Consider a cube of air with the dimension of 0.5 m (approximately 20 inches) at

Question

Consider a cube of air with the dimension of 0.5 m (approximately 20 inches) at an altitude of z_o = 0.25 km with a pressure of P_o = 100 kPa and a temperature of T_o = 20 C. If the volumetric fraction of oxygen is 0.21, how many moles of oxygen are in the cube? Consider a cube of the same dimension at the summit of Mount Everest. The elevation of the summit of Mount Everest is approximately 8.85 km. For a lapse rate of 4.5 C km^-1, what is the percent reduction in the number of moles of O_2 at the summit of Mount Everest? You can assume that the same volumetric fraction of oxygen.

Explanation / Answer

Volume of air is Va = (0.5m)3 = 0.125 m3

Volume of oxygen is Vo = 0.21XVa =  0.21X(0.125 m3) = 0.02625m3

We have,

PV = nRT

or, n = PV/RT = (100X103Pa)(0.02625m3)/[(8.31 J/mol-K)(273 + 20)K]

or, n = 1.078 mol is the number of moles of oxygen in the given cube.

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Pressure at the summit of the Mount Everest is P = 33.7X103 Pa

Temperature will be T = 20oC - (8.85 km - 0.25 km)(4.5oC/km) = -18.70C = 254.3 K

n = PV/RT = (33.7X103 Pa)(0.02625m3)/[(8.31 J/mol-K)(254.3)K]

n = 0.42 mol of oxygen at Mount Everest.

So, percentage reduction is n = [(1.078 mol - 0.42 mol)/1.078 mol]X100

or, n = 61%

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