Mixing of Ideal gas - Thermodynamic equilibrium

I always get confused what exactly happens when two ideal gases mix.

Consider the initial situation where two gases are in a box, separated by a rigid and adiabatic wall between them. Now when the wall between them is removed, they come to equilibrium (of course assuming the process is done quasi-statically). Initially the thermodynamic quantities of the gas be

$$U_i,\: S_i,\:T_i,\:P_i,V_i \:\:\: \:\:\: \:\:\: \:\:\: \:\:\: \:\:\: \:\:\:i=1,2$$ Now after the wall is removed, the condition for equilibrium is attained by the condition $$dS = 0$$ I am not able to apply this and find out exactly the final values of the thermodynamic quantities for an arbitrary ideal gas situation (I am not particularly able convince the situation physicaly)

(for instance if we consider and the two ideal gases to obey equation,

$$P_iV_i = c_iN_iRT_i \:\:\: \:\:\: \:\:\:i=1,2$$ where $c_i$ is the degrees of freedom of the gas, like $c_i = \frac{3}{2},\:\frac{5}{2}$ for monatomic and diatomic respectively.

Answer

The total initial internal energy is $U=U_1 + U_2=\frac{\nu_1}{2}c_1RT_1 + \frac{\nu_2}{2}c_2 RT_2$ where the last equality comes from Joules' first law for ideal gases and where $c_i$ is the number of moles of species $i$ and $\nu_i$ is the number of degrees of freedom of the molecule (3 for atoms, 5 for diatomic molecules etc..).

Now, once equilibrium is reached everybody should have the same temperature $T$. Since you are dealing with an ideal gas it implies that:

$$U= \frac{(\nu_1c_1+\nu_2c_2)RT}{2}$$ and hence since the whole system is isolated

$$T = \frac{\nu_1c_1T_1 + \nu_2c_2 T_2}{\nu_1c_1+\nu_2c_2}$$

Once the temperature is known, the rest follows easily. The pressure can be gotten straightforwardly as

$$P=\frac{(c_1+c_2)RT}{V_1+V_2}$$

because the ideal gas law is independent of the number of degrees of freedom of the different species.