In a closed system : `A(s)hArr 2B(g)+3C`, if the partial pressure of C is doubled, then partial pressure of B will be

Options

A. two times the original pressure
B. one half of its original value
C. `(1)/(2sqrt2)` times to the original value
D. `2sqrt2` times to the original value

Answer

To solve the problem, we need to analyze the given reaction and how the change in the partial pressure of one of the products affects the other product. The reaction is:

\[ A(s) \rightleftharpoons 2B(g) + 3C(g) \]

### Step-by-Step Solution:

1. **Identify the Reaction and Kp Expression**:
The equilibrium constant \( K_p \) for the reaction can be expressed as:
\[ K_p = \frac{(P_B)^2 (P_C)^3}{1} \]
where \( P_B \) is the partial pressure of gas B and \( P_C \) is the partial pressure of gas C. The solid A does not appear in the expression since its activity is considered to be 1.

2. **Initial Conditions**:
Let the initial partial pressures of B and C be \( P_B \) and \( P_C \) respectively.

3. **Change in Partial Pressure of C**:
According to the problem, the partial pressure of C is doubled:
\[ P_C' = 2P_C \]

4. **Set Up the New Kp Expression**:
With the new partial pressure of C, the equilibrium constant expression becomes:
\[ K_p = (P_B')^2 (P_C')^3 = (P_B')^2 (2P_C)^3 \]
Simplifying this gives:
\[ K_p = (P_B')^2 \cdot 8P_C^3 \]

5. **Equate the Two Kp Expressions**:
Since \( K_p \) remains constant, we can set the two expressions for \( K_p \) equal to each other:
\[ (P_B)^2 (P_C)^3 = (P_B')^2 \cdot 8P_C^3 \]

6. **Cancel Out Common Terms**:
Since \( P_C^3 \) is common in both sides, we can cancel it out (assuming \( P_C \neq 0 \)):
\[ (P_B)^2 = (P_B')^2 \cdot 8 \]

7. **Solve for \( P_B' \)**:
Taking the square root of both sides gives:
\[ P_B = P_B' \cdot \sqrt{8} \]
Rearranging this, we find:
\[ P_B' = \frac{P_B}{\sqrt{8}} = \frac{P_B}{2\sqrt{2}} \]

### Conclusion:
The new partial pressure of B, \( P_B' \), when the partial pressure of C is doubled, is:
\[ P_B' = \frac{P_B}{2\sqrt{2}} \]

### Final Answer:
The partial pressure of B will be \( \frac{1}{2\sqrt{2}} \) times its original pressure.

---