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 is the efficiency of the cycle?

Answer

`10.82`

Explanation

Efficiency,
`eta = ("Total W rk done in the cycle ")/("Heatabsorbed (positive heat)by the gas during the cycle ")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 %`