Question

At 1,013 bar total pressure propane boils at -42.1°C and n-butane boils at -0.5°C; the following vapor pressure data are available. T("C) Pp partial pressure of Propane (in kPa) Ph partial pressure of n-butane (in kPa) -31.2°C 160.0 26.7 -16.3°C 298.6 53.3 Assuming that these substances form ideal binary solutions with each other. (a) calculate the mole fractions of propane at which the solution will boil at 1.013 bar total pressure at -31.2 and -16.3°C. (b) Calculate the mole fractions of propane in the equilibrium vapor at these temperatures. (c) Sketch the temperature - mole fraction diagram at 1.013 bar using these data and label the regions. (1.01325 bar = 1.0 atm = 101325 Pascal. (Pa) a) Mole fractions of propane (in liquid) at -31.2 and -16.3°C b) Mole fractions of propane (in vapor) at -31.2 and -16.3°C c) Sketch of temperature - mole fraction diagram

Answer 1

I have used the Raoult's Law for the mixture of two gases.
I have taken the atmospheric pressure as it was given.

Here we use a very useful equation. x₁ = (y₁ X P_tot)/P_1. This is obtained by combining Raoult's law with Daltons Law.

In graph the two composition(solid and liquid) will have different composition line as we can see it from our numerical values, we have different composition of propane and butane in liquid and vapor phase.

Solution vapour Pressure of solution ley Rasults Law, pappy tbaar zonly for two component} het Pi i vip of propane (v.p is vapoure) be is vip of nodentane 4 is mole fraction of propane u s mole fraction of n-dentane Be total pressure of mixture, bo bipu + P2 (1-2y) 204 +22= 1) t - 31.2°C 101.3258 - (260)(x) + (2617) (1-2) a) 34 - 201.325 - 26.9)/(160-26.7) , au - 74-625)/(133.3) a = 0.56 (mrounding 9+) Mole fraction of propane will be 0.56 at -31-2°C

at 16.3°C 101.325 -(W) (298.6) + (---) (53.3) 1, 2 , (301.325 - 53.3)/(298.6–53.3) | ») ay - (98.025)/645-3) = 0.196 (manding) Mole fractions of propane will lee 0.1% at 16.30 at - 1.20 2 pts a total pressure at any temp. I pa pressure of porokate at any 0.56 x 160 temp. 9, - 101.3 9, 2 composition of propane in - vapour phase Y, = 0.8845 a composition of propane in - liquid phase more fraction of propane will lee 0.8845 at -31.2 cm vapour phase at - 16.3°C y, 0.196* 298.6 e) y,-0.527 313001.3 Mole fraction of broopane will see 0.587 at 16.30 in vapour phase

point boiling O . on line line vapour campoot in time temp (°C) comfort Composition Liquid - 42.195 - o mole fraction 1 lentane fig: diagram Mole fraction as temperature