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Using pV^gamma = c

  1. Feb 13, 2015 #1
    One mole of an ideal monatomic gas, initially at 1 atm pressure, is surrounded by adiabatic walls in a container of volume 0.01 m3 . The gas is compressed until the volume is 1/8 of the original volume. Calculate the final pressure and temperature.

    ^^ I'm having problems following this question. Its only a short answer, but i'm confused none the less!

    A: pV^5/3 = constant.
    pf = pi(Vi/Vf )^5/3 = 1(8)^5/3 = 25 = 32 atm.
    Tf = pfVf /nR = 32 × (0.01/8)101325/8.314 = 487.5 K

    I understand the last line in normal ideal gas equation, but the first two lines, which reference the equation: pV^5/3 = constant has thrown me, i don't know how the second line follows (where does pf=pi*(vi/vf)^5/3 come from? i see its similar to the constant equation
     
  2. jcsd
  3. Feb 13, 2015 #2

    Quantum Defect

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    For adiabatic processes (no heat exchanged withthe surroundings) in ideal gases, P*V^gamma = constant. Gamma is the ratio of the heat capacities at constant pressure and constant volume. For an ideal monatomic gas, C_v = 3/2 * R, and C_p = 5/2* R, so gamma = (5/2 * R) / (3/2 * R) = 5/3.

    http://en.wikipedia.org/wiki/Adiabatic_process
     
  4. Feb 13, 2015 #3
    Yeah, thanks

    The part i still don't understand is how the equation pf = pi*(vi/vf)^5/3 follow?
     
  5. Feb 13, 2015 #4

    Quantum Defect

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    Pf * Vf^gamma = constant = Pi * Vi^gamma
    Pf = Pi * (Vi/Vf)^gamma
     
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