Work Done in Thermodynamic Processes in PV Diagram

To determine the correct statements, we analyze each thermodynamic process shown in the $P-V$ diagram.

1. State Parameters:

From the given $P-V$ diagram:

• State $A$: $P_A = 100\text{ kPa} = 10^5\text{ Pa}$ $V_A = 0.80\text{ m}^3$

• State $B$: $P_B = 300\text{ kPa} = 3 \times 10^5\text{ Pa}$

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2. Process $A \rightarrow B$ (Adiabatic Compression):

Since the process $A \rightarrow B$ is adiabatic ($\gamma = 5/3$), the equation of state is: $P_A V_A^\gamma = P_B V_B^\gamma$

Substituting the given values: $100 \times (0.80)^{5/3} = 300 \times V_B^{5/3}$

$V_B^{5/3} = (0.80)^{5/3} \times \left(\frac{100}{300}\right) = (0.80)^{5/3} \times \frac{1}{3}$

$V_B = 0.80 \times \left(\frac{1}{3}\right)^{3/5} = 0.80 \times \left(\frac{1}{3}\right)^{0.6}$

Given that $\left(\frac{1}{3}\right)^{0.6} \simeq 0.5$: $V_B = 0.80 \times 0.5 = 0.40\text{ m}^3$

The work done in an adiabatic process is given by: $W_{A \rightarrow B} = \frac{P_A V_A - P_B V_B}{\gamma - 1}$

Substituting the values: $P_A V_A = 100\text{ kPa} \times 0.80\text{ m}^3 = 80\text{ kJ}$ $P_B V_B = 300\text{ kPa} \times 0.40\text{ m}^3 = 120\text{ kJ}$ $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$

$W_{A \rightarrow B} = \frac{80\text{ kJ} - 120\text{ kJ}}{2/3} = \frac{-40}{2/3} = -60\text{ kJ}$

The magnitude of the work done in process $A \rightarrow B$ is: $|W_{A \rightarrow B}| = 60\text{ kJ}$

Thus, Option C is correct.

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3. Process $B \rightarrow C$ (Isothermal Expansion):

From the $P-V$ diagram, State $C$ has the same volume as State $A$, so: $V_C = V_A = 0.80\text{ m}^3$

The work done during the isothermal expansion $B \rightarrow C$ at temperature $T_B$ is: $W_{B \rightarrow C} = n R T_B \ln\left(\frac{V_C}{V_B}\right) = P_B V_B \ln\left(\frac{V_C}{V_B}\right)$

Substituting $P_B V_B = 120\text{ kJ}$, $V_C = 0.80\text{ m}^3$, $V_B = 0.40\text{ m}^3$, and $\ln 2 \simeq 0.7$: $W_{B \rightarrow C} = 120\text{ kJ} \times \ln\left(\frac{0.80}{0.40}\right) = 120 \times \ln(2) \approx 120 \times 0.7 = 84\text{ kJ}$

The magnitude of the work done in process $B \rightarrow C$ is: $|W_{B \rightarrow C}| = 84\text{ kJ}$

Thus, Option B is correct.

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4. Total Work Done in $A \rightarrow B \rightarrow C$:

$W_{A \rightarrow B \rightarrow C} = W_{A \rightarrow B} + W_{B \rightarrow C} = -60\text{ kJ} + 84\text{ kJ} = +24\text{ kJ}$

The magnitude of the total work done is $|W_{A \rightarrow B \rightarrow C}| = 24\text{ kJ} \neq 144\text{ kJ}$.

Thus, Option A is incorrect.

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5. Process $C \rightarrow A$ (Isochoric Process):

From the diagram, the path $C \rightarrow A$ is a vertical line on the $P-V$ diagram, meaning $V_C = V_A = 0.80\text{ m}^3$ (volume is constant).

$W_{C \rightarrow A} = \int_{V_C}^{V_A} P \, dV = 0\text{ kJ}$

The magnitude of the work done in process $C \rightarrow A$ is zero.

Thus, Option D is correct.

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Conclusion:

The correct statements are B, C, and D.