Can Entropy Be Reduced in a Spontaneous Thermodynamic Process?

[jornrune]

Hi! I have a practical problem that causes me to wonder. It seems like I have to choose from gaining energy from nothing, or to get a reduction in entropy from a spontaneous process.

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!

 

[Curl]

If you open up the valve and wait for equilibrium, the gas inside your small chamber will be identical to the one in the large reservoir.