There are two compartments, each has half volume of the total volume, separated by an insulating partition. The whole setup is adiabatic. n mole of a monatomic gas with temperature T1 and pressure P in the left while in the right m mole of monatomic gas with termparture T2 and pressure P there. Now remove the partition so two gas mix, I am trying to find out the change entropy of the whole system.(adsbygoogle = window.adsbygoogle || []).push({});

Well, I assume the total change of the entropy is the sum of change of entropy of individual gas in free expand. So I calculate the change of entropy of each gas separately. It is easy to find out the final temperature of mixture, let's call it T^{f}. Obviously, we need to makeup an isotermal process to let the gas expand from V to 2V; and then makeup a isochoric process to increase/decrease the temperature to T^{f}. With this two processes, the change of entropy for each gas should be (assume T1<T2)

[tex]\Delta S_{left} = n R\ln 2 + nC_v\ln \frac{T_f}{T_1} [/tex]

[tex]\Delta S_{right} = m R\ln 2 + mC_v\ln \frac{T_2}{T_f} [/tex]

The total change of entropy is [tex]\Detal S= \Delta S_{left} + \Delta S_{right}[/tex].

My question is: we didn't tell if these two gas are same or not, and this result didn't tell any different even if there are two different gas? It is so confused! I think if we are considering the mixture of two different gas, the change of entropy should be different to that of mixing two identical gas, isn't it?

I found something about mixing of entropy in wiki, and there it gives a formular that tells the total change of entropy after mixture

[tex]\Delta S_m = -nR(x_1\ln x_1 + x_2\ln x_2)[/tex]

where n is the total number of moles, [tex]x_1, x_2[/tex] are the mole fraction of each of the mixed components. Again, it doesn't tell the any different for different substances being mixed ?

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# Mixing two gas

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