For an ideal gas, the ideal gas equation is given by $PV = nRT$, which implies $P = \frac{nRT}{V}$. Substituting this into the given process equation $PV^2 = C$: $\left(\frac{nRT}{V}\right) V^2 = C \implies nRTV = C \implies TV = \text{constant}$. This relation shows that temperature $T$ is inversely proportional to volume $V$ ($T \propto \frac{1}{V}$). Therefore, if $V_2 > V_1$, it must be that $T_2 < T_1$. Furthermore, since $P \propto \frac{1}{V^2}$, if $P_1 > P_2$, then $V_1 < V_2$, which in turn means $T_1 > T_2$. Thus, the only correct statement is 'If $V_2 > V_1$, then $T_2 < T_1$'.