The Molar Specific Heats of an Ideal Gas — Question 1

Question 1

The extra heat at fixed pressure pays for expansion work

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. An ideal gas has molar heat capacity $C_V$ independent of temperature. Compare reversible heating of $n$ moles through the same temperature increment $dT$ at fixed volume and fixed pressure.

1. Derive $C_p-C_V=R$ from the first law rather than assuming it.

2. For .400 mol with $C_V=\tfrac52R$ heated by 60.0 K at fixed pressure, find heat input, internal-energy increase, work and the fraction of heat converted to work.

Question 1 - Solution

The extra heat at fixed pressure pays for expansion work

(a) Method and calculation

$$\delta Q_V=dU=nC_VdT.$$$$p\,dV=nR\,dT\quad(p\text{ fixed}),\quad\delta Q_p=dU+p\,dV=n(C_V+R)dT.$$$$C_p=C_V+R.$$

(b) Interpretation and further calculation

$$\Delta U=.400(\tfrac52)(8.314)(60)=498.84\,\mathrm J,\quad W=.400(8.314)(60)=199.536\,\mathrm J.$$$$Q=\Delta U+W=698.376\,\mathrm J,\quad W/Q=R/C_p=2/7.$$Heat capacity depends on the heating constraint even though $\Delta U$ depends only on the ideal-gas temperature change.

Physical check

For a positive temperature rise, both energy storage and expansion require positive heat.