Question 7

For a monoatomic gas, $\gamma = 5/3$ and $C_v = \frac{3}{2}R$.

Process $ab$ (Isothermal at $T_a$): $Q_{ab} = nRT_a \ln(V_2/V_1)$. Since $V_2 > V_1$, $Q_{ab} > 0$ (Heat absorbed, $Q_{\text{in}}$).

Process $bc$ (Isochoric at $V_2$): $Q_{bc} = nC_v(T_c - T_b) = n\frac{3}{2}R(T_c - T_a)$. Since $T_c < T_a$, $Q_{bc} < 0$ (Heat released, $Q_{\text{out}} = |Q_{bc}|$).

Process $ca$ (Adiabatic): $T_c V_2^{\gamma-1} = T_a V_1^{\gamma-1} \implies T_a/T_c = (V_2/V_1)^{2/3}$.

Checking options:

(C) If $V_2/V_1 = 8$, $T_a/T_c = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4$. Thus $T_a = 4T_c$. (C) is correct.

(B) Efficiency $\eta = 1 - \frac{|Q_{bc}|}{Q_{ab}} = 1 - \frac{\frac{3}{2}nR(T_a - T_c)}{nRT_a \ln(V_2/V_1)} = 1 - \frac{3(1 - T_c/T_a)}{2 \ln(V_2/V_1)} = 1 - \frac{3(1 - (V_1/V_2)^{2/3})}{2 \ln(V_2/V_1)}$.

$\eta$ depends only on the ratio $V_2/V_1$ and is independent of $T_a$. (B) is correct.

(A) $|Q_{bc}| = n\frac{3}{2}R(T_a - T_a/4) = \frac{3}{2} \cdot \frac{3}{4} nRT_a = 1.125 nRT_a$.

$Q_{ab} = nRT_a \ln 8 = 3 nRT_a \ln 2 \approx 3 \times 0.7 nRT_a = 2.1 nRT_a$.

Since $1.125 < 2.1$, heat released in $bc$ is smaller than heat absorbed in $ab$. (A) is correct.

(D) For isothermal process $ab$, $P_a V_1 = P_b V_2 \implies P_a/P_b = V_2/V_1 = 8$. Statement (D) says 4 times, which is incorrect.