Thermodynamic Cycle with Semi Circle and Ellipse Paths

To determine the correct statements, we analyze each option step-by-step using the principles of thermodynamics and the given $P$-$V$ diagram.

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1. Analysis of Path $\mathrm{A} \rightarrow \mathrm{B}$ (Option A)

• From the given diagram, the coordinates in the $P$-$V$ plane are:
  • State $\mathrm{A}$: $P_{\mathrm{A}} = 3$, $V_{\mathrm{A}} = 2 \implies P_{\mathrm{A}}V_{\mathrm{A}} = 6$
  • State $\mathrm{B}$: $P_{\mathrm{B}} = 2$, $V_{\mathrm{B}} = 3 \implies P_{\mathrm{B}}V_{\mathrm{B}} = 6$
• An isothermal process for an ideal gas satisfies Boyle's Law, $PV = \text{constant}$. If the process were isothermal, the path between $\mathrm{A}$ and $\mathrm{B}$ would be a rectangular hyperbola:
  $P = \frac{6}{V}$
• However, the path $\mathrm{AB}$ is an arc of a circle centered at $(2, 2)$ with radius $R = 1$:
  $(V - 2)^2 + (P - 2)^2 = 1$
• Taking an intermediate state on this arc, for instance where $V = 2 + \frac{1}{\sqrt{2}} \approx 2.707$ and $P = 2 + \frac{1}{\sqrt{2}} \approx 2.707$:
  $PV \approx (2.707)^2 \approx 7.33 \neq 6$
  Since the product $PV$ (and hence the temperature $T$) is not constant along the path, the process is not isothermal.

Therefore, Option A is incorrect.

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2. Analysis of Path $\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}$ (Option B)

• According to the First Law of Thermodynamics:
  $Q = \Delta U + W$

• Work Done ($W$):
  Along the entire path $\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}$, the volume continuously decreases from $V_{\mathrm{B}} = 3$ to $V_{\mathrm{C}} = 2$, and further to $V_{\mathrm{D}} < 1$.
  Since $dV < 0$ throughout this path:
  $W_{\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}} = \int_{\mathrm{B}}^{\mathrm{D}} P \, dV < 0$

• Change in Internal Energy ($\Delta U$):
  The internal energy of an ideal gas depends solely on temperature:
  $\Delta U = n C_v (T_{\mathrm{D}} - T_{\mathrm{B}}) = \frac{C_v}{R} (P_{\mathrm{D}}V_{\mathrm{D}} - P_{\mathrm{B}}V_{\mathrm{B}})$
  From the diagram:
  $P_{\mathrm{B}}V_{\mathrm{B}} = 2 \times 3 = 6$
  $P_{\mathrm{D}}V_{\mathrm{D}} = 2 \times V_{\mathrm{D}} < 2 \times 1 = 2 \quad (\text{since } V_{\mathrm{D}} < 1)$
  Therefore:
  $P_{\mathrm{D}}V_{\mathrm{D}} < P_{\mathrm{B}}V_{\mathrm{B}} \implies T_{\mathrm{D}} < T_{\mathrm{B}} \implies \Delta U_{\mathrm{B} \rightarrow \mathrm{D}} < 0$

• Heat Transfer ($Q$):
  Since both $\Delta U < 0$ and $W < 0$:
  $Q = \Delta U + W < 0$
  A negative value of $Q$ implies that heat flows out of the gas.

Therefore, Option B is correct.

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3. Analysis of Work Done during $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$ (Option C)

• The work done along path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$ is given by the integral:
  $W_{\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}} = \int_{\mathrm{A}}^{\mathrm{B}} P \, dV + \int_{\mathrm{B}}^{\mathrm{C}} P \, dV$
• This equals the area under the curve $\mathrm{A} \rightarrow \mathrm{B}$ minus the area under the curve $\mathrm{C} \rightarrow \mathrm{B}$, which is geometrically the area enclosed by the semi-circle $\mathrm{ABC}$:
  $W_{\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}} = \text{Area of semicircle } \mathrm{ABC} = \frac{\pi R^2}{2} = \frac{\pi (1)^2}{2} = \frac{\pi}{2} > 0$
  Hence, the work done is positive and non-zero.

Therefore, Option C is incorrect.

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4. Analysis of Work Done in the Cycle $\mathrm{ABCDA}$ (Option D)

• In a $P$-$V$ indicator diagram, the net work done in a cyclic process is equal to the area enclosed by the cycle:
  • If the cycle is traced in a clockwise direction, the work done by the gas is positive ($W_{\text{net}} > 0$).
  • If the cycle is traced in an anticlockwise direction, the work done by the gas is negative ($W_{\text{net}} < 0$).
• Here, the cycle $\mathrm{ABCDA}$ is traversed in a clockwise direction. Thus, positive work is done by the gas during the complete cycle:
  $W_{\text{net}} = \text{Area enclosed by } \mathrm{ABCDA} > 0$

Therefore, Option D is correct.

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Conclusion:

The correct statements are B and D.