the degree of dissociation of `PCI_(5)` at a certain temperature and under one atmosphere pressure is 0.2. Calculate the pressure at which it is half dissociated at the same temperature.

Explanation

To solve the problem of finding the pressure at which the degree of dissociation of \( PCl_5 \) is 0.5, we can follow these steps:

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

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation. At a certain initial pressure \( P_0 = 1 \) atm, we are given that \( \alpha = 0.2 \).

### Step 3: Calculate the equilibrium constant \( K_p \)
The equilibrium constant \( K_p \) can be expressed in terms of partial pressures:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

At equilibrium, the partial pressures can be expressed as:
- \( P_{PCl_5} = P_0(1 - \alpha) \)
- \( P_{PCl_3} = P_0 \alpha \)
- \( P_{Cl_2} = P_0 \alpha \)

Substituting these into the expression for \( K_p \):
\[ K_p = \frac{(P_0 \alpha)(P_0 \alpha)}{P_0(1 - \alpha)} = \frac{P_0 \alpha^2}{1 - \alpha} \]

### Step 4: Substitute known values to find \( K_p \)
Substituting \( P_0 = 1 \) atm and \( \alpha = 0.2 \):
\[ K_p = \frac{1 \cdot (0.2)^2}{1 - 0.2} = \frac{0.04}{0.8} = 0.05 \]

### Step 5: Find the pressure at which \( \alpha = 0.5 \)
Now, we need to find the pressure \( P \) at which \( \alpha = 0.5 \). Using the same expression for \( K_p \):
\[ K_p = \frac{P \alpha^2}{1 - \alpha} \]

Substituting \( K_p = 0.05 \) and \( \alpha = 0.5 \):
\[ 0.05 = \frac{P (0.5)^2}{1 - 0.5} \]
\[ 0.05 = \frac{P \cdot 0.25}{0.5} \]
\[ 0.05 = \frac{P \cdot 0.25}{0.5} = \frac{P \cdot 0.25}{0.5} = \frac{P}{2} \]

### Step 6: Solve for \( P \)
Rearranging gives:
\[ P = 0.05 \cdot 2 = 0.1 \text{ atm} \]

### Final Answer
The pressure at which \( PCl_5 \) is half dissociated at the same temperature is:
\[ P = 0.1 \text{ atm} \]

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