For an ideal gas the internal energy depends on temperature only. We can consider the process in question to be one of simultaneous free expansion. Then the total energy $U=U_{1}+U_{2}$. Since
$U_{1}=C_{V} T_{1}, U_{2}=C_{V} T_{2}, U=2 C_{V} \frac{T_{1}+T_{2}}{2}$ and $\left(T_{1}+T_{2}\right) / 2$ is the final temperature. The
entropy change is obtained by considering isochoric processes because in effect, the gas remains confined to its vessel.
$$
\Delta S=\int_{T_{1}}^{\left(T_{1}+T_{2} / 2\right.} \frac{C_{V} d T}{T}-\int_{\left(T_{1}+T_{2}\right) / 2}^{T_{2}} C_{V} \frac{d T}{T}=C_{V} \ln \frac{\left(T_{1}+T_{2}\right)^{2}}{4 T_{1} T_{2}}
$$
Since $\left(T_{1}+T_{2}\right)^{2}=\left(T_{1}-T_{2}\right)^{2}+4 T_{1} T_{2}, \Delta S>0$