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Adiabatic process

  1. Nov 22, 2004 #1
    I have this problem and I do not have the answer, but I get an answer that I feel is probably wrong, so can someone please check my work and point out where I went wrong??

    Here is the problem:
    moles = 0.10 of O_2
    T(initial) = 150 C = 423 K
    P(initial) = 3.0 atm = 303.9 KPa

    The gas expands adiabatically until the pressure is halved, find the final volume and final pressure

    Since the pressure is halved, we know that
    P(final) = 1.5 atm = 151.95 KPa

    I need the initial volume, so I used the ideal gas equation, PV = nRT
    using the initial conditions with P in pascals, n in mols, T in kelvin, and R as 8.31 J/mol*K

    So I get a V(initial) = 1.156 x 10^(-3) m^3

    Then I need the final volume, and since this is adiabatic,

    Since O_2 is diatomic and we assume ideal conditions, gamma = 1.4

    So using the above equation, I find
    V(final) = 1.37 x 10^(-5) m^3 = answer to part a
    I don't know if this is right or wrong

    Then for part b, I used the idea gas equaion again, PV=nRT
    Using the final volume, final pressure, same n and same R, I get
    T = 2.50 K

    Obviously this is extremely COLD!! I don't think it makes sense that the temperature would drop from 423 K to 2.5 K....where did I go wrong?
  2. jcsd
  3. Nov 23, 2004 #2


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    The final volume can not be right, as it is much lower than the initial volume, and the gas has expanded.

    [tex] P_{initial}/P_{final}=(V_{final}/V_{initial})^{1.4}=2[/tex]


    [tex] V_{final}=1.970 \cdot10^{-3}\mbox{ } m^3[/tex]

  4. Nov 23, 2004 #3
    Yes, I just realized that at the same time you posted.....

    I used the wrong initial volume (actually I just used 1.156 instead of 1.156 x 10^-3



    btw..it is 1.90 x 10^-3 I think, not 1.970....probably just a typo though :)
    Last edited: Nov 23, 2004
  5. Nov 23, 2004 #4


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    Well, yes, it was 1.897 and I left out the "8" :)

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