# Statistical Mechanics, Simplifying dq/dT

1. May 24, 2013

### curio_city

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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 $\bar{ε}$ 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?

2. May 24, 2013

### 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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