Respuesta :
Explanation:
The given data is as follows.
   T = [tex]120^{o}C[/tex] = (120 + 273.15)K = 393.15 K, Â
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, Â Â Â Â Â [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]      Â
               = 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, Â Â Â Â Â Â Â Â [tex]y_{2} = \frac{p_{2}}{p}[/tex]
                = [tex]\frac{5.25}{9.85}[/tex]
                = 0.527 Â
Calculate the dew point as follows.
           [tex]y_{1}[/tex] = 0.5,    [tex]y_{2}[/tex] = 0.5 Â
     [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]       Â
               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
