Coefficient of Volume Expansion of Expanding Ideal Gas

To find the coefficient of volume expansion of the gas, we use the definition:
$\gamma = \frac{1}{V}\frac{dV}{dT}$

For an ideal gas, the equation of state is:
$PV = nRT \implies P = \frac{nRT}{V}$

The given process is:
$PT^2 = \text{constant}$

Substitute $P = \frac{nRT}{V}$ into the process equation:
$\left(\frac{nRT}{V}\right)T^2 = \text{constant}$

$\frac{T^3}{V} = \text{constant}' \implies V = C T^3$
where $C$ is a constant.

Differentiating both sides with respect to temperature $T$:
$\frac{dV}{dT} = 3 C T^2$

Now, substitute $\frac{dV}{dT}$ into the formula for the coefficient of volume expansion:
$\gamma = \frac{1}{V} \frac{dV}{dT} = \frac{1}{C T^3} \cdot (3 C T^2) = \frac{3}{T}$

Thus, the coefficient of volume expansion of the gas is $\frac{3}{T}$.

Therefore, the correct option is C.