There are two closed containers A and B Container A contains `PCl_3` gas at `25^@C` temperature and 1 bar pressure. Container B contain equilibrium mixture of gases obtained by partial decomposition of `PCl_(5)` (g) at the same temperature and pressure. Assuming that initially equal mole of `PCl_(5)` (g) at same temperature and pressure . Assuming that initially equal moles of `PCl_(5)` were taken into the containers ( before decomposition ), the gases are allowed to effuse through a pinhole in the containers, Which container will show higher rate of effusion ? Justify .

Options

A. container A
B. container B
C. both
D. none

Answer

To determine which container will show a higher rate of effusion, we need to analyze the gases present in each container and apply Graham's law of effusion.

### Step-by-Step Solution:

1. **Identify the Gases in Each Container**:
- Container A contains only `PCl3` gas.
- Container B contains an equilibrium mixture of gases formed from the partial decomposition of `PCl5`, which can be represented as:
\[ PCl5 \rightleftharpoons PCl3 + Cl2 \]
- Initially, we assume equal moles of `PCl5` were taken into both containers.

2. **Decomposition of `PCl5`**:
- Let’s denote the degree of dissociation of `PCl5` as α.
- Initially, we have 1 mole of `PCl5`, and after decomposition, the moles of gases will be:
- Moles of `PCl5` remaining: \( 1 - \alpha \)
- Moles of `PCl3` produced: \( \alpha \)
- Moles of `Cl2` produced: \( \alpha \)
- Total moles in Container B after decomposition:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

3. **Calculate the Molecular Weights**:
- The molecular weight of `PCl3`:
\[ M_{PCl3} = 30.97 \, (\text{P}) + 3 \times 35.45 \, (\text{Cl}) = 137.33 \, g/mol \]
- The molecular weight of `PCl5`:
\[ M_{PCl5} = 30.97 \, (\text{P}) + 5 \times 35.45 \, (\text{Cl}) = 208.24 \, g/mol \]
- The average molecular weight of the mixture in Container B can be calculated as:
\[ M_{\text{mixture}} = \frac{(1 - \alpha) \cdot M_{PCl5} + \alpha \cdot M_{PCl3} + \alpha \cdot M_{Cl2}}{1 + \alpha} \]
- The molecular weight of `Cl2` is:
\[ M_{Cl2} = 2 \times 35.45 = 70.90 \, g/mol \]

4. **Applying Graham's Law**:
- Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:
\[ \frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} = \sqrt{\frac{M_{\text{mixture}}}{M_{PCl3}}} \]
- Since the mixture in Container B contains lighter gases (`PCl3` and `Cl2`), its average molecular weight will be less than that of `PCl3`.

5. **Conclusion**:
- Since \( M_{\text{mixture}} < M_{PCl3} \), it follows that:
\[ \frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} < 1 \]
- Therefore, the rate of effusion in Container B is greater than that in Container A.

Thus, **Container B will show a higher rate of effusion**.