For a van der Waal's gas, determine Boyle Temperature (given a = 4.5 atm`L^2 mol^(-2) , b = 0.9 L mol^(-1) and R = 0.082 L atm K^(-1) mol^(-1)`]

Options

A. `609.8 K`
B. `6.09 K`
C. `273 K`
D. `60.98 K`

Answer

To determine the Boyle Temperature for a van der Waals gas, we can use the formula:

\[ T_B = \frac{a}{Rb} \]

where:
- \( a \) is the van der Waals constant related to attractive forces,
- \( b \) is the van der Waals constant related to the volume occupied by the gas molecules,
- \( R \) is the universal gas constant.

### Step-by-Step Solution:

1. **Identify the given values:**
- \( a = 4.5 \, \text{atm} \cdot \text{L}^2 \cdot \text{mol}^{-2} \)
- \( b = 0.9 \, \text{L} \cdot \text{mol}^{-1} \)
- \( R = 0.082 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)

2. **Substitute the values into the formula:**
\[ T_B = \frac{4.5}{0.082 \times 0.9} \]

3. **Calculate the denominator:**
- First, calculate \( R \times b \):
\[ R \times b = 0.082 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \times 0.9 \, \text{L} \cdot \text{mol}^{-1} = 0.0738 \, \text{L}^2 \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-2} \]

4. **Now, divide \( a \) by \( R \times b \):**
\[ T_B = \frac{4.5}{0.0738} \approx 60.98 \, \text{K} \]

5. **Conclude the result:**
The Boyle Temperature \( T_B \) for the van der Waals gas is approximately \( 60.98 \, \text{K} \).

### Final Answer:
\[ T_B \approx 60.98 \, \text{K} \]