= Solution
At fixed chain length, each <Normal mode>[harmonic mode] contributes a <Bose-Einstein distribution>[Bose–Einstein thermal occupation] and a temperature-independent <zero-point energy>. The latter contributes no <heat capacity>. With $x=\hbar\omega/(k_BT)$,
$$
\frac{C_V}{\mathcal L}=k_B\int_0^{\omega_m}g(\omega)
\frac{x^2e^x}{(e^x-1)^2}\,d\omega.
$$
At low <temperature> only the approximately linear part of the <Monatomic-chain dispersion relation>[acoustic dispersion] is excited, so $g(\omega)\simeq1/(\pi v_s)$. Using $\int_0^\infty x^2e^x/(e^x-1)^2dx=\pi^2/3$ gives
$$
\boxed{\frac{C_V}{\mathcal L}\simeq\frac{\pi k_B^2}{3\hbar v_s}T
=(4.732\times10^{-16}\ \mathrm{J\,K^{-2}\,m^{-1}}) T.}
$$
It is \b[linear in <temperature>], rather than the three-dimensional <Debye model>[Debye] $T^3$ law. At high <temperature> every scalar oscillator tends to $k_B$, hence
$$
\boxed{\frac{C_V}{\mathcal L}\longrightarrow\frac{k_B}{a}
=6.903\times10^{-14}\ \mathrm{J\,K^{-1}\,m^{-1}},\qquad C_V/(Nk_B)\to1.}
$$
The cutoff <temperature> is $\Theta_m=\hbar\omega_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; $\Theta_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(Nk_B)\simeq(2\pi/3)T/\Theta_m$. The low-temperature continuum limit also requires <temperature> above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode <energy> a finite chain does not retain a continuum linear law down to arbitrary <temperature>.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2016-monatomic-chain-heat-capacity.png]
{title=Single-branch chain heat capacity and its low- and high-temperature limits}
{description=Original numerical Bose–Einstein mode integral. The dashed line is the low-temperature linear asymptote; the plateau is one Boltzmann constant per atom for this scalar model.}
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