Calculate the volume percent of chlorine gas at equilibrium in the dissociation of `PCl_(5)(g)` under a total pressure of `1.5` atm. The `K_(p)` for its dissociation `=0.3`.

Explanation

To solve the problem of calculating the volume percent of chlorine gas at equilibrium in the dissociation of \( PCl_5(g) \), we will follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Define initial conditions and degree of dissociation
Let the initial amount of \( PCl_5 \) be 1 mole. At equilibrium, let \( \alpha \) be the degree of dissociation. Therefore:
- Initial moles of \( PCl_5 = 1 \)
- Moles of \( PCl_3 \) at equilibrium = \( \alpha \)
- Moles of \( Cl_2 \) at equilibrium = \( \alpha \)
- Moles of \( PCl_5 \) at equilibrium = \( 1 - \alpha \)

### Step 3: Calculate total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Use the total pressure to find partial pressures
The total pressure \( P \) is given as 1.5 atm. The partial pressures can be expressed as:
- \( P_{PCl_5} = \frac{1 - \alpha}{1 + \alpha} \times P \)
- \( P_{PCl_3} = \frac{\alpha}{1 + \alpha} \times P \)
- \( P_{Cl_2} = \frac{\alpha}{1 + \alpha} \times P \)

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
Substituting the expressions for the partial pressures:
\[ K_p = \frac{\left( \frac{\alpha}{1 + \alpha} \times P \right) \cdot \left( \frac{\alpha}{1 + \alpha} \times P \right)}{\frac{1 - \alpha}{1 + \alpha} \times P} \]
This simplifies to:
\[ K_p = \frac{\alpha^2 \cdot P^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 6: Substitute known values and solve for \( \alpha \)
Given \( K_p = 0.3 \) and \( P = 1.5 \):
\[ 0.3 = \frac{\alpha^2 \cdot (1.5)^2}{(1 - \alpha)(1 + \alpha)} \]
This simplifies to:
\[ 0.3 = \frac{2.25 \alpha^2}{(1 - \alpha)(1 + \alpha)} \]
Cross-multiplying gives:
\[ 0.3(1 - \alpha)(1 + \alpha) = 2.25 \alpha^2 \]
Expanding and rearranging leads to a quadratic equation in \( \alpha \).

### Step 7: Solve the quadratic equation
After solving the quadratic equation, we find:
\[ \alpha = \sqrt{\frac{1}{6}} \]

### Step 8: Calculate the volume percent of \( Cl_2 \)
The volume percent of \( Cl_2 \) is given by:
\[ \text{Volume percent of } Cl_2 = \frac{\alpha}{1 + \alpha} \times 100 \]
Substituting \( \alpha \):
\[ \text{Volume percent of } Cl_2 = \frac{\sqrt{\frac{1}{6}}}{1 + \sqrt{\frac{1}{6}}} \times 100 \]
Calculating this gives approximately:
\[ \text{Volume percent of } Cl_2 \approx 28.6\% \]

### Final Answer
The volume percent of chlorine gas at equilibrium is approximately **28.6%**.