The Adiabatic Expansion of an Ideal Gas — Question 10

Question 10

An adiabatic leg closes a three-process energy cycle

Use $pV=nRT$, $R=8.314$ J/(mol K), and $\Delta U=Q-W$ with $W$ positive for work done by the gas. A .100 mol monatomic ideal gas starts at state A with temperature 600 K and volume $V_0$. It expands reversibly and adiabatically to state B at $2V_0$, then is compressed at constant $p_B$ to state C at $V_0$, then heated at fixed volume back to A. Use $C_V=3R/2$ and $C_p=5R/2$.

1. Find temperatures at B and C, and work for each leg.

2. Find heat for each leg and verify that total heat equals net work over the closed cycle. State the sign of net work.

Question 10 - Solution

An adiabatic leg closes a three-process energy cycle

(a) Method and calculation

$$T_B=600(1/2)^{2/3}\approx377.976\,\mathrm K,\quad T_C=T_B(V_C/V_B)\approx188.988\,\mathrm K.$$$$W_{AB}=nC_V(T_A-T_B)\approx276.886\,\mathrm J,\quad W_{BC}=nR(T_C-T_B)\approx-157.125\,\mathrm J.$$$$W_{CA}=0,\quad W_{\mathrm{net}}\approx119.761\,\mathrm J.$$

(b) Interpretation and further calculation

$$Q_{AB}=0,\quad Q_{BC}=nC_p(T_C-T_B)\approx-392.812\,\mathrm J.$$$$Q_{CA}=nC_V(T_A-T_C)\approx512.573\,\mathrm J.$$$$Q_{\mathrm{net}}=512.573-392.812\approx119.761\,\mathrm J=W_{\mathrm{net}},\quad\Delta U_{\mathrm{cycle}}=0.$$The clockwise cycle produces positive net work; heat entering on the isochore exceeds the heat rejected on the isobar.

Physical check

Work and heat depend on the route, but returning to the initial state forces the net internal-energy change to vanish.