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(B) For path $A 	o B$, the process follows $PV^3 = RT$. Given $T = 300 \, K$:$W_{AB} = \int_{V_A}^{V_B} P \, dV = \int_{2}^{4} \frac{RT}{V^3} \, dV =

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$205$

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(B) For path $A 	o B$, the process follows $PV^3 = RT$. Given $T = 300 \, K$:$W_{AB} = \int_{V_A}^{V_B} P \, dV = \int_{2}^{4} \frac{RT}{V^3} \, dV = RT \int_{2}^{4} V^{-3} \, dV$$W_{AB} = 8 	imes 300 	imes \left[ \frac{V^{-2}}{-2} ight]_{2}^{4} = 2400 	imes \left( -\frac{1}{2} ight) \left( \frac{1}{16} - \frac{1}{4} ight) = -1200 	imes \left( \frac{1-4}{16} ight) = -1200 	imes \left( -\frac{3}{16} ight) = 225 \, J$.For path $B 	o C$, it is an isobaric process at $P = 10 \, Pa$ from $V = 4 \, m^3$ to $V = 2 \, m^3$:$W_{BC} = P(V_C - V_B) = 10(2 - 4) = -20 \, J$.For path $C 	o A$, it is an isochoric process at $V = 2 \, m^3$, so $W_{CA} = 0 \, J$.Total work $W_{net} = W_{AB} + W_{BC} + W_{CA} = 225 - 20 + 0 = 205 \, J$.

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