Question

A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both species 1 and 2. The pressure of this two- phase system is 1 bar; the temperature is 25°C. At this temperature, 1 = 200 bar and P2sat = 0.10 bar. Determine good estimates of x1 and y1. State and justify all assumptions.

Answer 1

Answer:X= 4.5x 10^-3 ; Y= 0.9  

Explanation:

Here, a binary solution which contains 2 species is in equilibrium with a vapor phase containing two species, 1 and 2

Given that the pressure of this 2 phase system is =1 bar

We will use the following assumptions to solve this problem

1. The vapor phase is ideal at a pressure

2.   that Henry’s law applies to dilute solutions

3. that Raoult law applies to concentrated solutions

from the question we have that  

Henry’s constant of species 1 , H1 =200bar,  

Saturated vapor phase of species 2, = 0.10 bar

Temp of system= 25°C = 298k

Let us apply henry’s law for specires 1

y1P= H1X1-------eqn 1

where y_1= mole fraction of species 1 in vapor phase

p=total pressure of system

x1=mole fraction of species 2 in liquid phase

Also applying Raoult's law for species 2

Y2P= 2 X2------Eqn 2

Combining  Eqn 1 and  2

P=H1X1 + $^p^{sat}$2 X2  

Then  Substituting The given variables  ie  200bar=H1, 0.10bar=2= in eqn 3 to solve for x

P=H1X1+(1+ X1)P2sat

1 bar= 200bar X x1+ (1-x)0.10bar

X=4.5X10-3

=mole fraction of species in  Liquid phase=4.5 x 10^-3

Substituting for xi=4.5x 10^-3 in eqn n1 becomes  

Y1P= H1X1-------eqn 1

y1 x 1 bar= 200bar x 4.5x 10^-3

y1=0.9

mole fraction of species  in vapour phase=0.9

Answer 2

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9