Question

In: Other

Given: Vapor–liquid equilibrium data in mole fractions for the system acetone–air–water at 1 atm (101.3 kPa)...

Given:

Vapor–liquid equilibrium data in mole fractions for the system
acetone–air–water at 1 atm (101.3 kPa) are as follows:

y acetone in air 0.004 0.008 0.014 0.017 0.019 0.020
x acetone in water 0.002 0.004 0.006 0.008 0.010 0.012

Find:

(a) Plot the data as (1) moles acetone per mole air versus moles
acetone per mole water, (2) partial pressure of acetone versus g acetone
per g water, and (3) y versus x. (b) If 20 moles of gas containing
0.015 mole-fraction acetone is contacted with 15 moles of water,
what are the stream compositions? Solve graphically. Neglect water/
air partitioning.

Solutions

Expert Solution

(a) 1. Here data is given between mole fractions so first we convert in the form of mole ratio

Where Y is the mole ratio that is

similarly

X Y
0.002004 0.00401
0.004016 0.00806
0.006036 0.0142
0.008064 0.0173
0.01010 0.0194
0.01214 0.0204

2. now we have to plot partial pressure of acetone verses g acetone per gram of water

Partial Pressure

Where PT is the total pressure = 101.3 kPa

To calculate the g acetone per g of water we multiply by molecular weight of acetone and divide by molecular weight of water

(g acetone /g water) X'=

P acetone (kPa) X'
0.4052 0.00645
0.8104 0.0129
1.4182 0.01936
1.722 0.0258
1.9247 0.032
2.026 0.03872

(3) For y verses x we can directly plot from the given data

(b)

20 mole of gas containing 0.015 mole fraction of acetone

Water = 15 moles

Let G = Moles of gas free from acetone = 20 - 20 0.015

G= 19.7 moles of acetone free gas

Pure water L = 15 moles

Let the counter current contact and

Y1 = moles ratio for gas at inlet = 0.0152

X1 = mole ratio for water outlet

Y2 = Mole ratio for Gas outlet

X2 mole ratio for water inlet = 0 (Water is pure initially)

We apply acetone mole balance

X1 and Y2 are in euillibrium

GY1+ LX2 =GY2 + LX1

G(Y1-Y2)=LX1

We put Y2 =Y

X1=X

Therfore Y=Y1-(L/G)X

When we draw this line in Graph A1 . Where this line intersect with the equillibrium line it gives us stram composition

Y= -(0.76)X + 0.0152

The Red line provides the by the equation of line

The stream composition

X= 0.005 (moles of acetone / mole of air)

Y= 0.012 (mole of acetone/ moles of water)


Related Solutions

1. Obtain the equilibrium mole fractions (liquid and vapor) of a mixture of water, methanol and...
1. Obtain the equilibrium mole fractions (liquid and vapor) of a mixture of water, methanol and ethanol at 100 kPa and 70°C. You can assume the mixture is ideal. 2. Create one plot where the equilibrium mole fraction of benzene in the vapor phase is plotted against the equilibrium mole fraction of benzene in the liquid phase for three different pressures (0.2 bar, 0.6 bar and 1 bar). Comment on the effect of pressure on the relative volatility
The vapor-liquid equilibrium data for heptane – ethylbenzene at 1 atm pressure is given below: x...
The vapor-liquid equilibrium data for heptane – ethylbenzene at 1 atm pressure is given below: x 0 0.10 0.20 0.40 0.50 0.60 0.70 0.90 1 y 0 0.25 0.45 0.65 0.74 0.82 0.86 0.96 1             A feed mixture containing 40 mol% heptane and 60 mol% ethylbenzene is to be fractionated at 1 atm pressure to produce a distillate containing 95 mol% heptane and a bottom product stream containing 96 mol% ethylbenzene. Determine the minimum reflux ratio if the feed...
The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic....
The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. H2​O(l)+40.7kJ H2​O(g) Assume at exactly 100.0°C and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62 L, respectively. A)Calculate the work done on or by the system when 2.45 mol of liquid H2O vaporizes. __________J Calculate the water's change in internal energy. ____________kj
The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic....
The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. Assume at exactly 100.0°C and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62 L, respectively. H20 (l) +40.7kJ ----> H20 (g) Please answer both parts of the question: 1. Calculate the work done on or by the system when 3.45 mol of liquid H2O vaporizes. 2. Calculate the water's change in internal...
Given the mole of air as 9.2x10-5 mol, the volume of water vapor, after correction, is...
Given the mole of air as 9.2x10-5 mol, the volume of water vapor, after correction, is 5.0 mL while the temperature is 75 oC, what is the vapor pressure of water in mmHg if the atmospheric pressure (Patm ) is 0.989 atm.
3- 1 mole of liquid benzene is in equilibrium with 1 mole of gaseous benzene at...
3- 1 mole of liquid benzene is in equilibrium with 1 mole of gaseous benzene at 80.1 oC in a isolated piston at 1.00 atm. Suppose the pressure of the piston is increased to 2.00 and kept constant, and no heat flows in and out of the system. What will be the final composition, volume, and the temperature inside the piston? Assume perfect gas behavior for the gas and molar volume of the liquid benzene (90.0 cm3/mol) does not change...
18. Explain why the vapor pressure of water in equilibrium with liquid water is temperature dependent?
18. Explain why the vapor pressure of water in equilibrium with liquid water is temperature dependent?
1) At 24 C the vapor pressure of water is 3 kPa. The vapor pressure of...
1) At 24 C the vapor pressure of water is 3 kPa. The vapor pressure of water over a sample is 2 kPa, what is the water activity of the sample? 2) The water activity above a saturated NaCl solution is 0.75. If a sample is enclosed in a jar with the NaCl solution and equilibrium is achieved, what is the water activity of the sample? NaCl and the sample are in separate containers within the jar.
Hot air containing 1 mole% water vapor flows into a textile dryer at 350°F and 6.2...
Hot air containing 1 mole% water vapor flows into a textile dryer at 350°F and 6.2 psig at a rate of 30,000 ft3/hr, and emerges at 220°F and 1 atm absolute pressure (i.e., it is vented outside the building) containing 10 mol% H2O. Calculate the following:a. The mass flow rate of the entering hot air stream in lbm/hrb. The rate of evaporation of the water from the textiles in lbm/hrc. The volumetric flow rate of the emerging air stream in...
When 5.00 g of acetone (C3H6O) burns in air, carbon dioxide gas and liquid water are...
When 5.00 g of acetone (C3H6O) burns in air, carbon dioxide gas and liquid water are formed. Enough heat is liberated to increase the temperature of 1.000 kg of water from 25.0◦C to 61.8◦C. The specific heat of water is 4.18 J/g-◦C 1. How many kJ of heat are liberated by the combustion described? 2. How many grams of acetone must be burned to liberate 5.00 kJ? 3. Write the thermochemical equation for the combustion of acetone. 4. What is...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT