Question

If a cylinder contains 12.0lb of liquid propane (C3H8), what volume does the propane occupy when it has completely vaporized at STP?

Answer 1

Ans. Mass of liquid propane = 12.0 lb = 12.0 x (453.592 g) = 5443.11 g

Moles of propane = Mass /Molar mass

                                    = 5443.11 g / (44.09652 g/ mol)

                                    = 123.4363 mol

# It’s assumed that vapor propane behave as ideal gas.

# Using Ideal gas equation:            PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation -

            1.00 atm x V = 123.4363 mol x (0.0821 atm L mol^-1K^-1) x 273.15 K

            Or, V = 2768.1349 atm L^-1 / 1.00 atm

            Hence, V = 2768.1349 L

Therefore, volume of propane vapor at STP = 2768.1349 L