Liquids A and B from ideal solution for all composition of A and B at `25 ^@C` Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 , respectively . What is the vapor pressure of pure B in bar ?

Explanation

To find the vapor pressure of pure B (P₀B) in bar, we can use the information provided about the two solutions and apply Raoult's Law. Here’s a step-by-step solution:

### Step 1: Understand Raoult's Law
Raoult's Law states that the vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. For a binary solution of A and B, we can express the total vapor pressure (P_total) as:
\[ P_{\text{total}} = P_A + P_B = P_0A \cdot X_A + P_0B \cdot X_B \]
where:
- \( P_A \) is the vapor pressure of component A,
- \( P_B \) is the vapor pressure of component B,
- \( P_0A \) and \( P_0B \) are the vapor pressures of pure A and B respectively,
- \( X_A \) and \( X_B \) are the mole fractions of A and B in the solution.

### Step 2: Set up equations for the two solutions
For the first solution with a mole fraction of A (X_A1 = 0.25):
\[ P_{\text{total1}} = P_0A \cdot 0.25 + P_0B \cdot 0.75 = 0.3 \text{ bar} \]
(1)

For the second solution with a mole fraction of A (X_A2 = 0.50):
\[ P_{\text{total2}} = P_0A \cdot 0.50 + P_0B \cdot 0.50 = 0.4 \text{ bar} \]
(2)

### Step 3: Rewrite the equations
From equation (1):
\[ 0.25 P_0A + 0.75 P_0B = 0.3 \]
(1)

From equation (2):
\[ 0.50 P_0A + 0.50 P_0B = 0.4 \]
(2)

### Step 4: Solve the equations simultaneously
We can rearrange equation (2) to express \( P_0B \) in terms of \( P_0A \):
\[ P_0B = 0.8 - P_0A \]
Substituting this into equation (1):
\[ 0.25 P_0A + 0.75(0.8 - P_0A) = 0.3 \]
Expanding and simplifying:
\[ 0.25 P_0A + 0.6 - 0.75 P_0A = 0.3 \]
Combining like terms:
\[ -0.5 P_0A + 0.6 = 0.3 \]
\[ -0.5 P_0A = 0.3 - 0.6 \]
\[ -0.5 P_0A = -0.3 \]
\[ P_0A = 0.6 \text{ bar} \]

### Step 5: Find \( P_0B \)
Now substituting \( P_0A \) back into the equation for \( P_0B \):
\[ P_0B = 0.8 - 0.6 = 0.2 \text{ bar} \]

### Final Answer
The vapor pressure of pure B (P₀B) is:
\[ P_0B = 0.2 \text{ bar} \]