Explanation:
The given data is as follows.
T =
= (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So,
= 0.5 and
= 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.

= 
= 4.6 bar
and, 
= 
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P =
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.

= 
= 0.467
and, 
= 
= 0.527
Calculate the dew point as follows.
= 0.5,
= 0.5


= 0.101966
P = 9.807
Composition of the liquid phase is
and its formula is as follows.

= 
= 0.5329

= 
= 0.467
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
Please refer to the attachment for a complete classification of your specified matter.