Answer :
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
- x₁= [tex]4.5 * 10^{-3}[/tex]
- y₁= 0.9
What is Binary solution?
A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2
If we consider the pressure of the 2 phase system is 1 bar.
The assumption are as follows:
- The vapor phase is ideal at pressure of 1 bar.
- Henry's law apply to dilute solution only.
- Raoult's law apply to concentrated solution only.
Where, these values are given:
Henry's constant for species 1 H= 200bar
Temperature = 25°C= 298.15K
P₂sat= 0.10 bar
Apply Henry's law for species 1
y₁P= H₁x₁.......... (i)
where 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= P₂sat. x₂...........(ii)
From (i) and (ii)
P=H₁x₁ + P₂sat. x₂
200bar= H₁
0.10= P₂sat
1 bar= P
Hence,
P=H₁x₁ + (1 - x₁) P₂sat
1bar= 200bar × x₁ + (1 - x₁) 0.10bar
x₁= [tex]4.5 * 10^{-3}[/tex]
The mole fraction of species 1 in liquid phase is [tex]4.5 * 10^{-3}[/tex]
To get y, substitute x₁=[tex]4.5 * 10^{-3}[/tex] in (i)
y × 1 bar = 200bar × [tex]4.5 * 10^{-3}[/tex]
y₁= 0.9
The mole fraction of species 1 in vapor phase is 0.9.
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