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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta W$. The work do

Vedclass · Accepted Answer

$-1.5 	ext{ kJ}$

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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta W$. The work done $\Delta W$ in one cycle is equal to the area enclosed by the $P-V$ diagram. The area of the triangle $ABC$ is $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Base $= (20 - 5) 	ext{ m}^3 = 15 	ext{ m}^3$. Height $= (30 - 10) 	ext{ N/m}^2 = 20 	ext{ N/m}^2$. Area $= \frac{1}{2} 	imes 15 	imes 20 = 150 	ext{ J}$. Since the cycle $ABCA$ is anti-clockwise,the work done by the gas is negative. Thus,$\Delta W_{	ext{cycle}} = -150 	ext{ J}$. For $10$ cycles,the total work done is $\Delta W_{	ext{total}} = 10 	imes (-150 	ext{ J}) = -1500 	ext{ J} = -1.5 	ext{ kJ}$. Therefore,the net heat absorbed is $\Delta Q = -1.5 	ext{ kJ}$.

$A$ gas in a closed container undergoes the cycle $ABCA$ as shown in the figure. The net heat absorbed by the gas after it has completed $10$ cycles is

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