Statistical Mechanics, Simplifying dq/dT

[curio_city]

Homework Statement

\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

Homework Equations

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

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 \bar{ε} to be the average molecular energy, since later they show that \bar{ε}_{trans}=\frac{3}{2}kT.

How can \sum_{i} g_i ε_i be the mean, without dividing by N? Isn't it the total molecular energy?

 

[DrClaude]

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.