Degrees of Freedom and Molar Specific Heats — Question 2

Question 2

Molecular geometry changes the number of vibrational normal modes

Use classical equipartition: each active independent quadratic energy term contributes $\tfrac12k_BT$ per molecule. Let $f$ count these terms, so for constant $f$, $C_V=fR/2$, $C_p=C_V+R$ and $\gamma=C_p/C_V$. Take $R=8.314$ J/(mol K). A linear triatomic molecule has two independent rigid rotations and $3N-5$ vibrational normal modes. A nonlinear triatomic molecule has three rigid rotations and $3N-6$ vibrational modes. Here $N=3$ atoms. Each fully active harmonic normal mode contributes two quadratic terms.

1. Find $C_V$ and $\gamma$ for both geometries with all vibrations frozen.

2. Find the fully classical $C_V$ and $\gamma$ when every vibrational mode is active, and explain why one vibrational mode contributes $R$ rather than $R/2$.

Question 2 - Solution

Molecular geometry changes the number of vibrational normal modes

(a) Method and calculation

$$f_{\mathrm{linear}}=3+2=5:\quad C_V=\tfrac52R,\quad\gamma=\tfrac75.$$$$f_{\mathrm{nonlinear}}=3+3=6:\quad C_V=3R,\quad\gamma=\tfrac43.$$

(b) Interpretation and further calculation

$$s_{\mathrm{linear}}=4,\quad f=5+2(4)=13:\quad C_V=\tfrac{13}2R,\quad\gamma=\tfrac{15}{13}.$$$$s_{\mathrm{nonlinear}}=3,\quad f=6+2(3)=12:\quad C_V=6R,\quad\gamma=\tfrac76.$$$$E_{\mathrm{mode}}=\frac{p_q^2}{2\mu}+\frac12\kappa q^2\ \Rightarrow\ \langle E_{\mathrm{mode}}\rangle=k_BT.$$A normal mode has one kinetic and one potential quadratic term.

Physical check

The supplied mode counts already include distinct bending modes; degeneracies must not be counted a second time.