Fall 2007 7 Sm

Molecules of an ideal gas have internal energy levels that are equidistant, $E_n = n\epsilon$, where n = 0, 1, … and $\epsilon$ is the level spacing. The degeneracy of nth level is n + 1. Find the contribution of these internal states to the energy of the gas of N molecules at temperature T.

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Answer

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Solution 1

For a non-interacting ideal gas,

(1)
\begin{align} E=-\frac{\partial}{\partial \beta}N \ln{\zeta} \end{align}

where $\zeta$ is the single-molecule partition function

(2)
\begin{align} \zeta=\sum_{n=0}^{\infty}(n+1)\exp{(-\beta n \epsilon)} \end{align}

This partition function can be evaluated as follows $(x\equiv\beta\epsilon)$:

(3)
\begin{align} \zeta=-e^x\frac{d}{dx}\sum_{n=0}^{\infty}\exp{\left(-(n+1)x\right)}=-e^x\frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\left[1-\exp{(-\beta\epsilon)}\right]^{-2} \end{align}

Hence, the sought contribution to the energy is

(4)
\begin{align} E=\frac{2N\epsilon}{\exp{(\epsilon/kT)}-1} \end{align}

Solution 2

Alternatively, one can reproduce this result as follows. One can imagine that every molecule has two independent internal degrees of freedom of harmonic oscilator type, with energy spacing $\epsilon$ each. It is easy to see that this model gives the same spectrum and degeneracies if the energy is counted from the ground state. With this convention, the average energy of a single harmonic oscillator is $\epsilon n_B(\epsilon)$, where $n_B(\epsilon)$ is the Bose-Einstein occupation number. Therefore, for the entire gas we get $E = 2N\epsilon n_B(\epsilon)$, in agreement with the first derivation.

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