Two solids dissociates as follows
`A(s) rarr B(g) + C(g), K_(P_(1)) = x atm^(2)`
`D(s) rarr C(g) + E(g), K_(P_(2)) = y atm^(2)`
The total pressure when both the solids dissociate simultaneously is :

Options

A. `2(sqrt(x + y))atm`
B. (x + y)atm
C. `(sqrt(x + y))atm`
D. `x^(2) + y^(2)`

Explanation

To solve the problem, we need to analyze the dissociation of the two solids and how they contribute to the total pressure when they dissociate simultaneously. Let's break it down step by step.

### Step 1: Write the dissociation reactions and equilibrium constants
The dissociation reactions are given as follows:
1. \( A(s) \rightleftharpoons B(g) + C(g) \) with \( K_{P1} = x \, \text{atm}^2 \)
2. \( D(s) \rightleftharpoons C(g) + E(g) \) with \( K_{P2} = y \, \text{atm}^2 \)

### Step 2: Define partial pressures at equilibrium
Let:
- The partial pressure of \( B \) be \( P_1 \)
- The partial pressure of \( C \) be \( P_C \)
- The partial pressure of \( E \) be \( P_2 \)

Since \( C \) is produced in both reactions, we can express its total pressure as:
\[ P_C = P_1 + P_2 \]

### Step 3: Write the expressions for \( K_{P1} \) and \( K_{P2} \)
From the first reaction:
\[ K_{P1} = P_B \cdot P_C = P_1 \cdot (P_1 + P_2) = x \]
This gives us our first equation:
\[ P_1(P_1 + P_2) = x \]

From the second reaction:
\[ K_{P2} = P_C \cdot P_E = (P_1 + P_2) \cdot P_2 = y \]
This gives us our second equation:
\[ (P_1 + P_2)P_2 = y \]

### Step 4: Expand and rearrange the equations
Expanding both equations:
1. \( P_1^2 + P_1 P_2 = x \)
2. \( P_1 P_2 + P_2^2 = y \)

### Step 5: Combine the equations
Adding both equations:
\[ P_1^2 + 2P_1 P_2 + P_2^2 = x + y \]
This can be rewritten as:
\[ (P_1 + P_2)^2 = x + y \]

### Step 6: Solve for \( P_1 + P_2 \)
Taking the square root of both sides:
\[ P_1 + P_2 = \sqrt{x + y} \]

### Step 7: Calculate the total pressure
The total pressure \( P_T \) is given by:
\[ P_T = P_B + P_C + P_E = P_1 + (P_1 + P_2) + P_2 \]
This simplifies to:
\[ P_T = 2P_1 + P_2 \]
Substituting \( P_C = P_1 + P_2 \):
\[ P_T = 2P_1 + P_C - P_1 = P_1 + P_C = P_1 + (P_1 + P_2) = 2P_1 + P_2 \]

Using \( P_1 + P_2 = \sqrt{x + y} \):
\[ P_T = 2\sqrt{x + y} \]

### Final Answer
Thus, the total pressure when both solids dissociate simultaneously is:
\[ P_T = 2\sqrt{x + y} \, \text{atm} \]