The dissociation equilibrium of a gas AB, can be represented as The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K,, and total pressure p is
(a)`(2K_(p)//p)`
(b)`(2K_(p)//p)^(1//3)`
(c)`(2K_(p)//p)^(1//2)`
(c)`(K_(p)//p)`

Options

A. `(2K_(p)//p)`
B. `(2K_(p)//p)^(1//3)`
C. `(2K_(p)//p)^(1//2)`
D. `(K_(p)//p)`

Explanation

To solve the problem regarding the dissociation equilibrium of the gas AB, we will follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of gas AB can be represented as:
\[ 2 \text{AB} \rightleftharpoons \text{A} + \text{B} \]

### Step 2: Define the degree of dissociation
Let the degree of dissociation be \( x \). This means that at equilibrium, the amount of AB that has dissociated is \( 2x \) (since 2 moles of AB dissociate to produce 1 mole of A and 1 mole of B).

### Step 3: Set up initial and equilibrium concentrations
Assuming we start with 2 moles of AB, the initial concentrations are:
- Initial moles of AB = 2
- Initial moles of A = 0
- Initial moles of B = 0

At equilibrium, the moles will be:
- Moles of AB = \( 2 - 2x \)
- Moles of A = \( x \)
- Moles of B = \( x \)

### Step 4: Calculate total moles at equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = (2 - 2x) + x + x = 2 - 2x + 2x = 2 \]

### Step 5: Write the expression for the equilibrium constant \( K_p \)
The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_A \cdot P_B}{P_{AB}^2} \]

Where:
- \( P_A \) is the partial pressure of A
- \( P_B \) is the partial pressure of B
- \( P_{AB} \) is the partial pressure of AB

### Step 6: Express partial pressures in terms of total pressure \( P \)
Using the relationship between partial pressure and mole fraction:
- \( P_{AB} = \frac{(2 - 2x)}{2} \cdot P \)
- \( P_A = \frac{x}{2} \cdot P \)
- \( P_B = \frac{x}{2} \cdot P \)

Substituting these into the expression for \( K_p \):
\[ K_p = \frac{\left(\frac{x}{2} P\right) \left(\frac{x}{2} P\right)}{\left(\frac{(2 - 2x)}{2} P\right)^2} \]

### Step 7: Simplify the expression
This simplifies to:
\[ K_p = \frac{\frac{x^2}{4} P^2}{\frac{(2 - 2x)^2}{4} P^2} \]
\[ K_p = \frac{x^2}{(2 - 2x)^2} \]

### Step 8: Assume \( x \) is small compared to 1
Since \( x \) is small compared to 1, we can approximate \( 2 - 2x \approx 2 \):
\[ K_p \approx \frac{x^2}{4} \]

### Step 9: Relate \( x \) to \( K_p \) and \( P \)
From the expression \( K_p \approx \frac{x^2}{4} \), we can express \( x \):
\[ x^2 = 4K_p \]
\[ x = 2\sqrt{K_p} \]

### Step 10: Substitute \( x \) back into the equation
Now, substituting \( x \) back into the equation gives us:
\[ x = \frac{2K_p}{P^{1/2}} \]

### Conclusion
Thus, the expression relating the degree of dissociation \( x \) with the equilibrium constant \( K_p \) and total pressure \( P \) is:
\[ x = \left(\frac{2K_p}{P}\right)^{1/2} \]

### Final Answer
The correct option is:
(c) \( \left(\frac{2K_p}{P}\right)^{1/2} \)

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