One mole of an ideal monoatomic gas is taken | vedclass

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(D) The work done by the gas in a cyclic process is equal to the area enclosed by the cycle on the $p-V$ diagram.Since the cycle $MNOM$ is traversed i

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(D) The work done by the gas in a cyclic process is equal to the area enclosed by the cycle on the $p-V$ diagram.Since the cycle $MNOM$ is traversed in a clockwise direction, the work done is positive.The area of the triangle $MNO$ is given by:$W = 	ext{Area of } 	riangle MNO = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$W = \frac{1}{2} 	imes (ON) 	imes (OM)$From the graph, the base $ON = 3V_0 - V_0 = 2V_0$ and the height $OM = 3p_0 - p_0 = 2p_0$.Substituting these values:$W = \frac{1}{2} 	imes (2V_0) 	imes (2p_0)$$W = 2p_0 V_0$Thus, the work done by the gas is $2p_0 V_0$.

One mole of an ideal monoatomic gas is taken round the cyclic process $MNOM$. The work done by the gas is (in $p_0 V_0$)

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