Question

In: Chemistry

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount...

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.

Solutions

Expert Solution

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


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