1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Final entropy of gas mixture

  1. Nov 3, 2011 #1
    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)

    An infinitely large reservoir of H2O gas has the following properties:
    1.0 Bar

    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

    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!
  2. jcsd
  3. Nov 4, 2011 #2
    Is there anyone here who would like to share their opinion?
  4. Nov 4, 2011 #3
    I thought this was an easy one... Should I expect more response if it was posted in the Classical Physics section?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook