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Prove non adiabatic reversible expansion and ΔS

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    One mole of a monatomic ideal gas is transformed from 0.00°C and 2.00 atm to
    -40.0°C and 0.400 atm. Show that this is NOT an adiabatic reversible expansion and
    calculate ΔS for the change. (Hint: you will need to consider a reversible path that
    includes a constant pressure step and a constant temperature step.)


    2. Relevant equations
    ΔU=q+w
    pV=nRT
    ΔS=nC(p,m)ln(Vf/Vi)

    3. The attempt at a solution
    For adiabatic reversible expansion, q=0, so i have to show that q= a number.
    Found Vi= 11.2L and Vf=47.8L via ideal gas law

    ΔS=(1mol)(5/2(8.314J/Kmol)ln(47.2/11.2)
    =30.15J/K
     
    Last edited: Nov 7, 2011
  2. jcsd
  3. Nov 8, 2011 #2

    I like Serena

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    Homework Helper

    Hi Nick_273! :smile:

    I'm afraid it does not suffice to calculate ΔS for the change in volume.
    You also need to account for the change in either pressure or temperature.

    There's a few more relevant equations that you could derive, but you can also find them here:
    http://en.wikipedia.org/wiki/Table_of_thermodynamic_equations

    In particular you can find the equations for ΔS here when p is constant, but also when T is constant.
    You need these 2 to follow the hint.
     
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