= Solution
For one mole, $pV=RT$, and constant <heat capacities> satisfy $C_p-C_V=R$. The specified diatomic-gas model has
$$
\gamma=1.4=\frac75,\qquad C_V=\frac{R}{\gamma-1}=\frac52R,
\qquad C_p=\frac72R.
$$
A reversible <adiabatic process> satisfies $TV^{\gamma-1}=\mathrm{constant}$. Constant-pressure expansion satisfies $T\propto V$. Applying these in order gives
$$
\boxed{\begin{aligned}
T_2&=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1},\\
T_3&=T_2\frac{V_3}{V_2}
=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}\frac{V_3}{V_2},\\
T_4&=T_3\left(\frac{V_3}{V_1}\right)^{\gamma-1}
=T_1\left(\frac{V_3}{V_2}\right)^\gamma,\\
T_{\rm final}&=T_1.
\end{aligned}}
$$
Every <volume (thermodynamics)> ratio is dimensionless, so the expressions have units of <temperature>. For the intended <volume (thermodynamics)> ordering, the minimum <temperature> is $T_1$ and the maximum is $T_3$.
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