Reversible & irreversible adiabatic processes Work done in reversible is more than that of irreversible

Suppose we are going from state $(P_1, V_1, T_1)$ to some other state say $(P_2, V_2, T_2)$ for reversible and $(P_2, V_2, T_2')$ for irreversible. Which should be greater, $T_2$ or $T_2'$?

My attempt

Let us say heat supplied is same i.e q for both reversible and irreversible. We already know that $U = q - W$

Reversible as work done is more so $U$ becomes less and in irreversible $W$ is less so $U'$ is more than $U$ i.e $U' &gt; U$

As $U'$ is greater $T_2'$ should be greater. Is this correct explanation?
For a given number of moles, once $P_1$ and $V_1$ are specified, that determines state 1 and $T_1$.  Once $P_2$ and $V_2$ are specified, that determines state 2 and $T_2$.  It doesn't matter what process is used to get from state 1 to state 2.  Only 2 intensive variables are required to specify the state of the system (assuming no phase change).  So $T_2=T_2"$.

Maybe you want to remove either the $P_2$ or the $V_2$ from the problem specification.