Question

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?

Answer 1

Correct Answer - A::B
(a) In a cyclic process, `DeltaU=0`
Therefore, `Q_(n et)=W_(n et)`
or `Q_1+Q_2+Q_3+Q_4=W_1+W_2+W_3+W_4`
Hence, `W_4=(Q_1+Q_2+Q_3+Q_4)-(W_1+W_2+W_3)`
`={(5960-5585-2980+3645)-(2200-825-1100)}`
or `W_4=765J`
(b) Efficiency,
`eta="Total work done in the cyle"/"Heat absorbed (positive heat)"xx100`
`=((W_1+W_2+W_3+W_4)/(Q_1+Q_4))xx100`
`={((2200-825-1100+765))/(5960+3645)}xx100`
=`(1040)/(9605)xx100`
`eta=10.82%`