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Statistical Mechanics, Simplifying dq/dT

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data

    [tex] \frac{dq}{dT}=\sum_{i} g_i \frac{dq}{dT} e^{-\frac{ε_i}{kT}} = \frac{1}{kT^2}\sum_{i} g_i ε_i e^{-\frac{ε_i}{kT}} = \frac{1}{kT^2} \bar{ε} q[/tex]

    2. Relevant equations

    [tex]q=\sum_{i} e^{-\frac{ε_i}{kT}}[/tex] or for degenerate states, [tex]q=\sum_{I} g_i e^{-\frac{ε_I}{kT}}[/tex]

    3. The attempt at a solution

    The equations in (1) are just set out in my notes. My problem is understanding the last step: I take [itex]\bar{ε}[/itex] to be the average molecular energy, since later they show that [itex]\bar{ε}_{trans}=\frac{3}{2}kT[/itex].

    How can [itex]\sum_{i} g_i ε_i [/itex] be the mean, without dividing by N? Isn't it the total molecular energy?
     
  2. jcsd
  3. May 24, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The probability for a particle to have energy ##\varepsilon_i## is
    $$
    \mathcal{P}(\varepsilon_i) = \frac{1}{q} g_i e^{-\varepsilon_i / kT}
    $$
    Therefore, the average energy can be calculated as
    $$
    \begin{align}
    \bar{\varepsilon} &= \sum_i \varepsilon_i \mathcal{P}(\varepsilon_i) \\
    &= \sum_i \varepsilon_i \frac{1}{q} g_i e^{-\varepsilon_i / kT}
    \end{align}
    $$
    Compare to your equation.
     
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