The Molar Specific Heats of an Ideal Gas — Question 5

Question 5

A temperature-dependent heat capacity must be integrated

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. For one mole, $C_V(T)=R[\tfrac32+T/(600\,\mathrm K)]$ between 300 K and 600 K. The gas is heated through that interval.

1. Find $\Delta U$ and the fixed-volume heat. Compare with using the initial heat capacity as a constant.

2. Find fixed-pressure heat and work over the same temperature interval, and identify the single capacity value that gives the exact internal-energy change here.

Question 5 - Solution

A temperature-dependent heat capacity must be integrated

(a) Method and calculation

$$\Delta U=R\int_{300}^{600}\left(\frac32+\frac T{600\,\mathrm K}\right)dT=R\left[\frac32(300)+\frac{600^2-300^2}{1200}\right]\mathrm K.$$$$Q_V=\Delta U=675R\,\mathrm K=5611.95\,\mathrm J.$$$$Q_{V,\mathrm{initial}}=C_V(300)(300)=600R\,\mathrm K=4988.40\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_p=R(600-300)=2494.20\,\mathrm J,\quad Q_p=\Delta U+W_p=8106.15\,\mathrm J.$$$$\overline C_V=\Delta U/\Delta T=2.25R=C_V(450\,\mathrm K).$$The midpoint value is exact because the supplied capacity is linear in temperature; that shortcut need not hold for a nonlinear capacity.

Physical check

At fixed volume, heat equals internal-energy change, but it still need not equal an endpoint capacity times the temperature interval.