Question

Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains 0.6 g of butane at −22∘C. How much butane is present as a liquid at the boiling point? If the butane is warmed to 25 ∘C, how much is present as a liquid?

Answer 1

Solution :-

Butane normal boiling point = -0.4 C + 273 = 272.6 K

Vapor pressure at boiling point = 1 atm

Delta H vap = 22.44 kJ/mol * 1000 J /* 1 kJ = 22440 J per mol

Leta calculate the Vapor pressure at -22 C + 273 = 251 K

Ln(P2/P1) = Delta vap / R [ (1/T1)-(1/T2)]

Ln(P2/1 atm) = 22440 J per mol / 8.314 J per mol K *[(1/272.6 )-(1/251)]

Ln(P2/1 atm) = -0.85205

P2/ 1 atm = anti ln [-0.85205]

P2/1atm = 0.426538

P2 = 0.42658 * 1atm

P2 = 0.426538 atm

Now leta calculate the moles of the butane at the boiling point using the ideal gas law equation

PV= nRT

PV/RT= n

0.426538 atm * 0.250 L / 0.08206 L atm per mol K * 272.6 K = n

0.004767 mol = n

Now lets convert the moles of butane to its mass

Mass= moles * molar mass

         = 0.004767 mol * 58.12 g per mol

         = 0.277 g

So the amount of butane in the gas phase = 0.277 g

Now lets calculate the amount of the butane in the liquid phase

Amount of butane in liquid phase = 0.60 g – 0.277 g

                                                               = 0.166 g

So at the boiling point the 0.166 g butane will be present in the liquid state.

Part2 ) At the 25 C temperature all the butane will be converted to gas so no butane will present in the liquid phase