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Thermodynamics- particle speed

  • 30-04-2012 7:39pm
    #1
    stefanG
    Closed Accounts Posts: 2 ✭


    I don't know if I'm just not seeing the answer in front of me but I've been stuck on this question for a while now and any help would be greatly appreciated.

    "Calculate the speed of argon atoms at standard tempeture(273.15 K) and standard pressure(1.01*10^5 Pa). You may assume that it behave as an ideal gas. The atomic mass of Argon = 40 u."
    Tagged:


Comments

  • #2
    brownacid
    Registered Users, Registered Users 2 Posts: 484 ✭✭brownacid


    First problem is your temperture is below absolute zero, you'd wanna check you took down the question right


  • #3
    stefanG
    Closed Accounts Posts: 2 ✭stefanG


    My bad, I put in the minus by accident; it's suppose to be 273.15 aka absolute zero.


  • #4
    citrus burst
    Registered Users, Registered Users 2 Posts: 147 ✭✭citrus burst


    Use this relationship

    1/2*m*v^2 = 3/2*n*R*T

    and rearrange to get the speed

    v = sqrt(3*n*R*T/m)

    You know T and m (temperature and mass) however you don't know n, the number of moles. Are you sure this is the full question? Usually they give the volume, or some other piece of information like the number of particles along with the pressure of the gas. That way you can use the relationship

    PV = nRT

    To figure out the missing variable, in this case n.

    Unless I am missing something, thats how I would do this question


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