Degrees of Freedom and Molar Specific Heats — Question 1

Question 1

A rigid linear molecule has two thermally active rotational terms

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 dilute diatomic gas is modeled as a rigid linear rotor with three translational terms and two active rotations perpendicular to its bond. Bond vibration and rotation about the bond are frozen. There are .600 mol of gas.

1. Find $f$, $C_V$, $C_p$ and $\gamma$, and the fixed-volume heat for a 50.0 K rise.

2. In a lower-temperature interval where both transverse rotations are also frozen, compare the heat required for the same amount and temperature increment. Treat each interval as having constant active terms.

Question 1 - Solution

A rigid linear molecule has two thermally active rotational terms

(a) Method and calculation

$$f=3+2=5,\quad C_V=\tfrac52R,\quad C_p=\tfrac72R,\quad\gamma=\tfrac75.$$$$Q_V=.600(\tfrac52)(8.314)(50.0)=623.55\,\mathrm J.$$

(b) Interpretation and further calculation

$$f_{\mathrm{low}}=3,\quad C_{V,\mathrm{low}}=\tfrac32R,\quad Q_{\mathrm{low}}=.600(\tfrac32)(8.314)(50.0)=374.13\,\mathrm J.$$$$Q_{\mathrm{low}}/Q_V=3/5.$$The comparison uses two separate temperature intervals; it does not model the transition between them as a discontinuous physical jump.

Physical check

The counted terms are active energy contributions, not simply the number of atoms.