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.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ A(s) \rightleftharpoons B(g) + C(g) \]
with equilibrium constant \( K_{P1} = x \, \text{atm}^2 \).
- The second reaction is:
\[ D(s) \rightleftharpoons C(g) + E(g) \]
with equilibrium constant \( K_{P2} = y \, \text{atm}^2 \).

2. **Establish Partial Pressures**:
- Let \( P_1 \) be the partial pressure of \( B \) and \( C \) from the first reaction.
- Let \( P_2 \) be the partial pressure of \( E \) from the second reaction.

3. **Relate Partial Pressures to Equilibrium Constants**:
- For the first reaction, the equilibrium expression is:
\[ K_{P1} = P_B \cdot P_C = P_1 \cdot (P_1 + P_2) = x \]
- For the second reaction, the equilibrium expression is:
\[ K_{P2} = P_C \cdot P_E = (P_1 + P_2) \cdot P_2 = y \]

4. **Combine the Equations**:
- From the first reaction:
\[ x = P_1 \cdot (P_1 + P_2) \]
- From the second reaction:
\[ y = (P_1 + P_2) \cdot P_2 \]

5. **Express Total Pressure**:
- The total pressure \( P_{total} \) when both solids dissociate is given by:
\[ P_{total} = P_B + P_C + P_E = P_1 + (P_1 + P_2) + P_2 = 2P_1 + P_2 \]

6. **Substitute for \( P_1 + P_2 \)**:
- From the combined equations, we can find \( P_1 + P_2 \):
\[ P_1 + P_2 = \sqrt{x + y} \]
- Thus, substituting this into the total pressure equation:
\[ P_{total} = 2P_1 + P_2 = 2\sqrt{x + y} \]

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