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(A) For a cyclic process, the change in internal energy is zero, so the net heat supplied is equal to the net work done: $\Delta Q = \Delta W$.The net

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(A) For a cyclic process, the change in internal energy is zero, so the net heat supplied is equal to the net work done: $\Delta Q = \Delta W$.The net work done in one cycle is equal to the area enclosed by the $P-V$ diagram.The area of the triangle $ABC$ is given by:$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$	ext{Base} = V_B - V_A = 20 \,m^3 - 5 \,m^3 = 15 \,m^3$$	ext{Height} = P_B - P_A = 30 \,N/m^2 - 10 \,N/m^2 = 20 \,N/m^2$$	ext{Area} = \frac{1}{2} 	imes 15 \,m^3 	imes 20 \,N/m^2 = 150 \,J$.Since the cycle is $A ightarrow B ightarrow C ightarrow A$ (counter-clockwise), the work done by the gas is negative, meaning work is done on the gas. Thus, heat is released.For $20$ cycles, the net heat released is:$\Delta Q = 20 	imes 150 \,J = 3000 \,J = 3 \,kJ$.

$A$ gas in a closed container undergoes the cycle $ABCA$ as shown in the figure. The net heat released by the gas after it has undergone $20$ cycles is (in $\,kJ$)

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