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

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(A) The work done in 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 ri

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$2PV$

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(A) The work done in 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 with vertices at $A(P, V)$,$C(P, 3V)$,and $B(3P, 3V)$.The base of the triangle is $AC = (3V - V) = 2V$.The height of the triangle is $BC = (3P - P) = 2P$.The area of the triangle $ABC$ is given by:$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$	ext{Area} = \frac{1}{2} 	imes (2V) 	imes (2P) = 2PV$.Since the cycle is traversed in the clockwise direction,the work done by the gas is positive.Therefore,the work done during the cycle is $2PV$.

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

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