At standard pressure, propane boils at -42.1degreeC and n-butane boils at -0.5de
ID: 971133 • Letter: A
Question
At standard pressure, propane boils at -42.1degreeC and n-butane boils at -0.5degreeC. You have collected the following vapor-pressure data: Assuming that propane and n-butane form an ideal binary solution. (a) Calculate the mole fraction of propane in solution above which the solution will boil at -31.2degreeC and at -16.3degreeC. (b) Calculate the mole fraction of propane in the vapor (in equilibrium with the solution) at each of these temperatures. (c) Based on the data and on your answers to (a)-(b), sketch the T-x (temperature - mole fraction) phase diaeram for the propane/pi-butane solution.Explanation / Answer
at -31.2 deg.c Vapor pressure n-propane= 1.6 bar and that of n-butane= 0.267 bar
Total pressure = x1P1sat +x2p2at, x1 and x2 are mole fractions of propane and butane respectively,
0.989 = x1*1.6 +x2*0.267= x1*1.6+(1-x1)*0.267
0.989= x1*1.6+0.267-0.267x1 =1.33x1+0.267
1.33x1= 0.989-0.267=0.722
x1= 0.722/1.33=0.5428, x2= 1-0.5428=0.4572
from y1P= x1p1sat
y1* 0.989= 0.5428*1.6
y1= 0.878 , y2= 1-0.878=0.122
at -16.3 also 0.989= x1*2.986+(1-x1)*0.533= x1*(2.986-0.533)+0.533
0.989= x1*2.453 +0.533
0.989-0.533= x1*2.453, x1 =0.19, x2=1-0.19=0.81
and y1= 0.19*2.453/0.989 =0.47 , y1= 1-0.47=0.53
Liquid phase mole fractions Vapor phase mole fractions
Temperature (deg.c) propane butane butane propane
-41.2 1 0 1 0
-31.2 0.5428 0.4572 0.878 0.122
-16.3 0.19 0.81 0.47 0.53
-0.5 0 1 0 1
The T-x diagram is shown below
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