I am currently stuck with the following problem:
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| V1 | V2 | ====> | V1+V2 |
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In a box there are two different gases in two compartments seperated by a solid wall. The Temperature $T$ of both gases is the same and both gases consist of $1mol$ particles. Therefore on the left side of the box, you have a volume of $V_1$ with a pressure of $p_1$ and on the right hand side we have volume $V_2$ and pressure $p_2$. No we remove the separating wall.
Lets first assume that we have to different gases on the left and on the right side. My goal is to calculate the change of entropy which occurs due to the removal of the wall. My first Ansatz was:
$\Delta S_1 = nR\ln(\frac{V_1+V_2}{V_1})$
$\Delta S_2 = nR\ln(\frac{V_1+V_2}{V_2})$
Where $\Delta S$ is the change of entropy of the system, $n$ the number of mols and $R$ the universal gas constant. The total change of entropy would now be the sum of both:
$\Delta S_{tot} = S_1 + S_2 = nR\ln(\frac{(V_1+V_2)^2}{V_1V_2})$
After a while I thought about the problem in a different way: First I move the wall isothermally so that $V_1=V_2$. The entropy change due to the movement would be:
$\Delta S_{mov1} = nR\ln(\frac{V_1+V_2}{2V_1})$
$\Delta S_{mov2} = nR\ln(\frac{V_1+V_2}{2V_2})$
$\Delta S_{mov} = S_{mov1} + S_{mov2} = nR\ln(\frac{(V_1+V_2)^2}{4V_1V_2})$
Now i remove the wall and get a entropy change due to the mixing of the gasses:
$\Delta S_{rem} = R(n_1 ln(\frac{n_1+n_2}{n_1}) + n_2 \ln(\frac{n_1+n_2}{n_2})) = 2nR\ln(2)$
so
$\Delta S_{tot} = \Delta S_{rem} + \Delta S_{mov} = nR\ln(\frac{(V_1+V_2)^2}{4V_1V_2}) + 2nR\ln(2)$
which is clearly different from the result I got from my approach in the first Ansatz. However since entropy is a state variable this should not happen. No I am stuck. I tend to the second result, but I am very far from sure. It would be great if you could explain which ansatz is correct (if any^^).
Moreover if I had identical gases in $V_1$ and $V_2$, how would the entropy behave? Since the pressure changes after the removal of the wall, I would say that the entropy also changes. Just like in my second Ansatz, however without the mixing term. Is this correct?
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EDIT: Oh, dear. I just realized that $2nRln(2)$ can be written as $nRln(4)$. Therefore
$nR\ln(\frac{(V_1+V_2)^2}{4V_1V_2}) + 2nR\ln(2) = nR\ln(\frac{(V_1+V_2)^2}{V_1V_2})$
so both ansatzes give the same result. Since this is cleared out, my second question remains. In case of two identical gases does the entropy change according to my second ansatz (without the mixing entropy)?
Thanks in advance
ftiaronsem