The volume to be excluded due to only two molecules of a gas in collision with a fixed point of impact is `0.09 l mol^(-1)`. If the value of 'a' is 3.6 atm `L^(2) mol^(-2)` , Calculate Boyle temperature. (R = 0.08 L atm `K^(-1) m^(-1)` )

Answer

To calculate the Boyle temperature (TB) for the given gas, we can use the formula:

\[ TB = \frac{A}{R \cdot B} \]

Where:
- \( A \) is the Van der Waals constant for attractive forces,
- \( R \) is the gas constant,
- \( B \) is the volume excluded due to the gas molecules.

### Step-by-step Solution:

1. **Identify the given values:**
- \( A = 3.6 \, \text{atm} \, \text{L}^2 \, \text{mol}^{-2} \)
- \( R = 0.08 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1} \)
- \( B = 0.09 \, \text{L} \, \text{mol}^{-1} \)

2. **Substitute the values into the formula:**
\[ TB = \frac{3.6}{0.08 \times 0.09} \]

3. **Calculate the denominator:**
\[ 0.08 \times 0.09 = 0.0072 \]

4. **Now substitute this value back into the equation:**
\[ TB = \frac{3.6}{0.0072} \]

5. **Perform the division:**
\[ TB = 500 \, \text{K} \]

### Final Answer:
The Boyle temperature \( TB \) is \( 500 \, \text{K} \).

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