Degrees of Freedom and Molar Specific Heats — Question 3

Question 3

A vibrational mode stores equal kinetic and potential energies

A classical harmonic bond coordinate $q$ has energy $E=p_q^2/(2\mu)+\kappa q^2/2$, reduced mass $\mu=1.00\times10^{-26}$ kg and stiffness $\kappa=100$ N/m. Take $T=500$ K and $k_B=1.381\times10^{-23}$ J/K. The mode is stipulated to be in its classical regime.

1. Find rms bond-coordinate displacement and rms rate $\dot q$.

2. Find the mean total mode energy and its heat-capacity contribution for .200 mol of such modes. How do the rms amplitudes change if temperature quadruples?

Question 3 - Solution

A vibrational mode stores equal kinetic and potential energies

(a) Method and calculation

$$\tfrac12\kappa\langle q^2\rangle=\tfrac12k_BT\ \Rightarrow\ q_{\mathrm{rms}}=\sqrt{k_BT/\kappa}\approx8.30963\times10^{-12}\,\mathrm m.$$$$\tfrac12\mu\langle\dot q^2\rangle=\tfrac12k_BT\ \Rightarrow\ \dot q_{\mathrm{rms}}=\sqrt{k_BT/\mu}\approx830.963\,\mathrm{m/s}.$$

(b) Interpretation and further calculation

$$\langle E\rangle=k_BT=6.905\times10^{-21}\,\mathrm J,\quad C_{\mathrm{sample}}=nR=.200(8.314)=1.6628\,\mathrm{J/K}.$$$$T\to4T:\quad q_{\mathrm{rms}}\to2q_{\mathrm{rms}},\quad\dot q_{\mathrm{rms}}\to2\dot q_{\mathrm{rms}}.$$The relative bond coordinate is not the center-of-mass displacement of the molecule.

Physical check

Equal energy in kinetic and potential terms does not imply equal numerical amplitudes; their coefficients have different units.