Heat and Thermodynamics

Concept:The cycle consists of isothermal, isochoric, and adiabatic processes involving a monoatomic ideal gas. Efficiency is computed from heat input and output.Explanation:For process $a \to b$ (isothermal expansion at temperature $T_a = T_b = T_0$): heat absorbed $Q_{ab} = nR T_0 \ln(V_2/V_1)$. For process $b \to c$ (isochoric): $Q_{bc} = \frac{3}{2}nR (T_c - T_0)$. For process $c \to a$ (adiabatic): $Q_{ca} = 0$. Using adiabatic relation $T_c V_2^{\gamma-1} = T_0 V_1^{\gamma-1}$ with $\gamma = 5/3$, so $\gamma-1 = 2/3$. Thus $T_0/T_c = (V_2/V_1)^{2/3}$.If $V_2/V_1 = 8$, then $T_0/T_c = 8^{2/3} = 4$, so $T_a = 4 T_c$. Option C is correct. From the ideal gas law for isothermal process: $P_a V_1 = P_b V_2 \Rightarrow P_a/P_b = V_2/V_1 = 8$, not 4, so option D is false. Heat released in $b \to c$: $|Q_{bc}| = \frac{3}{2}nR (T_0 - T_c) = \frac{3}{2}nR T_0(1-1/4) = \frac{9}{8}nR T_0 \approx 1.125 nR T_0$. Heat absorbed in $a \to b$: $Q_{ab} = nR T_0 \ln 8 = 3 nR T_0 \ln 2 \approx 3\cdot0.7 = 2.1 nR T_0$. Clearly $1.125 < 2.1$, so option A is correct. Efficiency $\eta = 1 - Q_{\text{out}}/Q_{\text{in}}$. Here $Q_{\text{in}} = Q_{ab}$ and $Q_{\text{out}} = |Q_{bc}|$. Substituting expressions, $\eta$ depends only on $V_2/V_1$, not on $T_0$. Hence option B is correct.Answer:The correct statements are A, B, and C.