In: Chemistry
Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 3.60 atm . Calculate the partial pressure of propane in the container.
Ans. Number of moles of methane = Mass/ molar mass = 8.0 g/ 16.0 gmol-1+ = 0.5 mol
Number of moles of ethane = 18.0 g/ 30.0 gmol-1+ = 0.6 mol
Let the number of moles of propane = X
Total number of moles of all gases = 0.5 mol (methane) + 0.6 mol (ethane) + X mol (propane)
= (1.1 + X) moles
Ideal gas Law: pV = nRT
Where, p = pressure in atm = 1 atm
V = volume in L
n = number of moles = 1 [standard condition]
R = universal gas constant= 0.082057338 atm L mol-1K-1
Assuming the gases (mixture of gases) to behave ideally,
Total pressure in the vessel, p = (nRT / V) , where n = (2 + X) moles
Or, 3.60 atm = [(1.1 + X) x 0.0821 atm L mol-1K-1 x 296.15 K ] / 10.0 L
Or, 3.60 atm = (1.1 + X) x 2.43 atm
Or, 3.60 atm / 2.43 atm = X + 1.1
Or, X = 1.48 – 1.1 = 0.38
Thus, number of moles of propane = 0.38
Total number of moles in the mixture = 1.1 + 0.38 = 1.48 moles
Mole fraction of propane = moles of propane / total moles
= 0.38 moles / 1.48 moles = 0.26
Partial pressure of propane = moles fraction of propane x total pressure
= 0.26 x 3.60 atm = 0.936 atm