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The ideal gas law for an adiabatic process

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi can someone please have a look at this question and let me know if I am on the right track, thanks.

    A diesel engine requires no spark plug; instead the air in the cylinder is compresses so highly the fuel ignites spontaneously on injection to the cylinder.

    Q. If the air is initially at 293K and then is compressed adiabatically by a factor of 15, what final temperature is attained ?( Just prior to injection, also consider air as an ideal gas)
    Variables:
    T1=293K
    V1= 1
    V2=1/15
    T2=?

    2. Relevant equations

    T1 *V1^(Ƴ-1) = T2* V2^(Ƴ-1)


    3. The attempt at a solution
    Above relationship applies as process is adiabatic?

    293× (1^(1.4-1)/ (1/15^(1.4-1) ))=866K
     
    Last edited: Feb 17, 2014
  2. jcsd
  3. Feb 17, 2014 #2

    BvU

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    Looks good to me. Why the doubtful question mark ?
     
  4. Feb 17, 2014 #3
    Confidence, I am learning to trust the equations.

    Another part of the question is; by what factor has the pressure increased? Can I use P V^Ƴ=constant, with p1=1, multiplied by ratio of V1/V2, both to the power of Ƴ? ans=2.95
     
  5. Feb 17, 2014 #4

    BvU

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    Re post 1: T1 *V1^(Ƴ-1) = T1* V1^(Ƴ-1) I read T1 *V1^(Ƴ-1) = T2* V2^(Ƴ-1) which looks good.

    Re post 3: numerically I get something different, but then I use ##\gamma##, not ##\gamma -1## as the exponent.... Big difference !
     
  6. Feb 17, 2014 #5
     
  7. Feb 17, 2014 #6

    Andrew Mason

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    Once you have the temperature worked out, just use the ideal gas law: PV=nRT

    P1V1/T1 = P2V2/T2

    P2/P1 = (V1/V2)(T2/T1)

    AM
     
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