For any `I^(st)` order gaseous reaction `A rarr 2B` Pressure devoloped after 20 min is 16.4 atm and after very long time is 20 atm. What is the total pressure devoloped after 10 min.

Options

A. 12
B. 13
C. 14
D. 15

Explanation

To solve the problem, we need to analyze the first-order gaseous reaction \( A \rightarrow 2B \) and the pressures involved at different time intervals.

### Step-by-step Solution:

1. **Identify Initial Conditions**:
- Let the initial pressure of gas \( A \) be \( P_0 \).
- At \( t = 0 \), pressure of \( A \) is \( P_0 \) and pressure of \( B \) is 0.

2. **Pressure at Infinite Time**:
- After a very long time, the total pressure is given as 20 atm. Since the reaction produces 2 moles of \( B \) for every mole of \( A \) that reacts, at infinite time, all of \( A \) will have converted to \( B \).
- Therefore, \( P_0 + P_B = 20 \) atm, where \( P_B \) is the pressure of \( B \) at infinite time.
- Since \( P_B = 2P_0 \) (because all of \( A \) converts to \( B \)), we can write:
\[ P_0 + 2P_0 = 20 \implies 3P_0 = 20 \implies P_0 = \frac{20}{3} \approx 10 \text{ atm} \]

3. **Pressure After 20 Minutes**:
- The pressure developed after 20 minutes is given as 16.4 atm.
- Let \( P' \) be the pressure of \( B \) after 20 minutes. The pressure of \( A \) at this time will be \( P_0 - P' \).
- The total pressure at this time can be expressed as:
\[ P_0 - P' + 2P' = P_0 + P' = 16.4 \text{ atm} \]
- Substituting \( P_0 = 10 \):
\[ 10 + P' = 16.4 \implies P' = 16.4 - 10 = 6.4 \text{ atm} \]

4. **Using the Geometric Progression Relation**:
- For a first-order reaction, the pressures follow a geometric progression:
\[ \frac{P_0 - P'}{P_0 - P} = \frac{P_0 - P}{P_0} \]
- Let \( P \) be the pressure of \( B \) after 10 minutes. Then:
\[ P_0 - P' = 10 - 6.4 = 3.6 \text{ atm} \]
- Substituting into the GP relation:
\[ \frac{3.6}{10 - P} = \frac{10 - P}{10} \]

5. **Cross Multiplying and Solving for \( P \)**:
- Cross-multiplying gives:
\[ 3.6 \cdot 10 = (10 - P)(10 - P) \]
\[ 36 = 100 - 20P + P^2 \]
\[ P^2 - 20P + 64 = 0 \]

6. **Using the Quadratic Formula**:
- Solving the quadratic equation:
\[ P = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
\[ P = \frac{20 \pm \sqrt{400 - 256}}{2} \]
\[ P = \frac{20 \pm \sqrt{144}}{2} \]
\[ P = \frac{20 \pm 12}{2} \]
- This gives two possible values for \( P \):
\[ P = \frac{32}{2} = 16 \quad \text{or} \quad P = \frac{8}{2} = 4 \]

7. **Calculating Total Pressure After 10 Minutes**:
- We take \( P = 4 \) atm (since \( P \) must be less than \( P_0 \)).
- The total pressure after 10 minutes is:
\[ P_0 + P = 10 + 4 = 14 \text{ atm} \]

### Final Answer:
The total pressure developed after 10 minutes is **14 atm**.