Degrees of Freedom and Molar Specific Heats — Question 6

Question 6

Internal modes divide stored energy without changing translational pressure

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 diatomic gas has three active translations, two rotations and one fully active harmonic vibration. Choose the thermal energy reference so that $U=f nRT/2$. At an equilibrium state $pV=1200$ J.

1. Find total thermal energy and its translational, rotational and vibrational portions.

2. For quasistatic fixed-pressure heating with all modes remaining active, find the fractions of supplied heat entering translation, rotation, vibration and expansion work.

Question 6 - Solution

Internal modes divide stored energy without changing translational pressure

(a) Method and calculation

$$f=3+2+2=7,\quad U=\tfrac72pV=4200\,\mathrm J.$$$$U_{\mathrm{tr}}=\tfrac32pV=1800\,\mathrm J,\quad U_{\mathrm{rot}}=pV=1200\,\mathrm J,\quad U_{\mathrm{vib}}=pV=1200\,\mathrm J.$$

(b) Interpretation and further calculation

$$Q=nC_p\Delta T=\tfrac92nR\Delta T.$$$$\frac{\Delta U_{\mathrm{tr}}}Q=\frac13,\quad\frac{\Delta U_{\mathrm{rot}}}Q=\frac29,\quad\frac{\Delta U_{\mathrm{vib}}}Q=\frac29,\quad\frac WQ=\frac29.$$$$\frac13+\frac29+\frac29+\frac29=1.$$Pressure is still determined by center-of-mass translation; internal modes change the energy required to reach a temperature.

Physical check

Total thermal energy exceeds $3pV/2$ because not all energy is translational.