`underset(Z)overset(M)()A(g)rarr underset(Z-4)overset(M-8)()B(g)+(alpha-` particles `)`
`( alpha-` particles are helium nuclei, so will form helium gas by trapping electrons )
The radioactive disintegration follows first `-` order kinetics Starting with 1 mol of A in a 1- litre closed flask at `27^(@)C` pressure developed after tw0 half- lives is approximately.

Options

A. 24 atm
B. 65 atm
C. 61.5 atm
D. 12 atm

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ A(g) \rightarrow B(g) + \alpha \text{ particles} \]
where \(\alpha\) particles are helium nuclei. The reaction follows first-order kinetics.

### Step 2: Determine the Half-Life
For a first-order reaction, the half-life (\(t_{1/2}\)) is constant and does not depend on the initial concentration. The half-life can be expressed as:
\[ t_{1/2} = \frac{0.693}{k} \]
where \(k\) is the rate constant.

### Step 3: Calculate Moles After Two Half-Lives
Starting with 1 mole of \(A\):
- After the first half-life, the amount of \(A\) remaining is:
\[ A_1 = \frac{1}{2} \text{ moles of } A \]
- After the second half-life, the amount of \(A\) remaining is:
\[ A_2 = \frac{A_1}{2} = \frac{1}{4} \text{ moles of } A \]

Thus, the amount of \(A\) that has reacted after two half-lives:
\[ \text{Reacted } A = 1 - A_2 = 1 - \frac{1}{4} = \frac{3}{4} \text{ moles of } A \]

### Step 4: Calculate Moles of Products Formed
From the stoichiometry of the reaction:
- For every mole of \(A\) that reacts, it produces:
- \(1\) mole of \(B\)
- \(2\) moles of \(\alpha\) particles (since each \(\alpha\) particle is a helium nucleus, we consider 2 moles of helium gas).

Thus, after the reaction:
- Moles of \(B\) formed:
\[ \text{Moles of } B = \frac{3}{4} \text{ moles} \]
- Moles of \(\alpha\) particles (which will form helium gas):
\[ \text{Moles of } \alpha = 2 \times \frac{3}{4} = \frac{3}{2} \text{ moles} \]

### Step 5: Calculate Total Moles in the System
Now, we can calculate the total moles in the system after two half-lives:
- Remaining moles of \(A\):
\[ \text{Remaining } A = \frac{1}{4} \text{ moles} \]
- Moles of \(B\) formed:
\[ \text{Moles of } B = \frac{3}{4} \text{ moles} \]
- Moles of \(\alpha\) (helium) formed:
\[ \text{Moles of } \alpha = \frac{3}{2} \text{ moles} \]

Total moles:
\[ N_{total} = \frac{1}{4} + \frac{3}{4} + \frac{3}{2} = \frac{1}{4} + \frac{3}{4} + \frac{6}{4} = \frac{10}{4} = 2.5 \text{ moles} \]

### Step 6: Calculate Pressure Using Ideal Gas Law
Using the ideal gas law:
\[ PV = nRT \]
Where:
- \(n = 2.5 \text{ moles}\)
- \(R = 0.0821 \text{ L atm/(K mol)}\)
- \(T = 27^\circ C = 300 \text{ K}\)
- \(V = 1 \text{ L}\)

Rearranging for pressure \(P\):
\[ P = \frac{nRT}{V} = \frac{2.5 \times 0.0821 \times 300}{1} \]

Calculating:
\[ P = 61.575 \text{ atm} \]

### Step 7: Final Answer
The pressure developed after two half-lives is approximately:
\[ P \approx 61.5 \text{ atm} \]