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Distillation of a Two Component Mixture (Part - 2) Video Lecture | Mass Transfer - Chemical Engineering
FAQs on Distillation of a Two Component Mixture (Part - 2) Video Lecture - Mass Transfer - Chemical Engineering
| 1. What is distillation and how is it used in chemical engineering? |  |
Ans. Distillation is a separation process used in chemical engineering to separate components of a mixture based on their different boiling points. It involves heating the mixture to vaporize the more volatile component, and then cooling and condensing the vapor to obtain a purified liquid. Distillation is commonly used in industries such as oil refining, petrochemicals, and pharmaceuticals.
| 2. How does distillation of a two-component mixture work? | |
Ans. In distillation of a two-component mixture, the mixture is heated in a distillation column. As the liquid mixture boils, the vapor formed contains a higher concentration of the more volatile component. The vapor rises in the column and is cooled at the top, where it condenses and is collected as a purified liquid. The less volatile component remains in the liquid phase and is collected separately.
| 3. What factors affect the efficiency of distillation in separating a two-component mixture? | |
Ans. Several factors can affect the efficiency of distillation in separating a two-component mixture. The difference in boiling points between the components is crucial, as a larger difference allows for easier separation. The composition of the mixture also plays a role, with mixtures closer to the azeotropic point being more challenging to separate. The design and operating conditions of the distillation column, such as the reflux ratio and column height, can also impact separation efficiency.
| 4. What is the role of azeotropes in distillation of a two-component mixture? | |
Ans. Azeotropes are mixtures of two or more components that have a constant boiling point and composition. They can make the separation of a two-component mixture by distillation more challenging, as they behave as single components with distinct boiling points. Azeotropes can form when the vapor pressures of the components are similar, leading to a minimum or maximum boiling point for the mixture. Special techniques, such as adding third components or using different distillation processes, may be required to separate azeotropic mixtures.
| 5. What are the advantages and limitations of distillation in chemical engineering processes? | |
Ans. Distillation has several advantages in chemical engineering processes. It is a widely used and versatile separation technique that can handle large-scale operations. Distillation can achieve high purity separations and is suitable for both binary and multicomponent mixtures. However, it also has limitations. It requires significant energy input, especially for mixtures with close boiling points. Some mixtures may form azeotropes or have temperature-sensitive components that are difficult to separate by distillation alone. In such cases, additional separation techniques may be necessary.
Text Transcript from Video
This is a continuation of the distillation
problem. Here we are looking for the temperature
of the reboiler and the composition of the
vapor assuming the pressure is 1 atm absolute.
Let me draw the picture again and let me show
you what it is that we are looking for. What
we are looking for is this part of the problem
and because the liquid and the vapor are in
equilibrium we can therefore use Raoult's
law. Again we know that the partial pressure
of the pentane has to equal the liquid mole
fraction times the vapor pressure at the temperature
that we are looking for as well as the hexane
partial pressure has to equal the mole fraction
of the liquid of the heaxane times the vapor
pressure of hexane at the temperature we are
looking for. We know that y sub P times P
plus y sub H times P equals the total pressure.
The sum of the partial pressures is going
to equal the total pressure. Then we can substitute
that this equals x sub P times P star at the
temperature we are looking for plus x sub
H times P star at the temperature that we
are looking at. We found earlier that this
xP was 0.04 therefore the xH is 0.96. Let's
write this out. 0.04, that is the x sub P,
times 10^(6.85221-1064.63/(T+232)). That is
the x sub P times P star of T. This all has
to equal 760 mmHg because that is the total
pressure. Again you are going to have to use
some kind of solver. The temperature equals
66.6 degrees C. How is that going to help
us find the composition of the vapor again
assuming the pressure is 1 atm. Again we are
going to use Raoult's Law. The composition
of the vapor, the y sub P, is going to equal
our x sub P times P star at 66.6 degrees C
divided by the total pressure which is 760.
What we do is we use Antoine's equation right
up here and we substitute for T. We substitute
66.6 degrees C and that allows us to find
the vapor pressure at that temperature. This
is 0.04 times 1905.5 mmHg divided by 760 mmHg.
That is going to equal 0.101. We can also
find our y sub H because that equals 1 minus
0.101 or 0.899. You would expect these kinds
of numbers because we had much higher mol
fractions in the liquid that came off at the top.
problem. Here we are looking for the temperature
of the reboiler and the composition of the
vapor assuming the pressure is 1 atm absolute.
Let me draw the picture again and let me show
you what it is that we are looking for. What
we are looking for is this part of the problem
and because the liquid and the vapor are in
equilibrium we can therefore use Raoult's
law. Again we know that the partial pressure
of the pentane has to equal the liquid mole
fraction times the vapor pressure at the temperature
that we are looking for as well as the hexane
partial pressure has to equal the mole fraction
of the liquid of the heaxane times the vapor
pressure of hexane at the temperature we are
looking for. We know that y sub P times P
plus y sub H times P equals the total pressure.
The sum of the partial pressures is going
to equal the total pressure. Then we can substitute
that this equals x sub P times P star at the
temperature we are looking for plus x sub
H times P star at the temperature that we
are looking at. We found earlier that this
xP was 0.04 therefore the xH is 0.96. Let's
write this out. 0.04, that is the x sub P,
times 10^(6.85221-1064.63/(T+232)). That is
the x sub P times P star of T. This all has
to equal 760 mmHg because that is the total
pressure. Again you are going to have to use
some kind of solver. The temperature equals
66.6 degrees C. How is that going to help
us find the composition of the vapor again
assuming the pressure is 1 atm. Again we are
going to use Raoult's Law. The composition
of the vapor, the y sub P, is going to equal
our x sub P times P star at 66.6 degrees C
divided by the total pressure which is 760.
What we do is we use Antoine's equation right
up here and we substitute for T. We substitute
66.6 degrees C and that allows us to find
the vapor pressure at that temperature. This
is 0.04 times 1905.5 mmHg divided by 760 mmHg.
That is going to equal 0.101. We can also
find our y sub H because that equals 1 minus
0.101 or 0.899. You would expect these kinds
of numbers because we had much higher mol
fractions in the liquid that came off at the top.
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