Question

Climbing Mt. Everest

At sea level the atmosphere contains about 78 molecules of nitrogen for every 21 molecules of oxygen. So, [N2]/[O2] ~ 3.71. what's the ratio of N2 to O2 at the peak of Mt. Everest, 8850 meters above sea level?

Assume air temperature is 300 K at all altitudes

Molecular weight of O2: 32g/mol

Molecular weight of Nitrogen: 20g/mol

kB = 1.38e-23 J/K. N_A = 6.02e23 molecules/mole

Answer 1

The concentration of a gas at an area is a proportion of the

discovering gas atoms at that area,

expected probabilities at an

height z to pursue a Boltzmann circulation is

CN_2(z) ~ exp(- mN₂ g z / k _B T ) CO_2(z) ~ exp ( - m O₂ g z / k_B T )

from these calculations,

fixations adrift dimension (z = 0), so we realize that

CN_2(z) = CN₂₍₀₎ exp ( - m N₂ g z / k_B T ) CO₂ ( z ) ~ CO₂₍₀₎ exp ( - m O₂ g z / k_B T )

Here C_N2(0) is the ocean level nitrogen fixation and CO₂₍₀₎ at sea level

for Oxygen

The fixation proportion at any elevation z is at that point

CN_2(z) / CO_2(z) = ( CN₂₍₀₎ / CO₂₍₀₎ ) exp ( - m N₂ g z / k_B T ) / exp( - m O₂ g z / k_B T )

here ( CN₂₍₀₎ / CO₂₍₀₎ ) = 3.71

C_N2(z) / CO_2(z) = 3.71 exp ( - ( m N₂ - m O₂ ) g z / k_B T )

and the values here we have is

m_N2 = 28.013 x 10^-3 kg / 6.022 x 10²³ particle

m_O2 = 32.000 x 10^-3 kg / 6.022 x 10²³ particle

and

z = 8850 m

g = 9.8 m / sec²

T = 300 K

CN_2(z) / CO_2(z) = ( CN₂₍₀₎ / CO₂₍₀₎ ) exp ( - m N₂ g z / k_B T ) / exp( - m O₂ g z / k_B T )

CN_2(z) / CO_2(z) = 3.71 x exp ( - 28.013 x 10^-3 / 6.022 x 10²³ x 9.8 x 8850 / 1.38 x 10^-23 x 300 ) / exp( - 32.000 x 10^-3 /

6.022 x 10²³ x 9.8 x 8850 / 1.38 x 10^-23 x 300 )

C_N2(z) / CO_2(z) = 4.260

so this can clarify why climbers convey oxygen containers to Mount Everest.