One mole of a monoatomic gas is carried along | vedclass

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(D) The net work done by the gas in a cyclic process is equal to the area enclosed by the $P-V$ curve.The cycle consists of two parts: the upper loop

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(D) The net work done by the gas in a cyclic process is equal to the area enclosed by the $P-V$ curve.The cycle consists of two parts: the upper loop $BCE$ and the lower loop $EDAB$.For the upper loop $BCE$ (clockwise,positive work):Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (3-1) 	imes (4-1) = \frac{1}{2} 	imes 2 	imes 3 = 3 \ J$.For the lower loop $EDAB$ (counter-clockwise,negative work):Area $= 	ext{length} 	imes 	ext{width} = (2-1) 	imes (2-1) = 1 \ J$.However,looking at the diagram,the cycle is $ABCDEA$. The area enclosed by the path $ABCDEA$ is the sum of the areas of the two loops.Alternatively,calculate work for each segment:$W_{BC} = P \Delta V = 4 	imes (2-1) = 4 \ J$$W_{CD} = 0$ (isochoric)$W_{DE} = P \Delta V = 1 	imes (3-2) = 1 \ J$$W_{EA} = 	ext{Area under line } EA = \frac{1}{2} 	imes (P_E + P_A) 	imes (V_A - V_E) = \frac{1}{2} 	imes (1+4) 	imes (1-3) = -5 \ J$$W_{AB} = 0$ (isochoric)Net work $W = 4 + 0 + 1 - 5 + 0 = 0 \ J$.

One mole of a monoatomic gas is carried along the process $ABCDEA$ as shown in the diagram. Find the net work done by the gas in $J$.

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