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

765

Explanation

In a cyclic process `Delta U = 0`
Therefore, `Q_("net") = W_("net")`
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 = 765 J`