A

Step 1: Understand the initial condition of the gas. The initial state is given by:
\( P_1 V_1^2 = k \) where \( k \) is a constant.
Given that initially the gas is at volume \( V_1 = V \), we can write the equation as:
\( P_1 V^2 = k \).

Step 2: When the gas expands to a volume \( V_2 = 2V \), we will denote the new pressure as \( P_2 \). Therefore, we can write:
\( P_2 (2V)^2 = k \). This simplifies to:
\( P_2 (4V^2) = k \) or \( P_2 = \frac{k}{4V^2} \).

Step 3: For an ideal gas, we know that \( PV = nRT \), where \( n \) is the number of moles and \( R \) is the ideal gas constant. The relationship for temperature will help us relate the initial and final states. From this we have:
\( P_1 V = nRT \ ext{ (initial state)} \) and \( P_2 (2V) = nRT_2 \ ext{ (final state)} \).

Step 4: Since we have the expressions for \( P_1 \) and \( P_2 \):
\( P_1 = \frac{nRT}{V} \) and \( P_2 = \frac{nRT_2}{2V} \)
Equating these gives us:
\( \frac{nRT}{V} V^2 = \frac{nRT_2}{2V} (2V)^2\) which simplifies to:
\( R T V = 4 R T_2 \)

Step 5: Therefore, we can simplify and solve for \( T_2 \):
\( T_2 = \frac{T}{4} \)

Conclusion: The final temperature when the gas expands to a volume of \( 2V \) is given by:\
\( T_2 = \frac{T}{4} \). Hence, the correct answer is that the temperature is \( \frac{T}{4} \) when the gas expands to a volume of \( 2V \). Therefore, assuming the options provided included the expression for \( T_2 \):
Thus, the correct answer is A.