You have a liquid-phase mixture of 0.550 mol fraction ethanol and 0.450 mol frac

ID: 956201 • Letter: Y

Question

You have a liquid-phase mixture of 0.550 mol fraction ethanol and 0.450 mol fraction water at 760.0 mmHg. You'd like to calculate the bubble-point temperature of the mixture, and the composition of the vapor formed at the bubble-point temperature two different ways. For the first method, assume that the solution behaves ideally and all components follow Raoult's law. What is the bubble-point temperature of the solution? What are the mole fractions of ethanol (e) and water (w) in the vapor phase at the bubble-point temperature?

Explanation / Answer

Antoine Equation can be written as , P= 10(A-B/C+T)

T in deg.c for water A= 8.14 B= 1810.94 and C= 244.485

For ethanol A=7.68 , B= 1332.04 and C= 199.2

Mole fraction of ethanol = 0.55 and mole fraction water= 0.45

From Raoults law x1p1sat+x2P2sat= 760 (1)

Where P1 sat and P2 sat are vapor pressures of ethanol and water.x1 =0.55 and x2= 0.45

Assume some temperature , Calculate P1sat and p2 sat at that temperature and match eq.1

Let T= 60 deg.c for ethanol P1sat = 10(7.68- 1332.04/60+199.2) =347 mm Hg

For water P2sat= = 10(8.14- 1810.94/60+244.485) =155.75 mm Hg

P=0.55*347+0.45*155.75= 261 mm Hg

So 60 deg.c is not correct since it is not mactching with 760mm Hg pressure

The trial and error calculatinos are done in an excel and the final results is shown which gives bubble point at 85.7 deg.c

7.68

8.14

1332.04

1810.94

199.2

244.485

85.73

B/(T+C)

4.674973

5.484124

A- B/(T+C)

3.005027

2.655876

1011.643

452.7683

mole fractions

0.55

0.45

Pressure

760.1493

At 85.7 deg.c , P1 sat= 1011.643 mm Hg and P2sat= 452.7 mm Hg

From y1P= x1p1sat, y1= 0.55*1011.643/760 =0.73 and y2= 1-0.73= 0.27