In the given P-V diagram,a monoatomic gas lef | vedclass

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(B) For the adiabatic process $(A ightarrow B)$:$P_A V_A^\gamma = P_B V_B^\gamma$$100 	imes (0.8)^{5/3} = 300 	imes (V_B)^{5/3}$$(V_B)^{5/3} = \fr

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(B) For the adiabatic process $(A ightarrow B)$:$P_A V_A^\gamma = P_B V_B^\gamma$$100 	imes (0.8)^{5/3} = 300 	imes (V_B)^{5/3}$$(V_B)^{5/3} = \frac{1}{3} 	imes (0.8)^{5/3}$$V_B = 0.8 	imes \left(\frac{1}{3}ight)^{3/5} = 0.8 	imes \left(\frac{1}{3}ight)^{0.6} \simeq 0.8 	imes 0.5 = 0.4 	ext{ m}^3$.Work done in process $A ightarrow B$:$W_{AB} = \frac{P_A V_A - P_B V_B}{\gamma - 1} = \frac{100 	imes 10^3 	imes 0.8 - 300 	imes 10^3 	imes 0.4}{5/3 - 1}$$W_{AB} = \frac{80000 - 120000}{2/3} = -40000 	imes \frac{3}{2} = -60000 	ext{ J} = -60 	ext{ kJ}$.Magnitude $|W_{AB}| = 60 	ext{ kJ}$. (Statement $C$ is correct).Work done in process $B ightarrow C$ (Isothermal process):$W_{BC} = nRT \ln\left(\frac{V_C}{V_B}ight) = P_B V_B \ln\left(\frac{V_C}{V_B}ight)$Since $V_C = V_A = 0.8 	ext{ m}^3$:$W_{BC} = (300 	imes 10^3) 	imes 0.4 	imes \ln\left(\frac{0.8}{0.4}ight) = 120000 	imes \ln(2) \simeq 120000 	imes 0.7 = 84000 	ext{ J} = 84 	ext{ kJ}$. (Statement $B$ is correct).Work done in process $C ightarrow A$ (Isochoric process):Since $\Delta V = 0$,$W_{CA} = 0$. (Statement $D$ is correct).Total work done $W_{ABC} = W_{AB} + W_{BC} + W_{CA} = -60 + 84 + 0 = 24 	ext{ kJ}$.Thus,statements $(B, C, D)$ are correct.

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