A binary mixture of benzene and styrene is kept in a closed vessel at equilibrium such that both liquid and vapor phases co-exist. The total amount of moles in the vessel is 100. The total moles of benzene that is distributed in the liquid and vapor phases is 35. The temperature is kept constant at 112.7°C and the pressure at 815.8 mmHg.
Raoult’s law can be assumed to describe the phase equilibrium for both species and Antoine constants can be used.
Calculate the mole fractions in liquid and vapor phases and then use material balance to find the amount of the liquid phase.
n = 100
To calculate the mole fractions in the liquid and vapor phases, we first need to calculate the partial pressures of benzene (P₁) and styrene (P₂) in the vapor phase using Raoult's law.
Raoult's law states that the partial pressure of a component in the vapor phase is equal to the product of its mole fraction in the liquid phase and its vapor pressure at that temperature.
Let's calculate P₁ and P₂:
P₁ = x₁ * P₁°
P₂ = x₂ * P₂°
where x₁ and x₂ are the mole fractions of benzene and styrene in the liquid phase, and P₁° and P₂° are the vapor pressures of benzene and styrene at 112.7°C, respectively.
To find x₁ and x₂, we need to use the given information that the total amount of moles in the vessel is 100 and the total moles of benzene in both phases is 35.
The mole fraction of benzene in the liquid phase (x₁) can be calculated using the formula:
x₁ = n₁ / (n₁ + n₂)
where n₁ is the moles of benzene in the liquid phase and n₂ is the moles of styrene in the liquid phase.
Given that the total amount of moles in the vessel is 100 and the total moles of benzene in both phases is 35, we can calculate n₂:
n₂ = 100 - 35
Now we can calculate x₁:
x₁ = 35 / (35 + n₂)
Next, we need to find P₁° and P₂°, which are the vapor pressures of benzene and styrene at 112.7°C. These can be obtained using the Antoine equation or from a reference source such as a handbook. For the purpose of this explanation, let's assume the values are:
P₁° = 300 mmHg
P₂° = 450 mmHg
Now we can calculate P₁ and P₂:
P₁ = x₁ * P₁°
P₂ = x₂ * P₂°
Now we have the partial pressures of benzene and styrene in the vapor phase.
To find the mole fraction of styrene in the liquid phase (x₂), we can use the fact that the sum of mole fractions in a phase is equal to 1:
x₂ = 1 - x₁
Now we have x₁ and x₂, the mole fractions in the liquid phase and vapor phase, respectively.
To use material balance to find the amount of the liquid phase, we can calculate the moles of benzene in the liquid phase (n₁) using the formula:
n₁ = x₁ * total moles
Substituting the values:
n₁ = x₁ * 100
Now you can calculate the value of n₁, which represents the moles of benzene in the liquid phase.
Please note that the actual values for P₁° and P₂° may differ based on the specific conditions and properties of benzene and styrene. It is important to check the literature or use reliable sources to obtain accurate values for the vapor pressures.