Here is a problem that is in my physical chemistry textbookthat is not on this s
ID: 76926 • Letter: H
Question
Here is a problem that is in my physical chemistry textbookthat is not on this site (Thermodynamics, StatisticalThermodynamics & Kinetics) that I am having a really hard timewith... Consider a 20.0 L-sample of moist air at 60C and one atm inwhich the partial pressure of water vapor is 0.120 atm. Assume that dry air has the composition 78.0 mole percent N2,21.0 mole percent O2 and 1 mole percent Ar. What are themole percentages of each of the gases in the sample? Any help would be appreciated! Thanks! Here is a problem that is in my physical chemistry textbookthat is not on this site (Thermodynamics, StatisticalThermodynamics & Kinetics) that I am having a really hard timewith... Consider a 20.0 L-sample of moist air at 60C and one atm inwhich the partial pressure of water vapor is 0.120 atm. Assume that dry air has the composition 78.0 mole percent N2,21.0 mole percent O2 and 1 mole percent Ar. What are themole percentages of each of the gases in the sample? Any help would be appreciated! Thanks!Explanation / Answer
PV = nRT n = PV/(RT) = (1 atm * 20 L)/(0.0821 L*atm/mol/K * 333 K) =7.38198059 total moles n_water = P_water *V /(R*T) = (0.120 atm)*(20 L)/(0.0821L*atm/mol/K * 333K) = 0.88583767 moles water moles air: 7.38198059 - 0.88583767 = 6.49614292 moles 0.78*(6.49614292) = 5.06699148 moles N2 0.21*(6.49614292) = 1.36419001 moles O2 0.01*(6.49614292) = 0.0649614292 moles Ar %N2 = 5.06699148 moles N2/7.38198059 total moles = 0.6864 =68.64% % O2 = 1.36419001 / 7.38198059 = 0.1848 = 18.48 % % Ar = 0.0649614292 / 7.38198059 = 0.0088 = 0.88 % % H2O = 0.88583767/ 7.38198059 = 12 %