)

P = Po [ 1 – (Lh/To)]^(gM/RL)

    = (0.2) [ 1- (0.0065x 10000 /253)]^(9.802 x 0.032) /( 8.314 x 0.0065)

= 0.036 bar

P= 0.036 bar.