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An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are `Q_1=5960J`, `Q_2=-5585J`, `Q_3=-2980J` and `Q_4=3645J` respectively. The corresponding quantities of work involved are `W_1=2200J`, `W_2=-825J`, `W_3=-1100J` and `W_4` respectively.
(a) Find the value of `W_4`.
(b) What are the efficiency of the cycle?

Text Solution

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In a cyclic process, `DeltaW=DeltaQ`
`W_1+W_2+W_3+W_4=Q_1+Q_2+Q_3+Q_4`
`2200-825-1100+W_4=5960-5585-2980`
`+3645`
`275+W_4=1040`
`W_4=1040-275=765J`
(b) Efficiency `eta=("work done")/("heat abosorbed")`
`=(DeltaW)/(Q_1+Q_4)=(1040)/(5960+3645)=(1040)/(9605)`
`=0.1083` i.e., `10.83%`
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