\( \newcommand\D{\mathrm{d}} \newcommand\E{\mathrm{e}} \newcommand\I{\mathrm{i}} \newcommand\bigOh{\mathcal{O}} \newcommand{\cat}[1]{\mathbf{#1}} \newcommand\curl{\vec{\nabla}\times} \newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\ZZ}{\mathbb{Z}} \)
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Temperature - Statistical Mechanics

Table of Contents

1. Molecular Model

We have a molecular model of gases, which has from

\begin{equation} \delta Q = nC_{V}\,\mathrm{d}T \end{equation}

And the increase in translational kinetic energy

\begin{equation} \delta K_{tr} = \frac{3}{2}nR\,\mathrm{d}T. \end{equation}

Setting these equal, since the total molecular energy increase is stipulated to be \(\delta K_{tr}\), we find

\begin{equation} C_{V} = \frac{3}{2}R. \end{equation}

This works for monatomic gases. 

(defvar gas-constant 8.31446261815324)

;; mono-atomic molecules
(* 1.5 gas-constant)

12.47169392722986

For diatomic gases, we need to account for rotational degrees of freedom, which changes \(3/2\) to \(5/2\) in the definition of total molecular energy:

\begin{equation} \delta K_{tr} = \frac{5}{2}nR\,\mathrm{d}T. \end{equation}

This gives us:

\begin{equation} C_{V} = \frac{5}{2}R. \end{equation} 
;; diatomic gases
(* 2.5 gas-constant)

20.7861565453831

Last Updated 2022-08-13 Sat 07:58.