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Isothermal compression of ideal gas

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    Attempted Answer
    I got this formula out of our textbook
    [tex]W_{b}[/tex]=([tex]P_{1}[/tex][tex]V_{1}[/tex])[tex]\textbf{ln}[/tex][tex]V_{2}/V_{1}[/tex]
    Because Isothermal:

    [tex]P_{1}[/tex][tex]V_{1}[/tex]=[tex]P_{2}[/tex][tex]V_{2}[/tex]

    Therefore the first equation can be written as:
    [tex]W_{b}[/tex]=([tex]P_{1}[/tex][tex]V_{1}[/tex])[tex]\textbf{ln}[/tex][tex]P_{1}/P_{2}[/tex]

    From this equation I worked the boundary work to be: 1364.4kJ. Using the following values: m=3kg, R = 0.4882, P(1) = 1400kpa, T(1) = 353K, V(1) = 0.369288 (meters cubed).

    Now I have boundary work to be 1364.4kJ and Q(in) = 495 (kJ/kg) x 3kg = 1485kJ

    To find the change in enthalpy dont I just add these values? I am not sure what I have done wrong.

    The answers are:
    (b)144.9kJ
    (c)106kJ
     
  2. jcsd
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