An ideal gas is found to obey and additional law `VP^2` = constant. The gas is initially at temperature T and volume V . When it expands to a volume 2 V , the temperature becomes

Options

A. `sqrt2T`
B. 2T
C. `2sqrt2T`
D. 4T

Answer

To solve the problem, we will use the relationship given for the ideal gas and the additional law \( VP^2 = \text{constant} \).

### Step-by-step Solution:

1. **Understanding the Given Law**:
We know that \( VP^2 = k \) (where \( k \) is a constant). This means that for any state of the gas, the product of volume \( V \) and the square of pressure \( P \) remains constant.

2. **Initial Conditions**:
Let the initial volume be \( V_1 = V \), the initial temperature be \( T_1 = T \), and the initial pressure be \( P_1 \).

3. **Final Conditions**:
When the gas expands to a volume \( V_2 = 2V \), we need to find the final temperature \( T_2 \) and the final pressure \( P_2 \).

4. **Using Ideal Gas Law**:
The ideal gas law states that \( PV = nRT \). Rearranging gives us \( P = \frac{nRT}{V} \).

5. **Substituting Pressure in the Given Law**:
Substitute \( P \) into the equation \( VP^2 = k \):
\[ V \left(\frac{nRT}{V}\right)^2 = k \]
Simplifying this gives:
\[ V \cdot \frac{n^2R^2T^2}{V^2} = k \implies \frac{n^2R^2T^2}{V} = k \]
This indicates that \( \frac{T^2}{V} \) is constant.

6. **Setting up the Ratio**:
From the above, we can say:
\[ \frac{T_1^2}{V_1} = \frac{T_2^2}{V_2} \]

7. **Substituting Known Values**:
Substitute \( T_1 = T \), \( V_1 = V \), \( V_2 = 2V \):
\[ \frac{T^2}{V} = \frac{T_2^2}{2V} \]

8. **Cross-multiplying**:
Cross-multiplying gives:
\[ 2T^2 = T_2^2 \]

9. **Solving for Final Temperature**:
Taking the square root of both sides:
\[ T_2 = \sqrt{2}T \]

### Final Answer:
The final temperature when the gas expands to a volume of \( 2V \) is:
\[ T_2 = \sqrt{2}T \]