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Speed of air molecules

  1. Feb 1, 2009 #1
    a) Show that the typical speed of air molecules in the room where you're sitting is about 500 m/s.

    b) Show that the typical speed is about 420 m/s at Mt. Everest. (Neglect the temperature difference between sea level and mountaintop, and assume that the air consists of only nitrogen molecules).


    [tex]\frac{3}{2}k_B T_R = \frac{1}{2}mv^2 \Rightarrow \sqrt{\left\langle v^2 \right\rangle} \approx 500 m/s[/tex] CORRECT


    At Mt. Everest (9000 m) some of the kinetic energy has been transported to potential energy. A [tex]N_2[/tex] molecule has mass 4.7E-26 kg, so that

    \Delta U = mgh = 4.7E-26 kg \cdot 9.81 m/s^2 \cdot 9000 m = 4.15E-21 J.

    Hence the average kinetic energy of a nitrogen molecules up there are 6.15E-21 J - 4.15E-21 J = 2.0E-21 J, and

    [tex]2.\cdot 10^{-21} J = \frac{1}{2}mv^2 \Rightarrow \sqrt{\left\langle v^2 \right\rangle} \approx 290 m/s[/tex] WRONG
    Last edited: Feb 1, 2009
  2. jcsd
  3. Feb 1, 2009 #2
    Try using a relation between pressure and average velocity since T is constant.
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