An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are`Q_1=5960J,Q_2=-5585J,Q_3= and -2980J Q_4 =3645 J,` respectively . The corresponding works involved are `W_1=2200J, W_2=-825 J, W_3= and -1100J W_4` respectively. The value of `W_4` is

Explanation

Since `W_(2)` and `W_(3)` are negative,its means that the work is done on the gas.Hence`Q_(2)`and `Q_(3)`are negative which implies that heat is evolved in processes 2 and 3. Since `Q_(1)` and `Q_(4)` are positive,heat is absorbed by the gas in processes 1 and 4. As `(Q_(1)+Q_(4))` is greater than `(Q_(2)+Q_(3))`, the gas absorbs a net amount of heat energyina complete cycle, whcih is given by
`DeltaQ=Q_(1)+Q_(2)+Q_(3)+Q_(4)`
`=5960-5585-2980+3645=1040` joule
The net worl done by the gas is
`DeltaW=W_(1)+W_(2)+W_(2)+W_(4)`
`=2200-825-1100+W_(4)=(275+W_(4))` joule
Since the process is cyclic, the change in internal energy`DeltaU=0`.From the first law of thermodynamics, we have
`DeltaW=DeltaQ-DeltaU=DeltaQ`
or` 275+W_(4)=1040` or `W_(4)=1040-275=765J`
b. Efficiencty of the cycle is defined as
`eta=("net work done by the gas")/("total heat absorbed by the gas")`
`=(DeltaW)/(Q_(1)+Q_(4))=(275+765)/(5960+3645)=1040/9605=0.1083=10.83%`