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(C) The work done in a cyclic process on a $p-V$ diagram is equal to the area enclosed by the cycle.Since the cycle $12341$ is traversed in a clockwis

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$6$

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(C) The work done in a cyclic process on a $p-V$ diagram is equal to the area enclosed by the cycle.Since the cycle $12341$ is traversed in a clockwise direction,the work done is positive.The area enclosed by the cycle is the area of the trapezoid formed by the vertices $1, 2, 3, 4$.Vertices are: $1(2V_0, 5P_0)$,$2(5V_0, 5P_0)$,$3(6V_0, 3P_0)$,and $4(3V_0, 3P_0)$.The shape is a trapezoid with parallel sides of lengths $(5V_0 - 2V_0) = 3V_0$ and $(6V_0 - 3V_0) = 3V_0$ at constant pressures $5P_0$ and $3P_0$ respectively.Alternatively,the area of the trapezoid is given by $	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.Here,the parallel sides are horizontal segments at $P = 5P_0$ (length $5V_0 - 2V_0 = 3V_0$) and $P = 3P_0$ (length $6V_0 - 3V_0 = 3V_0$).Actually,looking at the figure,the shape is a parallelogram with base $b = (5V_0 - 2V_0) = 3V_0$ and height $h = (5P_0 - 3P_0) = 2P_0$.Work done $W = 	ext{Area} = 	ext{base} 	imes 	ext{height} = (3V_0) 	imes (2P_0) = 6P_0V_0$.

An ideal monoatomic gas is taken round the cycle $12341$ shown in the figure ($p-V$ diagram). The work done in one cycle is (in $P_0V_0$)?

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