An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of `C, P_(C)^(@)=0.8` atm. Calculate the vapour pressure of pure A and B :

Options

A. `P_(A)^(@)=1.4atm,P_(B)^(@)=0.7atm`
B. `P_(A)^(@)=1.2atm,P_(B)^(@)=0.6atm`
C. `P_(A)^(@)=1.4atm,P_(B)^(@)=0.6atm`
D. `P_(A)^(@)=0.6atm,P_(B)^(@)=1.4atm`

Explanation

To solve the problem, we need to find the vapor pressures of pure liquids A and B using the information provided about their mixtures and the application of Raoult's Law.

### Step 1: Analyze the first mixture of A and B
We have a mixture with 2 moles of A and 2 moles of B. The total vapor pressure (P_T) of this mixture is given as 1 atm.

- Moles of A (n_A) = 2
- Moles of B (n_B) = 2
- Total moles (n_T) = n_A + n_B = 2 + 2 = 4

Now, we can calculate the mole fractions:
- Mole fraction of A (X_A) = n_A / n_T = 2 / 4 = 1/2
- Mole fraction of B (X_B) = n_B / n_T = 2 / 4 = 1/2

Using Raoult's Law:
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
Substituting the known values:
\[ 1 = P_A^0 \cdot \frac{1}{2} + P_B^0 \cdot \frac{1}{2} \]

Multiplying through by 2:
\[ 2 = P_A^0 + P_B^0 \]
This gives us our first equation:
\[ P_A^0 + P_B^0 = 2 \quad \text{(Equation 1)} \]

### Step 2: Analyze the second mixture of A, B, and C
In the second mixture, we have:
- 1 mole of A
- 3 moles of B
- 4 moles of C

The total vapor pressure of this mixture is greater than 1 atm, but when 4 moles of C are added, the vapor pressure drops to 1 atm. The vapor pressure of pure C (P_C^0) is given as 0.8 atm.

Total moles in the second mixture:
- Total moles (n_T) = 1 + 3 + 4 = 8

Calculating the mole fractions:
- Mole fraction of A (X_A) = 1 / 8
- Mole fraction of B (X_B) = 3 / 8
- Mole fraction of C (X_C) = 4 / 8 = 1/2

Using Raoult's Law again:
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot X_B + P_C^0 \cdot X_C \]
Substituting the known values:
\[ 1 = P_A^0 \cdot \frac{1}{8} + P_B^0 \cdot \frac{3}{8} + 0.8 \cdot \frac{4}{8} \]

Simplifying:
\[ 1 = P_A^0 \cdot \frac{1}{8} + P_B^0 \cdot \frac{3}{8} + 0.4 \]

Rearranging gives us:
\[ P_A^0 \cdot \frac{1}{8} + P_B^0 \cdot \frac{3}{8} = 0.6 \]
Multiplying through by 8:
\[ P_A^0 + 3P_B^0 = 4.8 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( P_A^0 + P_B^0 = 2 \) (Equation 1)
2. \( P_A^0 + 3P_B^0 = 4.8 \) (Equation 2)

We can solve these equations simultaneously. From Equation 1:
\[ P_A^0 = 2 - P_B^0 \]

Substituting this into Equation 2:
\[ (2 - P_B^0) + 3P_B^0 = 4.8 \]
\[ 2 + 2P_B^0 = 4.8 \]
\[ 2P_B^0 = 4.8 - 2 \]
\[ 2P_B^0 = 2.8 \]
\[ P_B^0 = 1.4 \, \text{atm} \]

Now substituting back to find \( P_A^0 \):
\[ P_A^0 = 2 - 1.4 = 0.6 \, \text{atm} \]

### Final Answer
- Vapor pressure of pure A, \( P_A^0 = 0.6 \, \text{atm} \)
- Vapor pressure of pure B, \( P_B^0 = 1.4 \, \text{atm} \)