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Basis of degree of freedom for monoatomic and diatomic mixture

  1. May 8, 2005 #1
    hi guys

    16 grams of helium gas is mixed with 16 grams of oxygen.....
    what will be the ratio Cp/Cv of the mixture.......

    how to calculate it??????? i thought on the basis of degree of freedom for monoatomic and diatomic mixture......but that can take me to the answer only by approximation......how do we arrive in a a formula that could get me answer........
     
  2. jcsd
  3. May 8, 2005 #2

    Andrew Mason

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    Aren't they both diatomic?

    AM
     
  4. May 8, 2005 #3
    how????

    y do u say that these two are diatomic???

    He...its completely satisfied without bonding
    He is a noble gas...so y do u say it is diatomic.....

    moreover someone help me solve the problem yaar........
     
  5. May 8, 2005 #4
    Monatomic Gases:
    Helium, Neon, Argon, Krypton, Xenon, Radon
    Single atom, or monatomic, gases have the smallest Specific Heat CV.

    Diatomic Gases:
    Oxygen, Nitrogen, Hydrogen
     
  6. May 8, 2005 #5

    Andrew Mason

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    Of course you are right. I was seeing He and thinking H.

    What is the ratio of the number of He atoms to O2 molecules?

    AM
     
  7. May 9, 2005 #6
    hi

    y isnt anyone helping me get the answer??????
    please if someone can explain the problem to me...it would do a world of good to me....


    please please please....the question again.....

    16 grams of helium gas is mixed with 16 grams of oxygen.....
    what will be the ratio Cp/Cv of the mixture.......

    how to calculate it??????? i thought on the basis of degree of freedom for monoatomic and diatomic mixture......but that can take me to the answer only by approximation......how do we arrive in a a formula that could get me answer........
     
  8. May 9, 2005 #7
    nomorevishnu Let helium be the gas 1 and oxygen be the gas 2 .

    No. Of moles of Helium: 4 = n1
    " " " " oxygen: 0.5 =n2

    Lamda factor for a mixture is given by:

    L= n1( Cp1) + n2 (Cp2) / n1(Cv1) + n2 (Cv2)

    Where Cp1/Cv1=5/3 ( for monoatomic helium)

    Cp2/Cv2=7/3 for diatomic oxygen


    Answer you get is: 1.62
     
  9. May 9, 2005 #8

    Andrew Mason

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    Have you worked out the proportion of numbers of atoms of He to number of molecules of O2?

    AM
     
  10. May 9, 2005 #9
    well...i knew that equation...and the answer


    but to find the effective lambda...how do we get to such an equation...any proof....its not given in Resnick and Halliday
     
  11. May 9, 2005 #10

    Andrew Mason

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    You have to go to basic principles:

    [tex]dU = (Cp - Cv)nT = nRT[/tex] so:

    [tex]C_p/C_v = \gamma = (C_v + R)/C_v = (1 + R/C_v)[/tex]

    So for the mixed gas:

    [tex](C_{peff} - C_{veff})n_{total}dT = n_{total}RdT[/tex]

    [tex]C_{peff} = (R + C_{veff}) [/tex]

    (1)[tex]C_{peff}/C_{veff} = \gamma_{eff} = (R/C_{veff} + 1)[/tex]

    Now:

    [tex]Vdp = (C_{vHe}n_{He} + C_{vO_2}n_{O_2})dT = C_{veff}n_{total}dT[/tex]

    (2) [tex]C_{veff} = (C_{vHe}n_{He} + C_{vO_2}n_{O_2})/n_{total}[/tex]

    Substitute from (2) into (1).

    I get 1.64

    AM
     
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