= Solution
Relative to the <ground state>, the <quantum harmonic oscillator> levels are $vhc\widetilde\omega$. Put $\theta_v=hc\widetilde\omega/k$. Summing the <geometric series> gives the <vibrational partition function> $q=(1-e^{-\theta_v/T})^{-1}$. Therefore the vibrational <internal energy> is
$$
\boxed{U=\frac{Nk\theta_v}{e^{\theta_v/T}-1}}
=\frac{Nhc\widetilde\omega}{e^{hc\widetilde\omega/kT}-1}.
$$
The omitted <zero-point energy> would add $Nhc\widetilde\omega/2$ if absolute oscillator energies were used.
Back to article page
