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Adiabatic compression of a gas

  1. Jul 13, 2009 #1
    An explosive liquid at temperature 300 K contains a spherical bubble of radius 5 mm, full of its vapour. When a mechanical shock to the liquid causes adiabatic compression of the bubble, what radius of the bubble is required for combustion of the vapour, given that the vapour ignites spontaneously at 1100 degrees C? That ratio of CV/(n*R) is 3.0 for the vapour.


    I'm going to combine relevant equations and attempt at solution together because I'm not sure if the equations I'm using are the relevant ones to begin with.

    So for reversible adiabatic changes in an ideal gas,

    PV[tex]^{\gamma}[/tex] = constant

    And if we put P = nRT/V into that equation, we get

    TV[tex]^{\gamma-1}[/tex] = constant

    The problem is, how am I supposed to know the n of the gas (how many particles)? I'm guessing I have to use the ratio CV/(n*R) somehow to also find [tex]\gamma[/tex]:

    CV/nR = 3 (from above)
    So
    CV = 3nR

    and

    CP = CV + R

    So

    CP = 3nR + R
    = (3n + 1)R

    [tex]\gamma[/tex] = CP/CV
    = [tex]\frac{(3n + 1)R}{3nR}[/tex]

    = [tex]\frac{3n+1}{3n}[/tex]

    And now I'm stuck again because I still don't know n

    Any help/guidance would be appreciated. Thanks.

    Also, this problem is from Introductory Statistical Mechanics 2nd ed. by Roger Bowley and Mariana Sanchez, Chapter 1, Problem 8
     
  2. jcsd
  3. Jul 13, 2009 #2
    Use the initial information in the beginning of the problem to give you the T and V of the bubble. Then use that equation. Set the initial and final equal to each other.

    As for the "n" part. I think you use the specific heat capacities instead.
     
  4. Jul 13, 2009 #3
    How do I find [tex]\gamma[/tex], though?

    [tex]\gamma[/tex] = CP/CV

    I don't know either values of CP or CV..

    BTW, I recently found the answer in the back of the book. The new radius is supposed to be 1.09 mm, but I still don't get how they got to this..
     
  5. Jul 13, 2009 #4
    The gamma is just the Cp/Cv but it is with specific heat capacities.

    Your equation with Cp = Cv + R is only for specific heat capacities. I wish that I could put the little bar above the Cp and Cv but I can't. Remember that it is specific heat capacities that have to be used in this problem.

    This means that the specific Cv would be 3R and the specific Cp would be 4R.
     
  6. Jul 13, 2009 #5
    Sorry, I'm still confused..:confused:

    How did you arrive to this?
     
  7. Jul 13, 2009 #6
    I arrived to this because the problem gave you Cv/(n*R) = 3.0 . Since specific Cv is just Cv/n, the specific Cv would be 3R.
    Since specific Cp is equal to specific Cv + R, then specific Cp would be 4R.
    This will give you the information needed for gamma and the rest of the problem.
     
  8. Jul 13, 2009 #7
    Thanks so much, bucher.

    I forgot the difference between heat capacity and molar heat capacity.

    Thanks again! :smile:
     
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