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Equalizing pressure between two vessels

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    There are two insulated air containers, first with a volume V1 and second with V2. In the first there is a pressure p1 and in the second p2. The temperatures are the same in both, thus T. Then both containers are connected with a hose so that pressure equalizes, but no heat is transferred between them.
    What are the final temperatures and final pressure?

    2. Relevant equations

    3. The attempt at a solution
    If there was heat transfer between containers, the final temperature would be the same in both and even the same as the temperature T in the beginning and because of that I could calculate the final pressure using pV = mRT (R = gas constant for air), where I'd get mass for one container from specific volumes.

    But this case with no heat transfer is a bit confusing. I can't get final pressure with the equation used above because I don't know the final temperatures. The book says the final pressure is the same as above, but I don't know why is it so.
  2. jcsd
  3. Nov 22, 2009 #2
    My results regarding the case where heat transfer occurs are that entropy actually decreases - using this formula:

    s = Cvln(pend/p1) + Cpln(vend/v1) +
    + Cvln(pend/p2) + Cpln(vend/v2),

    where v1 and v2 are specific volumes before mixing and vend is the final specific volume (the same for pressures).

    The actual numbers are:
    T = 293 K
    p1 = 1 bar
    p2 = 20 bar
    V1 = 5m3
    V2 = 2m3

    From that I got
    v1 = 0.8403 m3/kg)
    v2 = 0.0420 m3/kg)
    pend = 6.43 bar
    vend = 0.1308 m3/kg

    s = -206 J/kgK

    I know entropy cannot be decreased with mixing, so what's wrong here?
    Last edited: Nov 22, 2009
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