In the given $P$-$V$ diagram, a monatomic gas $\left(\gamma = \frac{5}{3}\right)$ is first compressed adiabatically from state $A$ to state $B$. Then it expands isothermally from state $B$ to state $C$. [Given $\left(\frac{1}{3}\right)^{0.6} = 0.5$, $\ln 2 = 0.7$]. Which of the following statement(s) is(are) correct?

Step 1: Identify the state variables for each state.

From the given $P$-$V$ diagram, we identify the coordinates of the states.

$$V_A = 0.80\,\text{m}^3,\quad P_A = 100\,\text{kPa}$$ $$P_B = 300\,\text{kPa}$$ $$V_C = 0.80\,\text{m}^3$$

Step 2: Find the volume at state $B$.

Since the process $A \to B$ is adiabatic, we use the adiabatic relation.

$$P_A V_A^{\gamma} = P_B V_B^{\gamma}$$ $$V_B = V_A \left(\frac{P_A}{P_B}\right)^{1/\gamma}$$

Substitute $\gamma = \frac{5}{3}$ and the given values.

$$V_B = 0.80 \times \left(\frac{100}{300}\right)^{0.6}$$ $$V_B = 0.80 \times \left(\frac{1}{3}\right)^{0.6}$$

Using the approximation $\left(\frac{1}{3}\right)^{0.6} = 0.5$:

$$V_B = 0.80 \times 0.5 = 0.40\,\text{m}^3$$

Step 3: Find the pressure at state $C$.

Since the process $B \to C$ is isothermal, the temperature remains constant.

$$P_B V_B = P_C V_C$$ $$P_C = \frac{P_B V_B}{V_C}$$ $$P_C = \frac{300 \times 0.40}{0.80} = 150\,\text{kPa}$$

Step 4: Calculate the work done during the adiabatic compression $A \to B$.

Use the formula for work done in an adiabatic process.

$$W_{AB} = \frac{P_A V_A - P_B V_B}{\gamma - 1}$$ $$W_{AB} = \frac{(100 \times 0.80) - (300 \times 0.40)}{\frac{5}{3} - 1}$$ $$W_{AB} = \frac{80 - 120}{2/3} = -60\,\text{kJ}$$

The magnitude of this work is:

$$|W_{AB}| = 60\,\text{kJ}$$

Step 5: Calculate the work done during the isothermal expansion $B \to C$.

Use the formula for work done in an isothermal process.

$$W_{BC} = P_B V_B \ln\left(\frac{V_C}{V_B}\right)$$ $$W_{BC} = (300 \times 0.40) \ln\left(\frac{0.80}{0.40}\right)$$ $$W_{BC} = 120 \ln 2$$

Using the approximation $\ln 2 = 0.7$:

$$W_{BC} = 120 \times 0.7 = 84\,\text{kJ}$$

Step 6: Calculate the work done during the process $C \to A$.

Since $V_C = V_A = 0.80\,\text{m}^3$, this is an isochoric process.

$$W_{CA} = 0\,\text{kJ}$$

Step 7: Calculate the total work done in the cycle.

Sum the work done in each individual step.

$$W_{\text{total}} = W_{AB} + W_{BC} + W_{CA}$$ $$W_{\text{total}} = -60 + 84 + 0 = 24\,\text{kJ}$$

Answer: The correct statements are:

• Magnitude of work done during $B \to C$ is $84\,\text{kJ}$
• Magnitude of work done during $A \to B$ is $60\,\text{kJ}$
• Work done during $C \to A$ is $0$