The van der Waals' constantes for a gas are `a=3.6 atmL^(2)mol^(-2),b=0.6Lmol^(-1)` .If `R=0.08LatmK^(-1)mol^(-1)` and the Boyle's temperature (K) is `T_(B)` of this gas, then what is the value of `(T_(B))/(15)`?

Explanation

To solve the problem, we need to find the Boyle's temperature \( T_B \) of the gas using the given van der Waals constants \( a \) and \( b \), and then calculate the value of \( \frac{T_B}{15} \).

### Step-by-Step Solution

1. **Identify the formula for Boyle's temperature**:
The Boyle's temperature \( T_B \) can be calculated using the formula:
\[ T_B = \frac{a}{Rb} \]
where \( a \) is the van der Waals constant for attraction, \( R \) is the ideal gas constant, and \( b \) is the van der Waals constant for volume.

2. **Substitute the given values**:
We are given:
- \( a = 3.6 \, \text{atm L}^2 \text{mol}^{-2} \)
- \( b = 0.6 \, \text{L mol}^{-1} \)
- \( R = 0.08 \, \text{L atm K}^{-1} \text{mol}^{-1} \)

Now substituting these values into the formula:
\[ T_B = \frac{3.6 \, \text{atm L}^2 \text{mol}^{-2}}{(0.08 \, \text{L atm K}^{-1} \text{mol}^{-1})(0.6 \, \text{L mol}^{-1})} \]

3. **Calculate the denominator**:
First, calculate \( Rb \):
\[ Rb = 0.08 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 0.6 \, \text{L mol}^{-1} = 0.048 \, \text{atm K}^{-1} \]

4. **Calculate \( T_B \)**:
Now substitute \( Rb \) back into the equation for \( T_B \):
\[ T_B = \frac{3.6 \, \text{atm L}^2 \text{mol}^{-2}}{0.048 \, \text{atm K}^{-1}} \]
The units of atm cancel out:
\[ T_B = \frac{3.6}{0.048} \, K \]

5. **Perform the division**:
\[ T_B = 75 \, K \]

6. **Calculate \( \frac{T_B}{15} \)**:
Finally, we need to find \( \frac{T_B}{15} \):
\[ \frac{T_B}{15} = \frac{75 \, K}{15} = 5 \, K \]

### Final Answer
Thus, the value of \( \frac{T_B}{15} \) is \( 5 \, K \).