In: Chemistry

For
ethanol/water (70;30) determine;

A) the bubble point temperature if p=10 bar

B) the bubble point temperature if p=100 bar

A) the bubble point temperature if p=10 bar

B) the bubble point temperature if p=100 bar

The bubble point is the temperature at which a liquid mixture will start to boil. As with a dew point calculation, we can use Dalton’s and Raoult’s Laws to calculate the bubble point. By combining the two equations, we can calculate the vapour mole fractions for a given liquid composition, i.e.:

Yi = (Xi * Pi°) / P

Calculating the bubble point is iterative.The vapour pressure of each component can be estimated using the Antoine Equation.

log (P°) = a - (b / (T (°C) +c))

For ethanol:

a = 5.33675

b= 1648.220

c= 230.918

For water:

a= 5.11564

b= 1687.537

c= 230.170

We have the liquid molar fractions X but we need the molar fractions of the vapour ‘Y’. Also, we know that Y1 + Y2 = 1.

A ) We need a first guess to the temperature, so let’s look into the water vapour tables. At 10 bar, the temperature of saturated water vapour is around 180 °C but we also know that the boiling point of ethanol at atmospheric pressure is under 80 °C. I’ll try 160 °C:

Ethanol

log (P°) = 5.33675 - (1648.220 / (160 + 230.918)) = 1.12

P° = 13.18 bar

Water

log (P°) = 5.11564 - (1687.537 / (160 + 230.170)) = 0.79

P° = 6.17 bar

Yet = 0.70 * 13.18 bar / 10 bar = 0.923

Y wat = 0.3 * 6.17 / 10 bar = 0.184

Y = 0.923 + 0.184 = 1.12 > 1 We need to try a lower T.

Let’s se 155°C:

Ethanol

log (P°) = 5.33675 - (1648.220 / (155 + 230.918)) = 1.07

P° = 11.75 bar

Water

log (P°) = 5.11564 - (1687.537 / (155 + 230.170)) = 0.73

P° = 5.37 bar

Yet = 0.70 * 11.75 bar / 10 bar = 0.822

Y wat = 0.3 * 5.37 / 10 bar = 0.161

Y = 0.983 < 1

Use both data to interpolate the result:

T = 160 °C, Y = 1.12

T = 155 °C, Y = 0.983

**T = 155.6 °C**

B) Do exactly the same procedure and:

**T = 273 °C**

*If you have a scientific calculator, you also can
input all the equations and let a program solve
it.*

A solution of ethanol CH3CH2OH and water that is 10.% ethanol by
mass is boiling at 98.7°C. The vapor is collected and cooled until
it condenses to form a new solution.
Calculate the percent by mass of ethanol in the new solution.
Here's some data you may need:
normal boiling point
density
vapor pressure at
98.7°C
ethanol
78.°C
0.79gmL
1563.torr
water
100.°C
1.00gmL
725.torr
Be sure your answer has the correct number of significant
digits.
Note for advanced students: you...

Calculate the following: a. The bubble point temperature of a
liquid mixture with a molar composition of 60.0 mol% n-hexane and
40.0 mol% n-heptane at 1.00 atm and the composition (mole
fractions) of the vapor in equilibrium with this mixture.
Bubble Point: °C
yHexane:
yHeptane:
b. The dew-point temperature of a gas mixture with a molar
composition of 25.0% n-hexane, 15.0% n-heptane, and 60.0% air at
1.00 atm and the composition (mole fractions) of the liquid in
equilibrium with this...

Calculate the following:
a. The bubble point temperature of a liquid mixture with a molar
composition of 45.0 mol% n-hexane and 55.0 mol% n-heptane at 1.00
atm and the composition (mole fractions) of the vapor in
equilibrium with this mixture.
Bubble Point: _____°C
yHexane: _____
yHeptane: _____
b. The dew-point temperature of a gas mixture with a molar
composition of 30.0% n-hexane, 35.0% n-heptane, and 35.0% air at
1.00 atm and the composition (mole fractions) of the liquid in
equilibrium with this mixture....

a. The bubble point temperature of a liquid mixture with a molar
composition of 35.0 mol% n-hexane and 65.0 mol% n-heptane at 1.00
atm and the composition (mole fractions) of the vapor in
equilibrium with this mixture.
Find bubble point
Find x Hexane
Find x Heptane
b. The dew-point temperature of a gas mixture with a molar
composition of 30.0% n-hexane, 15.0% n-heptane, and 55.0% air at
1.00 atm and the composition (mole fractions) of the liquid in
equilibrium with...

3.
Ethanol (MW = 46) is soluble in water and is a liquid at room
temperature, but propane (MW = 44) is insoluble in water and is a
gas at room temperature. Explain why these two compounds have such
different properties despite their similar size.

For ethanol liquid with 50% excess of air. Determine adiabatic
flame temperature

What is the freezing point of a 50% by volume ethanol solution?
Assume water is the solvent. (Use data from Table 12.7 in the
textbook.)
Solvent
Normal Freezing Point (°C)
Kf (°C m−1)
Normal Boiling Point (°C)
Kb (°C m−1)
Benzene (C6H6)
5.5
5.12
80.1
2.53
Carbon tetrachloride (CCl4)
−22.9
29.9
76.7
5.03
Chloroform (CHCl3)
−63.5
4.70
61.2
3.63
Ethanol (C2H5OH)
−114.1
1.99
78.3
1.22
Diethyl ether (C4H10O)
−116.3
1.79
34.6
2.02
Water (H2O)
0.00
1.86
100.0
0.512
For water:...

Determine the temperature, volume, and quality for one kg of
water at H=3500kJ/kg, P =0.1MPa. (Using steam tables)

An ideal gas is expanded from 10 bar to 1.0 bar at constant
temperature. Calculate deltaU, deltaH, and deltaS. CP =
5/2 R.

During isobaric process m=64 kg of wet water stream at
temperature t=262.7?C and pressure p=49 bar changed its specific
volume from v1 =0.0018 m3/kg to v2=0.035 m3/kg. Find initial and
final quality of steam, change of internal energy of steam, work
done by steam, amount of heat exchanged with surroundings. Draw the
process at the enthalpy-entropy diagram. From tables for p=49 bar:
v’=0.0013 m3/kg, v”=0.0403 m3/kg, h’= 1148.4 KJ/kg, h”= 2794.7
KJ/kg

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