The Molar Specific Heats of an Ideal Gas — Question 6

Question 6

A prescribed pressure-volume path has its own process heat capacity

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 quasistatic process obeys $pV^k=\text{constant}$, with constant $k\ne1$, and the gas has constant $C_V=\tfrac32R$. Define $C_{\mathrm{path}}=\delta Q/(n\,dT)$.

1. Derive $C_{\mathrm{path}}$ as a function of $k$.

2. Evaluate it for $k=0$, $k=2$ and $k=4/3$. For $k=4/3$, explain how temperature can increase while heat leaves the gas.

Question 6 - Solution

A prescribed pressure-volume path has its own process heat capacity

(a) Method and calculation

$$T V^{k-1}=\text{constant}\ \Rightarrow\ \frac{dT}T=(1-k)\frac{dV}V.$$$$p\,dV=\frac{nR}{1-k}dT,\quad C_{\mathrm{path}}=C_V+\frac R{1-k}.$$

(b) Interpretation and further calculation

$$C_{\mathrm{path}}(0)=\tfrac52R,\quad C_{\mathrm{path}}(2)=\tfrac12R,\quad C_{\mathrm{path}}(4/3)=-\tfrac32R.$$$$k=4/3:\quad W=-3nR\Delta T,\quad Q=-\tfrac32nR\Delta T,\quad\Delta U=\tfrac32nR\Delta T.$$During compression, positive work on the gas exceeds the heat rejected, leaving a positive internal-energy increase.

Physical check

A negative process heat capacity does not imply a negative equilibrium $C_V$; it reflects simultaneous work transfer.