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I My Case of Total Translational Kinetic Energy of Ideal Gas

  1. Mar 6, 2016 #1
    I have really problem with equation of the root – mean – square (rms) speed of the gas molecules when we use the universal gas constant (R) mentioned in all references.

    Accordance to all reference books, we have below equation for the total translational kinetic energy:

    Vrms = (3RT/ M)^0.5

    Where: "M" is the molar mass

    While I think above formula is wrong because for total KE, we have:

    KEtotal translational = N (1/2 mo V^2) = 3/2 nRT

    n = mo/M and n = N/NA then mo . NA = M . N

    Therefore, we will have:

    Vrms = (3RT/ M .n)^0.5

    Could you please help me where I am taking a mistake?
     
  2. jcsd
  3. Mar 6, 2016 #2
    What is mo? Mass of a molecule or mass of the gas?
     
  4. Mar 6, 2016 #3
    mo is total mass of gas.
     
  5. Mar 6, 2016 #4

    Dale

    Staff: Mentor

    I am not sure where you went wrong, but this is clearly incorrect. It would say that if you had 1 mole of helium at 300 K then the RMS velocity would be half that of 2 moles of helium at 300 K. In other words, the more gas you have the faster it would be moving at a fixed temperature.

    What is the difference between a 2 mole container of helium and two 1 mole containers? If you dropped a barrier half way inside a 2 mole container, what would possibly make all of the gas slow down?
     
  6. Mar 6, 2016 #5
    Let me tell you above mentioned in another view.

    According to all references books, we have:

    Vrms = (3kBT/mo)^0.5 = (3RT/ M)^0.5

    There is a discrepancy between (3kBT/mo)^0.5 and (3RT/ M)^0.5. In the case of kB, we have two variables of "T" and "mo" but the case of "R" tells us that there is one variable of "T" because molar mass (M) is the constant for each gas.

    Now, let me have an example:

    Assume, we have a tank with constant volume filled with 2 mole of hydrogen in temperature of 20.0 Celsius degree. By using of both formulas, we can calculate Vrms as follows:

    Vrms = (3kBT/mo)^0.5 = 1345m/s

    Vrms = (3RT/ M)^0.5 = 1902 m/s

    How can I explain this discrepancy?

    We as well as know three laws of ideal gas including T – constant, P – constant and V – constant. In the case of Gay-Lussac's law, assume we have a system which is able to prevent increase of temperature when we increase pressure. In this status, when we increase number of molecules but temperature and volume are constant, Vrms will decrease. In fact, I can say that three factors are controlling internal energy of gas: 1) Molar mass, lighter molecules have more velocity. 2) Temperature 3) The number of molecules in system. when "N" increase, then Vrms decrease because increase of temperature distribute among all molecules.

    Is my explanation true?
     
  7. Mar 6, 2016 #6

    Dale

    Staff: Mentor

    In this formula mo is the molecular mass. It is constant for each gas and it is not the same as your mo above which was the total mass of the gas. I think you are misreading the formulas.
     
  8. Mar 6, 2016 #7
    Do you mean that mo = M?
     
  9. Mar 6, 2016 #8

    Dale

    Staff: Mentor

    No mo is the mass of one molecule, and M is the mass of one mole of molecules. So ##M = N_A m_0##

    Edit: at least in the standard formulas that you cited. In your own formula in the original post you defined it differently.
     
  10. Mar 6, 2016 #9
    Thanks. I understood my mistake.
     
  11. Mar 6, 2016 #10
    Regarding to mo (equal to mass of one molecule), could you please tell me how you calculate the mass of 0.5 mole hydrogen?
    Is below calculation true?
    0.5 u = (1.6766E-24)/2 g = 8.383E-28 kg
     
  12. Mar 6, 2016 #11
    No. One mol of molecular hydrogen is 2 grams. Half a mol will be 1g.
     
  13. Mar 6, 2016 #12
    oh. sorry. I mixed everything. thanks
     
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