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Work done in an adiabatic process

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data
    An adiabatic process is performed on 15 moles of an ideal gas. The initial temperature is 320 K and the initial volume is
    0.80 m3. The final volume is 0.40 m3. The adiabatic constant for the gas is 1.44. How much work is done by the gas?

    The answer is -32kJ

    2. Relevant equations
    W = Integrate[p dv, vInitial, vFinal] for adiabatic processes
    pV = nRT

    3. The attempt at a solution
    If p(v^y) initial == p(v^y) final where y = 1.44, and v varies, then p also varies, and I don't know how to use a varying pressure with the formula for work for adiabatic processes.

    I try to find the initial pressure by
    pV = nRT
    p*V == 15*8.31*320
    With Vinitial, p = 49860Pa
    With Vfinal, p = 99720Pa

    And I'm stuck.
     
  2. jcsd
  3. Nov 28, 2009 #2
    Here is my solution: for adiabatic process, you can write: PV^(gamma) = C => P = C / V^(gamma). Plug it into W = int(PdV). After that, replace C by P and V.
     
  4. Nov 29, 2009 #3
    p*((0.8)^1.44) = c
    Which gives
    p = 1.37896c

    Integrating 1.37896 c dV from 0.8 to 0.4 gives
    -0.551582 c

    If I use the original p (that I found) and v (given) for c,
    -0.551582(p*v) = -0.551582(49860*0.8) = -22kJ, which isn't the answer. Where did I go wrong?
     
  5. Nov 29, 2009 #4
    No.

    We have [tex]PV^ \gamma = C ==> P = C/V^ \gamma[/tex]

    [tex]W = \int PdV = \int CV^- ^\gamma dV = \frac{C V^ 1 ^- ^\gamma}{1 - \gamma}[/tex]
     
  6. Nov 29, 2009 #5
    Oh, I understand, thanks for your help!

    The solution for future reference:
    p*0.8 = 15*8.31*320
    p -> 49860 Pa (Initial pressure)

    p*(0.8^1.44) = c (0.8 is the initial volume)
    c -> 36157.8

    w = Integrate[c (v^-1.44), dv]
    w = -(2.27273 c)/v^0.44
    -> wf - wi = -32327.8 J
     
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