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Mixing two gas

  1. Jan 3, 2009 #1


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    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.

    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 Tf. 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 Tf. 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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  3. Jan 5, 2009 #2


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    Right. For example, if the two parts of the container separate gasses that are identical and in identical states, then when you remove the partition, there is no change in entropy, despite the fact that your equations might say otherwise. This is known as the Gibbs paradox.
  4. Jan 5, 2009 #3


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    You're right, KFC, the problem must specify whether the two compartments contain the same gas or different gases. In the first case the entropy is constant; in the second the entropy increases. The reason is that "mixing" is a meaningless term when discussing identical gases in identical states. It's impossible to identify or quantify mixing in this scenario because the particles are taken to be indistinguishable.
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