Mass Balance: $$n_1+n_2=n_0$$where ##n_0## is the number of moles of gas initially in the left chamber, ##n_1## is the final number of moles of gas in the left chamber, and ##n_2## is the final number of moles in the right chamber. From the ideal gas law, this gives: $$\frac{PV}{RT_1}+\frac{PV}{RT_2}=\frac{P_0V}{RT_0}$$where P is the final pressure in both chambers, V is the volume of each of the two chambers, ##P_0## is the initial pressure in the left chamber, ##T_1## is the final temperature in the left chamber, ##T_2## is the final temperature in the right chamber, and ##T_0## is the initial temperature of the gas in the left chamber. So, $$\frac{T_0}{T_1}+\frac{T_0}{T_2}=\frac{P_0}{P}\tag{1}$$Energy Balance: $$\Delta U=n_1C_V(T-T_0)+n_2C_v(T_2-T_0)=0$$, or, combining with the ideal gas law:$$1-\frac{T_0}{T_1}+1-\frac{T_0}{T_2}=0$$or$$\frac{T_0}{T_1}+\frac{T_0}{T_2}=2\tag{2}$$So, combining Eqns. 1 and 2, we have $$P=\frac{P_0}{2}\tag{3}$$Thus, the final pressure in both chambers is half the initial pressure in the left chamber.
The gas finally remaining in the left chamber has experienced an adiabatic reversible expansion in gradually pushing the gas ahead of it into the stopcock, and into the right chamber. Thus, $$\frac{T_1}{T_0}=\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{1}{2}\right)^{\frac{\gamma-1}{\gamma}}\tag{4}$$Next, combining Eqns. 2 and 4 yields:$$\frac{T_2}{T_0}=\frac{1}{2-2^{\frac{\gamma-1}{\gamma}}}>1$$