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Reversible Adiabatic Process Question

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data

    An Ideal Gas at 300K has a volume of 15L and a pressure of 15 atm. Calculate the change when the system goes under a reversible adiabatic expansion to a pressure of 10 atm. Gamma = 5/3. Cv = 1.5R. q = 0(definition of adiabatic processes).


    2. Relevant equations

    dw = P DV (lowercase means squiggly d and uppercase means straight d)
    P = nRT/V



    3. The attempt at a solution

    The answer in the book says w = 5130 J, which is the opposite of what I got for the internal energy.

    First, I calculated n: n = PV/RT = 15 atm * 15 L / .08206 (L*atm / K * mole) * 300K

    n = 9.14 moles

    Then, I calculated Vf: Vf = [(Pf * Vi ^ (5/3)) / (Pf)] ^ 3/5

    Vf = 19.13 L

    dW = P DV
    dw = NRT


    (dw = P DV so then work = NRT * Integral [DV/V]) ****This is wrong since T changes temperature between states*****

    w = nRT ln(Vf/Vi)

    Plugged in everything and the answer didn't turn out.
     
    Last edited: Aug 24, 2010
  2. jcsd
  3. Aug 23, 2010 #2
    Your method looks to be fine, so you might want to consider the units. Rather than L, convert to m^3, and instead of atm, convert to pascals.
     
  4. Aug 24, 2010 #3
    Thanks for taking a look at the problem Gear300. The problem I found with my logic is that P, V, and T all are changing from state 1 to state 2 so I have to use the Internal Energy formula reversing the temperature. (i.e. Ti - Tf).
     
  5. Aug 25, 2010 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    Not sure what the question is. Are you trying to find the work done? Find the change in temperature and use:

    [tex]W = \Delta U = nC_v\Delta T[/tex]

    AM
     
  6. Aug 25, 2010 #5
    andrew,

    That's what I ended up doing to get the answer.
     
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