= Solution
The <Boltzmann entropy> is
$$
S=k_B\ln\Omega=k_B\ln\Omega_0+\frac{5N}{2}k_B\ln(E/E_0).
$$
The thermodynamic definition of <temperature> gives
$$
\frac1T=\left(\frac{\partial S}{\partial E}\right)_{N,\ldots}=\frac{5Nk_B}{2E},\qquad \boxed{T=\frac{2E}{5Nk_B}.}
$$
Holding the other system parameters fixed, $E=(5N/2)k_BT$, so its <heat capacity> is \b[$C=5Nk_B/2$]. The constants $\Omega_0$ and $E_0$ set an additive reference for <entropy>; neither affects this derivative or the resulting <heat capacity>.
Back to article page
