# 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.

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Are the initial pressures of the gases the same? If they are you just have the entropy of mixing to calculate. If not you have to allow the higher pressure gas to do work on the lower pressure gas. –  John Rennie Jan 31 at 8:27
In all generality they are not the same –  user38249 Jan 31 at 8:32

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.

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Thanks, but my point was if one was monoatomic and the other diatomic, then it will not be $\frac{3}{2}$ in both cases. –  user38249 Jan 31 at 9:53
Also what about the final pressure, and of course the volume will be the final volume. –  user38249 Jan 31 at 9:55
I have edited my answer to account for the general case and put the pressure as well. –  gatsu Jan 31 at 12:35
Thanks for the last point, that was something that I was looking for. –  user38249 Jan 31 at 16:37