Law of Partial Pressures?

Question

A sample of air is collected over water at 30.00°C. At equilibrium, the total pressure of the moist air is

0.9970 atm. The equilibrium vapor pressure of water at 30.00 °C is 31.824 Torr. The composition of dry

air is 78.08 mol% N2, 20.95 mol% O2, 0.93 mol% Ar and 0.04 mol% CO2. a) Calculate the partial

pressures in atm of N2, O2, Ar and CO2 in the wet mixture assuming ideal gas behavior. b) Calculate the

mole fractions of H2O, N2, O2, Ar and CO2 in the wet mixture, again assuming ideal gas behavior. Use

the appropriate number of significant figures.

Answer 1

31.824 Torr @ 1 atm / 760 torr = 0.04187 atm H2O vapor

total 0.9970 atm. - 0.04187 atm H2O vapor = 0.9551 atm "dry air"

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a) Calculate the partial pressures in atm in the wet mixture assuming ideal gas behavior:

N2 78.08% of 0.9551 = 0.7458 atm

O2 20.95% of 0.9551 = 0.2001 atm

Ar  0.93% of 0.9551 = 0.0089 atm

CO2 0.04% of 0.9551 = 0.0004 atm

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b) Calculate the mole fractions  in the wet mixture, again assuming ideal gas behavior.

H2O  0.0419 atm = 0.0419 mole fraction

N2    0.7458 atm  =  0.7458 molefraction

O2   0.2001 atm  =  0.2001 mole fraction

Ar    0.0089 atm  =  0.0089 mol fraction

CO2  0.0004 atm =  0.0004 mol fraction