An ideal gas undergoes a cyclic transformatio | vedclass

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(C) The work done in a cyclic process is equal to the area enclosed by the cycle in the $P-V$ diagram.For a $V$ vs $P$ graph,the area enclosed is give

Vedclass · Accepted Answer

$200$

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(C) The work done in a cyclic process is equal to the area enclosed by the cycle in the $P-V$ diagram.For a $V$ vs $P$ graph,the area enclosed is given by the formula for the area of a triangle: $Area = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,the base along the $P$-axis is $(30 - 10) \ kPa = 20 \ kPa = 20 	imes 10^3 \ Pa$.The height along the $V$-axis is $(30 - 10) \ dm^3 = 20 \ dm^3 = 20 	imes 10^{-3} \ m^3$.Since the cycle $A$ $ightarrow B$ $ightarrow C$ $ightarrow A$ is clockwise in a $V$ vs $P$ graph,the work done is positive.$W = \frac{1}{2} 	imes (20 	imes 10^3 \ Pa) 	imes (20 	imes 10^{-3} \ m^3) = \frac{1}{2} 	imes 20 	imes 20 \ J = 200 \ J$.

An ideal gas undergoes a cyclic transformation starting from the point $A$ and coming back to the same point by tracing the path $A$ $\rightarrow B$ $\rightarrow C$ $\rightarrow A$ as shown in the diagram. The total work done in the process is . . . . . . $J$.

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