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.

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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons

c (diamond),

sio2 (quartz)