Answer :
The given problem involves a benzene (1)/toluene (2) system at a temperature of 110 °C and a pressure of 130 kPa. We need to determine the following: the molar ratio of liquid to vapor (L/V) is approximately 1.316.
In summary: (a) x₁ = 1.8 and y₁ = 0.76
(b) The molar ratio of liquid to vapor (L/V) is approximately 1.316.
(a) x₁ and y₁, which represent the mole fractions of benzene and toluene in the liquid and vapor phases, respectively.
To solve for x₁ and y₁, we need to use the Antoine equation to calculate the saturation pressure for benzene (P₁ˢᵃᵗ) and toluene (P₂ˢᵃᵗ) at the given temperature of 110 °C.
At 110 °C, P₁ˢᵃᵗ = 234.3 kPa for benzene and P₂ˢᵃᵗ = 99.3 kPa for toluene.
Next, we can use the equations:
P₁ = x₁ * P
P₂ = y₁ * P
where P₁ and P₂ are the partial pressures of benzene and toluene, respectively, and P is the total pressure of the system (130 kPa).
We can rearrange the equations to solve for x₁ and y₁:
x₁ = P₁ / P
y₁ = P₂ / P
Substituting the values, we get:
x₁ = 234.3 kPa / 130 kPa
y₁ = 99.3 kPa / 130 kPa
Calculating these values, we find:
x₁ = 1.8
y₁ = 0.76
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