Entropy of mixing, non-isolated system

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SUMMARY

The discussion centers on the entropy changes in a non-isolated system when two gas partitions are removed, allowing for mixing. The key equation used is the entropy change formula, ΔS = -R (x ln x + (1-x) ln (1-x)). The analysis concludes that while the entropy of the system increases due to the mixing of gases A and B, the entropy of the surroundings remains unchanged since there is no net energy exchange with the environment. Thus, the total entropy of the system increases, reflecting the second law of thermodynamics.

PREREQUISITES
  • Understanding of thermodynamic principles, particularly entropy.
  • Familiarity with the Gibbs free energy concept.
  • Knowledge of statistical mechanics and the role of molecular interactions.
  • Basic proficiency in calculus for manipulating logarithmic functions.
NEXT STEPS
  • Study the implications of the second law of thermodynamics in non-isolated systems.
  • Explore Gibbs free energy calculations in various thermodynamic processes.
  • Investigate the role of temperature in entropy changes during gas mixing.
  • Learn about statistical mechanics and its application to entropy and molecular behavior.
USEFUL FOR

This discussion is beneficial for students of thermodynamics, physicists, and chemists interested in the behavior of gases and entropy in non-isolated systems.

pieterb
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Homework Statement


Take a container of volume V, which has two separated 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?
 
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pieterb said:
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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