The Boyle temperature of a van der Waal gas is `-246^(@)C`. Its critical temperature on absolute temperature scale is :

Options

A. `5K`
B. `7K`
C. `8K`
D. `10K`

Explanation

To solve the problem, we need to find the critical temperature (Tc) of a van der Waals gas given its Boyle temperature (TB) of -246°C. Let's break down the steps:

### Step 1: Convert Boyle Temperature to Kelvin
The Boyle temperature (TB) is given in degrees Celsius. To convert it to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given value:
\[ TB = -246 + 273 = 27 \, K \]

### Step 2: Use the Relationship Between TB and Tc
For a van der Waals gas, the relationship between Boyle temperature (TB) and critical temperature (Tc) is given by:
\[ TB = \frac{A}{Rb} \]
and
\[ Tc = \frac{8A}{27Rb} \]
where \( A \) is the van der Waals constant for pressure correction, and \( b \) is the van der Waals constant for volume correction.

### Step 3: Derive the Relationship Between TB and Tc
From the equations above, we can express the ratio of TB to Tc:
\[ \frac{TB}{Tc} = \frac{A/Rb}{8A/(27Rb)} = \frac{27}{8} \]
This implies:
\[ Tc = \frac{8}{27} \cdot TB \]

### Step 4: Substitute the Value of TB
Now, we can substitute the value of TB we calculated:
\[ Tc = \frac{8}{27} \cdot 27 \, K \]
\[ Tc = 8 \, K \]

### Step 5: Conclusion
The critical temperature (Tc) of the van der Waals gas is:
\[ Tc = 8 \, K \]

### Final Answer
The critical temperature on the absolute temperature scale is **8 K**.

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