An ideal gas undergoes a cyclic process ABCDA | vedclass

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(B) The work done by the gas in a cyclic process is equal to the area enclosed by the $p-V$ loop.In the given $p-V$ diagram, the process $ABCDA$ forms

Vedclass · Accepted Answer

$4P_0V_0$

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(B) The work done by the gas in a cyclic process is equal to the area enclosed by the $p-V$ loop.In the given $p-V$ diagram, the process $ABCDA$ forms a rectangle.The area of the rectangle is given by $	ext{Area} = 	ext{width} 	imes 	ext{height}$.The width of the rectangle along the $V$-axis is $(5V_0 - 3V_0) = 2V_0$.The height of the rectangle along the $p$-axis is $(4P_0 - 2P_0) = 2P_0$.Therefore, the magnitude of the work done is $	ext{Area} = (2V_0) 	imes (2P_0) = 4P_0V_0$.Since the cycle is traversed in a clockwise direction, the work done by the gas is positive.

An ideal gas undergoes a cyclic process $ABCDA$ as shown in the given $p-V$ diagram. What is the magnitude of the amount of work done by the gas?

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