Ten moles of an ideal monoatomic gas, initially in state $a$ at atmospheric pressure and temperature $T_a=27^\circ\text{C}$, is enclosed in a metal cylinder of volume $V_0$ fitted with a frictionless piston. The gas is suddenly compressed to state $b$ with volume $V_0/3$. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature $11^\circ\text{C}$ until the gas reaches the temperature of the water bath, which is denoted as state $c$. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state $f$. If $R$ is universal gas constant, then the correct option(s) is/are: [Given: $9^{1/3}=2.08$]

Step 1: Identify the parameters of the gas and the states.

For a monoatomic ideal gas, the heat capacity at constant volume $C_V$ and the adiabatic index $\gamma$ are:

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

The initial state $a$ parameters are:

$$n = 10\text{ mol}$$ $$T_a = 27^\circ\text{C} = 300\text{ K}$$ $$V_a = V_0$$

Step 2: Analyze the adiabatic compression from state $a$ to state $b$.

Since the gas is suddenly compressed, the process $a \to b$ is adiabatic. The volume at state $b$ is $V_b = \frac{V_0}{3}$.

Find the temperature $T_b$ using the adiabatic relation:

$$T_b = T_a \left(\frac{V_a}{V_b}\right)^{\gamma - 1}$$ $$T_b = 300 \times 3^{2/3}$$

Using the given value $9^{1/3} = 3^{2/3} = 2.08$:

$$T_b = 300 \times 2.08 = 624\text{ K}$$

Step 3: Calculate the change in internal energy for the process $a \to b$.

$$\Delta U_{a\to b} = n C_V (T_b - T_a)$$ $$\Delta U_{a\to b} = 10 \times \frac{3}{2}R \times (624 - 300)$$ $$\Delta U_{a\to b} = 15R \times 324 = 4860R$$

Thus, option (2) is correct.

Step 4: Analyze the processes $b \to c$ and $c \to f$.

In process $b \to c$, the piston is stationary (isochoric process) and the gas cools to the water bath temperature:

$$T_c = 11^\circ\text{C} = 284\text{ K}$$

In process $c \to f$, the gas undergoes isothermal expansion back to the initial volume:

$$T_f = T_c = 284\text{ K}$$ $$V_f = V_0$$

Step 5: Calculate the net change in internal energy for the entire process.

Since internal energy is a state function, the net change depends only on the initial and final temperatures:

$$\Delta U_{\text{net}} = n C_V (T_f - T_a)$$ $$\Delta U_{\text{net}} = 10 \times \frac{3}{2}R \times (284 - 300)$$ $$\Delta U_{\text{net}} = 15R \times (-16) = -240R$$

Thus, option (3) is correct.

Step 6: Calculate the pressure at state $b$.

Using the adiabatic relation for pressure:

$$P_b = P_a \left(\frac{V_a}{V_b}\right)^{\gamma}$$ $$P_b = 1\text{ atm} \times 3^{5/3}$$ $$P_b = 3 \times 3^{2/3} = 3 \times 2.08 = 6.24\text{ atm}$$

Since $P_b = 6.24\text{ atm} \neq 2.08\text{ atm}$, option (4) is incorrect.

Step 7: Verify the $P-V$ diagram.

The processes are:

• $a \to b$: Adiabatic compression (curve rising to the left)
• $b \to c$: Isochoric cooling (vertical line downwards)
• $c \to f$: Isothermal expansion (curve going to the right)

This sequence matches the described $P-V$ diagram, so option (1) is correct.

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