The Molar Specific Heats of an Ideal Gas — Question 7

Question 7

Subtracting the vessel heat capacity changes the inferred gas result

Use $pV=nRT$, $R=8.314$ J/(mol K), and the sign convention $\Delta U=Q-W$, with $W$ the work done by the gas. Only pressure–volume work occurs. A rigid calorimeter contains .250 mol of ideal gas. Its empty vessel has heat capacity 12.0 J/K. A 20.0 W heater operates for 30.0 s; 15.0% of its energy leaks to the room. Gas and vessel warm together by 30.0 K. No other energy storage occurs.

1. Infer the gas molar $C_V$.

2. Find the apparent capacity obtained if both leakage and vessel heating were ignored. Quantify the error relative to the inferred gas value.

Question 7 - Solution

Subtracting the vessel heat capacity changes the inferred gas result

(a) Method and calculation

$$E_{\mathrm{heater}}=Pt=600\,\mathrm J,\quad Q_{\mathrm{retained}}=.850(600)=510\,\mathrm J.$$$$Q_{\mathrm{gas}}=510-(12.0)(30.0)=150\,\mathrm J,\quad C_V=\frac{150}{(.250)(30.0)}=20.0\,\mathrm{J/(mol\,K)}.$$

(b) Interpretation and further calculation

$$C_{V,\mathrm{apparent}}=\frac{600}{(.250)(30.0)}=80.0\,\mathrm{J/(mol\,K)}.$$$$\frac{C_{V,\mathrm{apparent}}-C_V}{C_V}=3.00=300\%.$$The measured temperature rise belongs to both gas and apparatus, while the source energy includes heat that never remains in either.

Physical check

The retained energy must exceed the vessel share to infer a positive gas heat capacity.