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The parameter \(\beta\) will be determined by noting that \(\beta\) in the partition function depends only on the heat bath and not on the system. Thus \(\beta\) will be shown to be equal to \(1/kT\) by demanding that the average energy \(U\) as computed from partition function be equal to the value \(\frac{3}{2}kT\) given by equipartition law
Internal energy of an ideal gas
The average energy of a system \(U\), in contact with a heat bath, is \begin{equation} U=-\frac{\partial}{\partial\beta} \log Z. \end{equation} where \(Z\) is the canonical partition function. The partition function for an ideal gas is \begin{equation} Z = V^N\left(\frac{2\pi}{\beta m}\right)^{3N/2}, \end{equation} Therefore for an ideal gas the average energy is \begin{align*} U=&-\frac{\partial}{\partial\beta}\ln Z = -\frac{3N}{2}\frac{\partial}{\partial\beta}\Big(\ln\frac{2\pi}{\beta m} \Big) = \frac{3N}{2\beta}. \end{align*}
Identification of $\beta$
Since $\beta$ is independent of system, we shall consider perfect gases, we compare the average energy obtained from the partition function with the expression \[U= \frac{3}{2}N k T,\] given by the equipartition law and where \(k\) is Boltzmann constant \(( =1.380\times10^{16} {\rm ergs/deg})\). Thus we have \begin{equation} \frac{3}{2}NkT = U = \frac{3N}{2\beta} \end{equation} This leads to identification of \(\beta\) as \begin{equation} \label{Beta} \boxed{\beta=\frac{1}{kT}} \end{equation}
