For an ideal solution, if a graph is plotted between `(1)/(P_(T))` and `y_(A)` (mole fraction of A in vapour phase) where `p_(A)^(0)gtp_(B)^(0)` then

Options

A. intercept of the graph `=(1)/(p_(B)^(0))`
B. slope of the graph `=((1)/(p_(A)^(0))+(1)/(p_(B)^(0)))`
C. slope of the graph `((1)/(p_(B)^(0))-(1)/(p_(B)^(0)))`
D. the graph is linear

Explanation

To solve the problem, we need to analyze the relationship between the total pressure \( P_T \) and the mole fraction of component A in the vapor phase \( y_A \) for an ideal solution. We will derive the equation and identify the characteristics of the graph plotted between \( \frac{1}{P_T} \) and \( y_A \).

### Step 1: Understand the relationship for an ideal solution
For an ideal solution, the total vapor pressure \( P_T \) is given by Raoult's Law:
\[ P_T = P_A^0 \cdot x_A + P_B^0 \cdot x_B \]
where:
- \( P_A^0 \) and \( P_B^0 \) are the vapor pressures of pure components A and B, respectively.
- \( x_A \) and \( x_B \) are the mole fractions of components A and B in the liquid phase.

### Step 2: Express \( x_A \) in terms of \( y_A \)
Since \( x_A + x_B = 1 \), we can express \( x_B \) as:
\[ x_B = 1 - x_A \]
Now, substituting \( x_B \) in the equation for \( P_T \):
\[ P_T = P_A^0 \cdot x_A + P_B^0 \cdot (1 - x_A) = P_A^0 \cdot x_A + P_B^0 - P_B^0 \cdot x_A \]
This simplifies to:
\[ P_T = (P_A^0 - P_B^0) \cdot x_A + P_B^0 \]

### Step 3: Relate \( y_A \) and \( P_T \)
The mole fraction of A in the vapor phase \( y_A \) is given by:
\[ y_A = \frac{P_A}{P_T} \]
where \( P_A = P_A^0 \cdot x_A \) (from Raoult's Law).

### Step 4: Substitute \( P_A \) in terms of \( y_A \)
From the expression for \( P_T \), we can express \( P_A \):
\[ P_A = P_A^0 \cdot x_A \]
Substituting this into the equation for \( y_A \):
\[ y_A = \frac{P_A^0 \cdot x_A}{P_T} \]

### Step 5: Rearranging the equation
Now, we can express \( \frac{1}{P_T} \):
\[ \frac{1}{P_T} = \frac{1}{P_A^0 \cdot x_A + P_B^0 \cdot (1 - x_A)} \]
This leads to the equation:
\[ \frac{1}{P_T} = \frac{1}{P_B^0} + \frac{1}{P_A^0 - P_B^0} \cdot y_A \]
This is a linear equation of the form \( y = mx + c \), where:
- \( y = \frac{1}{P_T} \)
- \( x = y_A \)
- The slope \( m = \frac{1}{P_A^0 - P_B^0} \)
- The intercept \( c = \frac{1}{P_B^0} \)

### Step 6: Analyze the graph
From the derived equation, we can conclude:
1. The intercept of the graph is \( \frac{1}{P_B^0} \).
2. The slope of the graph is \( \frac{1}{P_A^0 - P_B^0} \).
3. The graph is linear.

### Conclusion
Based on the analysis:
- The first option (intercept is \( \frac{1}{P_B^0} \)) is correct.
- The second option (slope is \( \frac{1}{P_A^0 + P_B^0} \)) is incorrect.
- The third option (slope is \( \frac{1}{P_A^0 - P_B^0} \)) is correct.
- The fourth option (the graph is linear) is correct.

Thus, the correct options are 1, 3, and 4.