A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is
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a
P0V0
b
2P0V0
c
P0V02
d
Zero
answer is D.
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Detailed Solution
In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise. As is clear from figure, WAEDA=+ area of ∆AED=+12P0V0 WBCEB=-Area of ∆BCD=-12P0V0The network done by the system isWnet =WAEDA+WBCEB=+12P0V0-12P0V0= zero