For a first order homogeneous gaseous reaction
`Ato 2B+C`
if the total pressure after time t was `P_(1)` and after long time `(t to infty)` was `P_(infty)` then K in terms of `P_(t_(1))` `P_(infty)` t is

Options

A. `K=(2.303)/(t)log((P_(infty))/(P_(infty)-P_(t)))`
B. `K=(2.303)/(t)log((2P_(infty))/(P_(infty-P_(t))))`
C. `K=(2.303)/(t)log((2P_(infty))/(3(P_(infty-P_(t)))))`
D. none of these

Explanation

To solve the problem, we need to derive the expression for the rate constant \( K \) in terms of the total pressure at time \( t \) (denoted as \( P_t \)) and the total pressure at equilibrium (denoted as \( P_{\infty} \)) for the reaction \( A \to 2B + C \).

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- At \( t = 0 \), only reactant \( A \) is present. Let the initial pressure of \( A \) be \( P_A \) and the pressures of \( B \) and \( C \) be \( 0 \).
- Therefore, \( P(0) = P_A \).

2. **Define Pressures at Time \( t \)**:
- At time \( t \), let \( P_{A} \) decrease by \( P_d \) (the pressure of \( A \) that has reacted).
- The pressure of \( B \) formed will be \( 2P_d \) (since 2 moles of \( B \) are formed for every mole of \( A \) that reacts), and the pressure of \( C \) formed will be \( P_d \).
- The total pressure at time \( t \) can be expressed as:
\[ P_t = (P_A - P_d) + 2P_d + P_d = P_A + 2P_d \]
- Rearranging gives:
\[ P_d = \frac{P_t - P_A}{2} \]

3. **Define Pressures at Equilibrium**:
- At \( t \to \infty \), all of \( A \) will have reacted, so:
- \( P_A = 0 \)
- The total pressure will be:
\[ P_{\infty} = 2P_d + P_d = 3P_d \]
- Therefore, we can express \( P_d \) in terms of \( P_{\infty} \):
\[ P_d = \frac{P_{\infty}}{3} \]

4. **Substituting \( P_d \) into the Expression for \( P_t \)**:
- Substitute \( P_d \) back into the equation for \( P_t \):
\[ P_t = P_A + 2\left(\frac{P_{\infty}}{3}\right) = P_A + \frac{2P_{\infty}}{3} \]
- Rearranging gives:
\[ P_A = P_t - \frac{2P_{\infty}}{3} \]

5. **Using the First Order Reaction Rate Expression**:
- For a first-order reaction, the rate constant \( K \) can be expressed as:
\[ Kt = \ln\left(\frac{P_A}{P_t - P_A}\right) \]
- Substitute \( P_A \) into this equation:
\[ Kt = \ln\left(\frac{P_t - \frac{2P_{\infty}}{3}}{P_t - \left(P_t - \frac{2P_{\infty}}{3}\right)}\right) = \ln\left(\frac{P_t - \frac{2P_{\infty}}{3}}{\frac{P_{\infty}}{3}}\right) \]

6. **Final Expression for \( K \)**:
- Rearranging gives:
\[ K = \frac{1}{t} \ln\left(\frac{3(P_t - \frac{2P_{\infty}}{3})}{P_{\infty}}\right) \]
- This can be simplified to:
\[ K = \frac{1}{t} \ln\left(\frac{3P_t - 2P_{\infty}}{P_{\infty}}\right) \]

### Final Answer:
\[ K = \frac{1}{t} \ln\left(\frac{3P_t - 2P_{\infty}}{P_{\infty}}\right) \]