Thermodynamics: Diagram-Based Questions (3)

A → B (isochoric, $V$ constant): $W_{AB} = 0$.

B → C (straight line on $P$-$V$): work = area under the curve = trapezoidal area between $V = V_0$ and $V = 2V_0$.

$W_{BC} = \tfrac{1}{2}(2P_0 + P_0)(2V_0 - V_0) = \tfrac{3}{2}P_0 V_0$

C → A (isobaric at $P_0$, $V$ goes from $2V_0$ to $V_0$):

$W_{CA} = P_0(V_0 - 2V_0) = -P_0 V_0$

$W_{net} = 0 + \tfrac{3}{2}P_0 V_0 - P_0 V_0 = \tfrac{1}{2}P_0 V_0$

This matches the enclosed triangular area on the $P$-$V$ diagram (vertices at $A$, $B$, $C$): $\tfrac{1}{2}\cdot V_0 \cdot P_0 = \tfrac{1}{2}P_0V_0$. ✓

Compute temperatures: $T_A = P_0V_0/R = T_0$, $T_B = 2P_0V_0/R = 2T_0$, $T_C = 2P_0V_0/R = 2T_0$.

So heating happens during A → B (and partially in B → C, where $T$ rises and falls). Heat absorbed only counts where $dQ > 0$.

For monoatomic gas: $C_V = \tfrac{3}{2}R$, $C_P = \tfrac{5}{2}R$.

$Q_{AB} = nC_V \Delta T = 1 \cdot \tfrac{3}{2}R \cdot T_0 = \tfrac{3}{2}P_0V_0$

For B → C, careful: $\Delta U = nC_V(T_C - T_B) = 0$ (since $T_C = T_B$). $Q_{BC} = \Delta U + W_{BC} = 0 + \tfrac{3}{2}P_0V_0 = \tfrac{3}{2}P_0V_0$. Heat absorbed.

So total heat absorbed $Q_{in} = \tfrac{3}{2}P_0V_0 + \tfrac{3}{2}P_0V_0 = 3P_0V_0$.

$\eta = \frac{W_{net}}{Q_{in}} = \frac{P_0V_0/2}{3P_0V_0} = \frac{1}{6} \approx 16.7\%$

Final answer: $W_{net} = \tfrac{1}{2}P_0V_0$, $\eta = \tfrac{1}{6}$.