What happens to entropy when mixing two gases with different properties?

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SUMMARY

The discussion centers on the mixing of two H2O gas samples with differing properties in a closed chamber, leading to an equilibrium state at 1 Bar. The initial conditions are 0.5 Bar, 200°C, and 2850 kJ/kg for the small chamber, and 1.0 Bar, 240°C, and 2950 kJ/kg for the reservoir. The resulting properties after mixing are predicted to be 1.0 Bar, 220°C, and 2900 kJ/kg. The participant questions the implications for entropy, suggesting that the process appears to lower entropy, which raises concerns regarding compliance with the second law of thermodynamics.

PREREQUISITES
  • Understanding of thermodynamics principles, particularly the second law of thermodynamics.
  • Familiarity with properties of gases, including pressure, temperature, and enthalpy.
  • Knowledge of adiabatic processes and their implications on energy transfer.
  • Ability to interpret steam tables and thermodynamic diagrams.
NEXT STEPS
  • Research the implications of the second law of thermodynamics on spontaneous processes.
  • Study the concept of entropy in thermodynamic systems and its behavior during mixing.
  • Examine adiabatic processes in greater detail, focusing on energy conservation and state changes.
  • Explore steam tables and their application in calculating properties of gases during phase changes.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, particularly those studying gas behavior, engineers working with thermal systems, and anyone interested in the principles governing energy transfer and entropy in physical processes.

jornrune
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Hi, I am new here, so please direct me if I am posting completely wrong here.

Imagine a small enclosed chamber filled with dry H2O gas. The properties are:
0.5 Bar
8.0 kJ/(kg*K)
200C
2850kJ/kg

An infinitely large reservoir of H2O gas has the following properties:
1.0 Bar
8.0kJ/(kg*K)
240C
2950kJ/kg

Here is a reference chart:
http://www.steamtablesonline.com/images/steam tables p-h diagram (large).png

A valve between the large reservoir and the small chamber is opened and the two gasses are mixed until they reach equilibrium inside the chamber and the pressure is stabilized at 1 bar. Am I correct in assuming that no work has been done and that the properties will be as follows?:
1.0 Bar
7.9kJ/(kg*K)
220C
2900kJ/kg

This seems to be correct as far as temperature and pressure goes. It also seems to add up with the enthalpy as the process is thought to be adiabatic. However, the entropy is lowered, and the process seems to be a so called spontaneous process. The question is if this goes along with the second law of thermodynamics. (It can be read here if you need it refreshed: http://en.wikipedia.org/wiki/Second_law_of_thermodynamics)

Also, I see no room for different results. The pressure is given by the infinite source. This means we can not leave the 1 bar line. There is no way we can get any condensation, so we have the following options:
1: The entropy is actually reduced
2: Enthalpy is lost or gained, and so is temperature

...or am I overlooking something here?

Thanks for any response!
 
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Is there anyone here who would like to share their opinion?
 
I thought this was an easy one... Should I expect more response if it was posted in the Classical Physics section?
 

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