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One mole of an ideal gas undergoes a cyc...

One mole of an ideal gas undergoes a cyclic change ABCDA as shown in (figure). What is the net work done (in joule) in the process? Take `1 atm = 10^(5)Pa`.

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The correct Answer is:
4
Work done= area ABCDA `= ABxxBC`
`=[(0.3-0.1)xx10^(-3)]xx(0.4-0.2)10^(5)= 4J`
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