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The Internal Energy of Neon Gas

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    The internal energy of a monoatomic ideal gas such as neon is simply the total kinetic energy of all its atoms.

    What is the internal energy of 2 liters of neon at a temperature of 200 K and pressure of 0.7 atm?

    2. Relevant equations

    PV = nRT
    KE(ave) = 3/2kT
    U = 3/2nRT

    3. The attempt at a solution

    KE = 1/2mv^2

    From a previous problem, I figured out that v(rms) = 499.227 m/s.

    And, 2 liters of neon X 0.9002 g/L (density of neon) = 0.0018004 kg

    KE = 1/2(0.0018004 kg)(499.227 m/s) = 224.3546833 J (wrong answer)

    I feel like I am approaching this problem in the wrong manner. Please help!
     
  2. jcsd
  3. Jun 16, 2009 #2

    LowlyPion

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    That's the density of Ne at STP. What is the density at 200K and .7 atm?
     
  4. Jun 16, 2009 #3
    So, d = P X MM/RT

    d = (0.7 atm)(20 g/mol)/(8.31 J/mol*K)(200 K)
    d= 0.008423586

    2 L * 0.008423586 = 1.684717208E-5 kg

    KE = 1/2(1.68E-5)(499.227 m/s)^2 = 2.09 J (WRONG ANSWER)

    AM I USING THE WRONG UNITS OR DID I DO A MATH ERROR???
     
  5. Jun 16, 2009 #4

    LowlyPion

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    I'd say your numbers are wrong, because without even looking you had .0018 kg using STP. I wouldn't expect such a small number after accounting for the 200/273 ratio and the .7 ratio.

    Won't the approach work out to be more like (P1/T1)/(P2/T2) = d1/d2 ?
     
  6. Jun 17, 2009 #5
    I may be missing something but it looks from the revelant equations that
    U=1.5*(R/M)*T where R is the universal gas constant, M is the molecular weight of neon and T is temperature in deg K. Am I over simplifing this?
     
  7. Jun 17, 2009 #6

    ideasrule

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    The solution to the problem is startlingly simple; you were less than a hair away from getting it. You wrote, as a relevant equation, U = 3/2nRT. You also wrote PV = nRT. So if U=(3/2)PV, and you have both P and V.

    Remember this neat result: the internal energy of an ideal gas depends only on its pressure and volume.
     
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