Step 1: Understanding the Concept:
We need to derive the expression for the equilibrium constant \( K_p \) in terms of partial pressures, which depend on the mole fractions (determined by the degree of dissociation \(\alpha\)) and the total pressure \( P \).
Step 2: Key Formula or Approach:
1. Reaction: \( \text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2 \).
2. Let initial moles be 1. At equilibrium:
- \( n_{\text{N}_2\text{O}_4} = 1 - \alpha \)
- \( n_{\text{NO}_2} = 2\alpha \)
- Total moles \( n_{total} = 1 + \alpha \).
3. Partial Pressure \( p_i = \left( \frac{n_i}{n_{total}} \right) P \).
4. \( K_p = \frac{(p_{\text{NO}_2})^2}{p_{\text{N}_2\text{O}_4}} \).
Step 3: Detailed Explanation:
Calculate partial pressures:
\[ p_{\text{NO}_2} = \left( \frac{2\alpha}{1+\alpha} \right) P \]
\[ p_{\text{N}_2\text{O}_4} = \left( \frac{1-\alpha}{1+\alpha} \right) P \]
Substitute these into the \( K_p \) expression:
\[ K_p = \frac{\left[ \frac{2\alpha P}{1+\alpha} \right]^2}{\left[ \frac{(1-\alpha)P}{1+\alpha} \right]} \]
\[ K_p = \frac{4\alpha^2 P^2}{(1+\alpha)^2} \times \frac{1+\alpha}{(1-\alpha)P} \]
\[ K_p = \frac{4\alpha^2 P}{(1+\alpha)(1-\alpha)} = \frac{4\alpha^2 P}{1-\alpha^2} \]
Now, solve for \( \alpha \):
\[ K_p (1 - \alpha^2) = 4\alpha^2 P \]
\[ K_p - K_p \alpha^2 = 4\alpha^2 P \]
\[ K_p = \alpha^2 (4P + K_p) \]
\[ \alpha^2 = \frac{K_p}{4P + K_p} \]
Divide numerator and denominator by \( P \):
\[ \alpha^2 = \frac{K_p/P}{4 + K_p/P} \]
Taking the square root:
\[ \alpha = \left( \frac{K_p/P}{4 + K_p/P} \right)^{1/2} \]
Step 4: Final Answer:
The relation is \( \alpha = \left( \frac{K_p/P}{4 + K_p/P} \right)^{1/2} \).