Some Special Cases of the First Law of Thermodynamics — Question 6

Question 6

A rectangular cycle partitions heat among four special processes

Use $\Delta U=Q-W$, with work by the gas positive. A supplied gas model has $U=\tfrac32pV+U_0$. It follows A $(1\,\mathrm L,100\,\mathrm{kPa})$ to D $(1\,\mathrm L,300\,\mathrm{kPa})$ to C $(3\,\mathrm L,300\,\mathrm{kPa})$ to B $(3\,\mathrm L,100\,\mathrm{kPa})$ to A. All legs are quasistatic; only pressure-volume work occurs. Use $1\,\mathrm{kPa\,L}=1\,\mathrm J$.

1. Calculate $Q$ on each leg from its work and internal-energy change.

2. Find total inward heat, total outward heat and net work. Verify energy closure over the complete cycle.

Question 6 - Solution

A rectangular cycle partitions heat among four special processes

(a) Method and calculation

$$Q_{AD}=\tfrac32(300-100)+0=300\,\mathrm J.$$$$Q_{DC}=\tfrac32(900-300)+300(3-1)=1500\,\mathrm J.$$$$Q_{CB}=\tfrac32(300-900)=-900\,\mathrm J,\quad Q_{BA}=\tfrac32(100-300)+100(1-3)=-500\,\mathrm J.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{in}}=1800\,\mathrm J,\quad Q_{\mathrm{out}}=1400\,\mathrm J.$$$$W_{\mathrm{net}}=600-200=400\,\mathrm J=Q_{\mathrm{net}}.$$$$\Delta U_{\mathrm{cycle}}=300+900-900-300=0.$$Vertical legs contribute heat even though they contribute no displacement work.

Physical check

The clockwise area gives positive net work; the individual heat inputs need not equal the individual work outputs.