Converting volume to L, we note the conversion factor, 1 L = 1 times 10^-3 s m^3

Question

Converting volume to L, we note the conversion factor, 1 L = 1 times 10^-3 s m^3, so 1 m^3 is 1000 L. Noting also that -51 degree C will be 222 K, we can substitute: n = (230/760 atm) (1000 L)/(0.082057 L. atm/K.mol)(227 K) = 16.6 mol air Given the density as 480 g/m^3, we can calculate the molar mass: 480 g/16.6 mol = 28.9 = 29 g/mol The density of air 18 km above Earth's surface is 480 g/m^3. The pressure of the atmosphere is 230 mm Hg, and the temperature is -51 degree C. If the atmosphere at this altitude consists of only O_2 and N_2, what is the mole fraction of each gas?

Explanation / Answer

b)

H = 18 km

D = 480 g/m3

P = 230 mm Hg

T = -51° C

Find mol fraction of each:

MW = 29 g/mol

so..

MW = MW-N2 * x-N2 + MW-O2 * x-O2

29 = 28 * (1-x-O2) + 32 * x-O2

29 = 28 - 28*x-O2 + + 32 * x-O2

29-28 = 4-XO2

XO2 = 1/4 = 0.25

so

YO2 = 1-0.25 = 0.75

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