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Kinetic Theory Of Temperature Problem

  1. Oct 31, 2005 #1
    The rms speed of the molecules in a gas is 600ms^-1 and the mass of the molecules is 4.6*10^-26kg. What is the temperature of the gas in degrees Celcius?
    [tex]
    v_{r.m.s.}\=\600\ms^-1 \\
    m\=\4.6x10^-26kg \\
    n\=\6.022x10^23 \\
    r\=8.31\j\k\mol^-1 \\
    \\
    \\
    rt\=\frac{\2}{3}\(\frac{\1}{2}m<v^2>)\\
    t\=\frac{\frac{\2}{3}\(\frac{\1}{2}m<v^2>)}{r}\\
    t\=\frac{\frac{\2}{3}\(\frac{\1}{2}\x\4.6x10^-26\x\600^2)}{8.31}\\
    t\=\frac{3324.14}{8.31}\\
    t\=\400K\\
    t(c)\=\400-273\=\127c\\
    [/tex]
    Does this seem correct? I have no way of checking the answer and I feel really unsure, especially using the Vr.m.s.
    Thanks
    EDIT: I can't understand why my TEX doesn't work :(

    My tex code:

    V_{r.m.s.}\=\600\ms^-1 \\
    m\=\4.6x10^-26kg \\
    N\=\6.022x10^23 \\
    R\=8.31\J\K\mol^-1 \\
    \\
    \\
    RT\=\frac{\2}{3}\(\frac{\1}{2}m<v^2>)\\
    T\=\frac{\frac{\2}{3}\(\frac{\1}{2}m<v^2>)\\}{R}\\
    T\=\frac{\frac{\2}{3}\(\frac{\1}{2}\x\4.6x10^-26\x\600)\\}{8.31}\\
    T\=\frac{3324.14}{8.31}\\
    T\=\400K\\
    T(C)\=\400-273\=\127C\\
     
    Last edited: Oct 31, 2005
  2. jcsd
  3. Oct 31, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    You need to write the tex tag using lowercase letters.
     
  4. Oct 31, 2005 #3
    1/2m(rms) = 3/2RT or possibly 1/2m(rms) = 3/2kT depends on the ammount of gas molecules for one mole it would be the first one i think.
     
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