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Compression (and temperature) of a Jaguar XK8 Cylinder helpp

  1. Jan 24, 2007 #1
    1. A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 cm^3 of air at atmospheric pressure (1.01 x 10 ^5 Pa) and a temperature of 27.0 C. At the end of the stroke, the air has been compressed to a volume of 46.2 cm^3 and the gauge pressure has increased to 2.72 x 10 ^6 Pa .



    2. I am kind of lost how to start this...



    3. none as of yet...
     
  2. jcsd
  3. Jan 24, 2007 #2

    Kurdt

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    So what are you trying to work out?
     
  4. Jan 24, 2007 #3
    The final temperature that it is at...

    here:

    Hint 1. How to approach the problem
    Use the ideal gas law to relate the initial pressure, temperature, and volume to their final values. Calculate the final temperature given the initial and final values in the introduction. Also, be very careful about the units in your calculations.

    Hint 2. Mass of air in the cylinder
    Because the air in the cylinder is trapped and cannot enter or leave, the mass of the air in the cylinder must be constant. Therefore, the number of moles (n) is a constant for both the initial and final states of the cylinder.

    Hint 3. Relation between the initial and final states
    From the ideal gas equation, after a little algebraic manipulation, we get nR = pV/T. This will be true in both the final and initial states of the cylinder, and, as explained in the previous hint, since (n) is constant, the two states are related by
    .

    PiVi/Ti = PfVf/Tf


    Hint 4. Gauge pressure
    Recall that the gauge pressure is the difference between the absolute pressure and the atmospheric pressure. Thus, if you measure a gauge pressure pg , then the absolute pressure (p) is given by pa+pg = p, where pa is the atmospheric pressure.
     
  5. Jan 24, 2007 #4

    Kurdt

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    With all those hints you should really be making a better attempt at the solution before seeking help.
     
  6. Jan 24, 2007 #5
    i have been...i just haven't found the right answer
     
  7. Jan 24, 2007 #6
    anyone help? no one? this can't be that hard!
     
  8. Jan 24, 2007 #7
    You spell it out in Hint 3. Ignore all the crap about the car and look at what you have to use.

    At the beginning, one of the cylinders contains 499 cm^3 of air at atmospheric pressure (1.01 x 10 ^5 Pa) and a temperature of 27.0 C.

    At the end, the air has been compressed to a volume of 46.2 cm^3 and the gauge pressure has increased to 2.72 x 10 ^6 Pa .
     
  9. Jan 24, 2007 #8

    Kurdt

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    You can't receive any help until you've shown a reasonable attempt at the solution yourself. This is clearly spelled out in the guidelines. With the hints you have been given there is really no reason why you can't attempt this yourself and if your result is not correct, to post your attempt so we can give you guidance as to where you've gone wrong.
     
  10. Jan 24, 2007 #9
    Yes...and i have tried that in relation to plugging in the values that i do know in order to find the final temperature but my answers keep saying wrong!

    If the air has been compressed than we know that the temperature is greater. I don't really know what to do with the guage pressure though...

    im trying to sort out a few differnet functions that the teacher threw at us.
     
  11. Jan 24, 2007 #10
    Well would the temperature be greater? im thinking it won't be now...
     
  12. Jan 24, 2007 #11
    I've tried to take:

    2.82 x 10 ^ 6 Pa * 499 cm ^ 3 / 27.0 C = 2.82 x 10 ^ 6 Pa * 46.2 cm ^ 3 / ?

    the question mark would be the final temperature.

    or am i way off with this? You guys gotta understand that our teacher like doesn't teach us and i would learn way more if someone could give me a much better guiding step as i have spent way too much time on this problem.
     
  13. Jan 24, 2007 #12

    NoTime

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    First do the algebra on the equation to solve for Tf.

    Second, where did 2.82 x 10 ^ 6 Pa come from?
    Its not one of your givens.
     
  14. Jan 25, 2007 #13
    ok well ive tried it a few times. I added the two amounts based on hint two but i guess that is wrong. Do i even incorporate guage pressure?

    I canceled the atm pressure out since it'll be the same at the start and finish.

    So i took 499/27 = 46.2/?

    So I had (1/46.2)*(499/27) = .400

    But this is far from the right answer...so i am still stuck...ahhh i need some direction people, please!
     
  15. Jan 25, 2007 #14

    NoTime

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    First take this equation from hint 3 -> PiVi/Ti = PfVf/Tf
    Use algebra to rework the equation to solve for Tf.
    Just work with the symbols (Pi, Vi, Ti, Pf, Vf and Tf ) only, no number substitution.
     
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