Thermo A single vapor-liquid equilibrium point for the water (1) + ethanol (2) s

Question

Thermo A single vapor-liquid equilibrium point for the water (1) + ethanol (2) system is experimentally measured at 30 degree C. The experiment provides the following information: x_1 = 0.30, y_1 = 0.23, and P = 10.1 kPa. Use this information to estimate the system pressure and vapor-phase mole fraction when x_1 = 0.8. Use Antoine's equation to estimate the pure components vapor pressure. Assume that the excess molar Gibbs free energy and the activity coefficients can be correlated using the following equations: G^E/RT = Ax_1, x_2, ln gamma_1 = Ax_2^2, ln gamma_2 = Ax_1^2

Explanation / Answer

Antone constants for water

Logpsat (mm Hg)= 8.07131- 1730.63/(t (deg.c)+233.426)

At 30 deg,c, logpsat= 8.07131- 1730.63/(30+233.426)

Psat= 31.7402 mm Hg =31.7402*0.133 Kpa=4.22 Kpa

Antoine constants for ethanol : Logpsat (mm Hg)= 8.20417- 1642.89/(t (deg.c)+230.3)

At 30 deg.c, Psat= 78.0989 mm Hg =78.0989*0.133 Kpa =10.4 KPa

Modified Raoults law is

y1P= x1Y1psat and y2P= x2Y2P2sat, P= x1Y1P1sat +x2Y2P2sat, P=101.3

x1=0.3,x2=1-0.3= 0.7

lnY1= Ax22 = A*(0.7)2= 0.49A,   Y1= e(0.49A) lnY2= Ax12= A*(0.3)2= 0.09A, Y2= e(0.09A)

10.1 = 0.3*e(0.49A)*10.4 +0.7*e(0.09A)*4.22

Assume some value and solve by trial and error. When solved, A= 1.56

Hence LnY1= 1.56x22 and LnY2= 1.56x12

for x1= 0.8, x2= 1-0.8= 0.2, lnY1= 1.56*(0.2)2= , Y1= 1.064 , lnY2= 1.56*(0.8)2, Y2= 2.713

hence P= 0.8*1.064*10.4+0.2*2.713*4.22 =11.14 Kpa

y1= x1Y1P1sat/P= 0.8*1.064*10.4/11.14 =0.8   y2= 1-0.8= 0.2

Azeotrope is formed at this composition.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.