A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (ab), followed by an isochoric process (bc) and an adiabatic process (ca) as shown in the figure. The volumes of the gas are $V_1$ and $V_2$ at $a$ and $b$, respectively. If the cycle has heat input $Q_{\text{in}}$ and output $Q_{\text{out}}$, then the efficiency of the cycle is defined as $\eta = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}}$. The correct statement(s) is/are: [Given: $\ln 2 \approx 0.7$]

Step 1: Identify the properties of the gas and the processes.

For a monoatomic ideal gas, the ratio of specific heats $\gamma$ and the molar heat capacity at constant volume $C_v$ are:

$$\gamma = \frac{5}{3}$$ $$C_v = \frac{3}{2}R$$

The processes in the cycle are:

• $a \to b$: Isothermal expansion at temperature $T$ from volume $V_1$ to $V_2$.
• $b \to c$: Isochoric cooling at volume $V_2$ from temperature $T$ to $T_c$.
• $c \to a$: Adiabatic compression from volume $V_2$ to $V_1$.

Step 2: Analyze the heat exchange in each process.

For the isothermal process $a \to b$, the heat absorbed is:

$$Q_{ab} = nRT\ln\left(\frac{V_2}{V_1}\right)$$

Since $V_2 > V_1$, $Q_{ab} > 0$, which represents the heat input $Q_{\text{in}}$.

For the isochoric process $b \to c$, the heat released is:

$$Q_{bc} = nC_v(T - T_c)$$

This represents the heat output $Q_{\text{out}}$.

For the adiabatic process $c \to a$:

$$Q_{ca} = 0$$

Step 3: Relate the temperatures using the adiabatic process.

For the adiabatic process $c \to a$:

$$T_c V_2^{\gamma-1} = T V_1^{\gamma-1}$$

Substitute $\gamma = \frac{5}{3}$:

$$T_c = T\left(\frac{V_1}{V_2}\right)^{2/3}$$

Step 4: Evaluate the statements for $V_2/V_1 = 8$.

Let $r = \frac{V_2}{V_1} = 8$.

Calculate the heat input $Q_{\text{in}} = Q_{ab}$:

$$Q_{ab} = nRT\ln 8 = 3nRT\ln 2$$

Using $\ln 2 \approx 0.7$:

$$Q_{ab} \approx 2.10\,nRT$$

Calculate the temperature $T_c$:

$$T_c = T\left(\frac{1}{8}\right)^{2/3} = \frac{T}{4}$$

This gives:

$$T_a = 4T_c$$

Thus, Statement (3) is correct.

Calculate the heat output $Q_{\text{out}} = Q_{bc}$:

$$Q_{bc} = \frac{3}{2}nR\left(T - \frac{T}{4}\right) = \frac{9}{8}nRT \approx 1.125\,nRT$$

Comparing the heat values:

$$Q_{bc} < Q_{ab}$$

Thus, Statement (1) is correct.

Step 5: Analyze the efficiency of the cycle.

The efficiency $\eta$ is given by:

$$\eta = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{Q_{bc}}{Q_{ab}}$$

Substitute the expressions for $Q_{bc}$ and $Q_{ab}$:

$$\eta = 1 - \frac{\frac{3}{2}nRT\left(1 - r^{-2/3}\right)}{nRT\ln r}$$ $$\eta = 1 - \frac{\frac{3}{2}\left(1 - r^{-2/3}\right)}{\ln r}$$

Since $\eta$ depends only on the volume ratio $r$ and is independent of the temperature $T$, Statement (2) is correct.

Step 6: Analyze the pressure ratio.

For the isothermal process $a \to b$:

$$P_a V_1 = P_b V_2$$ $$\frac{P_a}{P_b} = \frac{V_2}{V_1} = 8$$

Thus, Statement (4) is incorrect.

Answer: The correct statements are (1), (2), and (3).