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Need help understanding sceanario only, is it a closed reversible system?

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    @= alpha= thermal expansion coefficent= 5.1x10^-5
    Kt= compression= 3.2x10^-5
    B=beta= -(dE/dV)t = 0.5
    Po=1 atm
    To=293K
    Vo=1L
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Solid held in a piston and cylinder at equilibrium is in a constant temperature bath at 293k. Two irreversible steps are done at constant temp: I) pressure is suddenly increased to 100 atm and this pressure is kept constant while solid is irreversibly compressed. II)The pressure is suddenly dropped to zero and the solid expands till V=Vo=1L at which point the pressure is quickly restored to 1 atm.


    2. Relevant equations

    V=Vo(1-Kt(P-Po)+@(T-To) solid equation of state

    3. The attempt at a solution

    I attempted to draw the graph came up w/ 5 irreversible steps. increasing pressure to 100 atm then going at constant pressure to a decreased volume then dropping the tempreture to zero then going to the initial volume then finally raising the pressure back to the initial pressure.

    By doing all of this the system appears to be closed and reversible correct? So the delta S universe should equal zero but it isn't....I believe that is it reversible because of the fact that equilibrium is achieved by going up and down the isotherm but specifically in the question it has been done irreversibly. But according to the notes I've taken in class the delta S of the universe is not equal to zero and even though there is a tempreture bath, there is heat loss that escapes the bath...why is that so? The tempreture stays constant through out the whole process...nothing in this question makes sense T_T I'd really appreciate any kind of help or ideas. Thank you.
     
  2. jcsd
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