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

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Mixing two gases with different properties in a closed chamber leads to a new equilibrium state with a stabilized pressure of 1 bar. The user calculates the resulting properties as 1.0 Bar, 7.9 kJ/(kg*K), 220C, and 2900 kJ/kg, assuming an adiabatic process with no work done. Despite the temperature and pressure aligning with expectations, the entropy appears to decrease, raising questions about compliance with the second law of thermodynamics. The discussion highlights the complexities of entropy changes in spontaneous processes and the implications of mixing gases from different sources. Further insights or confirmations from experts in the field are sought to clarify these thermodynamic principles.
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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