Adiabatic Mixing of Gases

I see answers to similar questions like this and this that confuse me. I feel like they ignore how the expansion of the gases occurs.

Consider two chambers of gases at $ n_{1i}, P_{1i} $ and $ n_{2i}, P_{2i} $ each with constant initial temperature $ T_i $. Put them in adiabatic contact such that the total volume $ V = V_1 + V_2 $ is constant, no heat exchange $ dT_i = 0 $, and the chambers are allowed to expand against each other based on their pressures. Final pressure will be equilibrium $ P_f $

As they expand, they should be subject to Adiabatic relation for P-T,

$$
P_{1i}^{1-gamma} T_i^gamma = P_f^{1-gamma} T_{1f}^gamma
P_{2i}^{1-gamma} T_i^gamma = P_f^{1-gamma} T_{2f}^gamma
$$

Dividing those, you can get a relation,
$$
T_{1f} = left(frac{P_{1i}}{P_{2i}}right)^{(1 – gamma)/gamma} T_{2f}
$$

Only now do gas mixing according to the formulas in the linked posts,

$$begin{align}
T_{mixed} &= frac{n_1 left(frac{P_{1i}}{P_{2i}}right)^{(1 – gamma)/gamma} + n_2}{n_1 + n_2} T_{2f}
&= frac{n_1 left(frac{P_{1i}}{P_f}right)^{(1 – gamma)/gamma} + n_2 left(frac{P_{2i}}{P_f}right)^{(1 – gamma)/gamma}}{n_1 + n_2} T_{i}
end{align}$$

$ gamma = 7/5 $ for diatomic, so $ (1 – gamma)/gamma < 0 $ implying that we would expect cooling $ T_{mixed} < T_i $ if combining a low pressure gas with a high pressure gas ($ P_{2i} > P_{1i} $) in equal volumes (implying $ n_2 > n_1 $) which intuitively makes sense.

Am I wrong? If I am correct, then this would imply that this is either incorrect or incomplete because they predict no temperature change.