Two solids `A` and `D` dissociates into gaseous products as follows
`C(s)hArrB(g)+C(g),K_(P_(1))=300 , D(s)hArrE(g)+C(g)K_(P_(2)=600`
at `27^(@)C`, then find the total pressure of the solid mixture.

Answer

To solve the problem, we need to analyze the dissociation of the solids A and D into their gaseous products and calculate the total pressure of the system. Let's break it down step by step.

### Step 1: Write the dissociation reactions
The dissociation reactions for the solids A and D are given as follows:
1. \( A(s) \rightleftharpoons B(g) + C(g) \) with \( K_{P1} = 300 \)
2. \( D(s) \rightleftharpoons E(g) + C(g) \) with \( K_{P2} = 600 \)

### Step 2: Set up the equilibrium expressions
For the first reaction involving solid A:
- Let the partial pressure of \( B \) be \( P_1 \).
- The partial pressure of \( C \) will also be \( P_1 \) since one mole of \( C \) is produced for each mole of \( B \).

The equilibrium expression for the first reaction is:
\[ K_{P1} = \frac{P_B \cdot P_C}{1} = P_1 \cdot P_1 = P_1^2 \]
Thus, we have:
\[ P_1^2 = 300 \quad \text{(Equation 1)} \]

For the second reaction involving solid D:
- Let the partial pressure of \( E \) be \( P_2 \).
- The partial pressure of \( C \) will also be \( P_2 \).

The equilibrium expression for the second reaction is:
\[ K_{P2} = \frac{P_E \cdot P_C}{1} = P_2 \cdot P_2 = P_2^2 \]
Thus, we have:
\[ P_2^2 = 600 \quad \text{(Equation 2)} \]

### Step 3: Relate the pressures
Since \( C \) is a common product in both reactions, we can express the total pressure in terms of \( P_1 \) and \( P_2 \):
\[ P_{total} = P_B + P_C + P_E = P_1 + P_1 + P_2 = 2P_1 + P_2 \]

### Step 4: Solve the equations
From Equation 1:
\[ P_1 = \sqrt{300} = 10\sqrt{3} \approx 17.32 \]

From Equation 2:
\[ P_2 = \sqrt{600} = 10\sqrt{6} \approx 24.49 \]

### Step 5: Calculate the total pressure
Now substituting \( P_1 \) and \( P_2 \) into the total pressure equation:
\[ P_{total} = 2P_1 + P_2 = 2(10\sqrt{3}) + 10\sqrt{6} \]
Calculating this gives:
\[ P_{total} \approx 2(17.32) + 24.49 \approx 34.64 + 24.49 \approx 59.13 \text{ atm} \]

### Final Answer
Thus, the total pressure of the solid mixture is approximately \( 60 \, \text{atm} \).

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