Respuesta :
Explanation:
The given data is as follows.
T = [tex]120^{o}C[/tex] = (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So, [tex]x_{1}[/tex] = 0.5 and [tex]x_{2}[/tex] = 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
[tex]p^{sat}_{1}[/tex] (393.15 K) = 9.2 bar
[tex]p^{sat}_{1}[/tex] (393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.
[tex]p_{1} = x_{1} \times p^{sat}_{1}[/tex]
= [tex]0.5 \times 9.2 bar[/tex]
= 4.6 bar
and, [tex]p_{2} = x_{2} \times p^{sat}_{2}[/tex]
= [tex]0.5 \times 10.5 bar[/tex]
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P = [tex]p_{1} + p_{2}[/tex]
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.
[tex]y_{1} = \frac{p_{1}}{p}[/tex]
= [tex]\frac{4.6}{9.85}[/tex]
= 0.467
and, [tex]y_{2} = \frac{p_{2}}{p}[/tex]
= [tex]\frac{5.25}{9.85}[/tex]
= 0.527
Calculate the dew point as follows.
[tex]y_{1}[/tex] = 0.5, [tex]y_{2}[/tex] = 0.5
[tex]\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}[/tex]
[tex]\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}[/tex]
[tex]\frac{1}{P}[/tex] = 0.101966 [tex]bar^{-1}[/tex]
P = 9.807
Composition of the liquid phase is [tex]x_{i}[/tex] and its formula is as follows.
[tex]x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}[/tex]
= [tex]\frac{0.5 \times 9.807}{9.2}[/tex]
= 0.5329
[tex]x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}[/tex]
= [tex]\frac{0.5 \times 9.807}{10.5}[/tex]
= 0.467
