Entropy of mixing, non-isolated system

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Homework Statement


Take a container of volume V, which has two seperated partitions, each with volume V/2. Each partition contains a gas of N molecules, with respectively mass [tex]\m_{a}[/tex] and [tex]\(m_{b}\)[/tex]. The temperature in the partitions is equal to that of its surroundings, and energy exchange between the molecules in the partitions with the outside is possible.

When the partition between gases A and B are removed, what happens to the quantities U, P and S?


Homework Equations


[tex]\Delta S = -R (x ln x + (1-x) ln (1-x) )[/tex]


The Attempt at a Solution



I'm guessing this should be done with the Gibbs free energy distribution. The problem seems trivial in the case of an isolated system, the only increase in entropy is because of the increase in potential volume for each molecule. What happens however, when there exchange of energy with the environment is allowed?
 

Answers and Replies

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Andrew Mason
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What happens however, when there exchange of energy with the environment is allowed?
There is no net exchange of energy with the surroundings since the gas and the surroundings are at the same temperature. So there is no change in entropy of the surroundings.

AM
 

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