An ideal gas is taken around the cycle ABCA a | vedclass

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(D) The net work done by the gas during a cyclic process is equal to the area enclosed by the cycle in the $P-V$ diagram.In the given diagram,the cycl

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$2{P_1}{V_1}$

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(D) The net work done by the gas during a cyclic process is equal to the area enclosed by the cycle in the $P-V$ diagram.In the given diagram,the cycle $ABCA$ forms a right-angled triangle.The base of the triangle is the segment $AC$,which represents the change in volume: $\Delta V = 3{V_1} - {V_1} = 2{V_1}$.The height of the triangle is the segment $AB$,which represents the change in pressure: $\Delta P = 3{P_1} - {P_1} = 2{P_1}$.The area of the triangle is given by: $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.$	ext{Work done} = \frac{1}{2} 	imes (2{V_1}) 	imes (2{P_1}) = 2{P_1}{V_1}$.Since the cycle is traversed in a clockwise direction,the work done by the gas is positive.

An ideal gas is taken around the cycle $ABCA$ as shown in the $P-V$ diagram. The net work done by the gas during the cycle is equal to

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