On heating vapours of `S_(8)(g)` decomposes to `S_(2)(g)`. Due to this, the van-der Waal's constant 'b' for the resulting gas

Options

A. increases
B. decreases
C. remains same
D. change unpredictably

Answer

To solve the question regarding the change in the van der Waals constant 'b' when the vapors of \( S_8(g) \) decompose into \( S_2(g) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Van der Waals Constant 'b':**
The van der Waals constant 'b' represents the volume occupied by one mole of gas particles. It accounts for the finite size of gas molecules and is related to the volume available to the gas.

2. **Analyzing the Reaction:**
The reaction given is:
\[ S_8(g) \rightarrow 4 S_2(g) \]
This means that one mole of \( S_8 \) decomposes into four moles of \( S_2 \).

3. **Comparing the Molar Volumes:**
- The volume occupied by one mole of \( S_8 \) is greater than the total volume occupied by the four moles of \( S_2 \) because \( S_8 \) is a larger molecule.
- When \( S_8 \) decomposes into \( S_2 \), the total number of moles increases from 1 mole to 4 moles.

4. **Effect on 'b':**
Since 'b' is related to the volume occupied by the gas molecules, and the decomposition leads to a decrease in the effective volume occupied by the gas (as the volume of \( S_8 \) is greater than the volume of \( S_2 \)), the value of 'b' will decrease.

5. **Conclusion:**
Therefore, the van der Waals constant 'b' for the resulting gas after the decomposition of \( S_8 \) to \( S_2 \) will decrease.

### Final Answer:
The value of 'b' will **decrease**.

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