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Thermodynamics - reversible isothermal cycle

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Untitled.png

    2. Relevant equations



    3. The attempt at a solution

    I can do everything else except work for part one going from A to B.

    What I did was

    -∫PdV = -nRT ∫(1/v)dV = -nRT ln (Vf/Vi)

    I can solve for T because everything is given, T in Kelvin is 882.
    From PV=nRT
    T=(5atm)(10L)/1mol*R = 882K

    where R = .0820582

    Therefore w= -1mol * R * 882 * ln (5) = - 11.8kJ

    where R = 8.314

    But this is incorrect. It is supposed to be -8.15kJ
     
  2. jcsd
  3. Mar 24, 2014 #2

    ehild

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    Redo the calculation in red.

    ehild
     
  4. Mar 24, 2014 #3
    Hmm...

    50 / (.0820582) = 609 K

    Argh nevermind I thought it gave me the T in Celcius! Thank you Ehild!
     
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