An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are `Q_(1) = 5960 J, Q_(2) = -5585 J, Q_(3) = -2980 J and Q_(A)=3635J` respectively. The corresponding works involved are `w_(1) = 2200 J, W_(2) = -825 J, W_(3) =-1100 J, and W_(4)` respectively (i) Find the value of W, (ii) What is the efficiency of the cycle ?

Explanation

In a cyclic process `dU=0`
Net heat absorbed by the system,
`Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)`
`=5960-5585-2980+3645=1040J`
Net work done `W=W_(1)+W_(2)+W_(3)+W_(4)`
`=2200-825-1100+W_(4)=275+W_(4)`
From 1st law of thermodynamics,
`Q=dU+W or 1040=0+275 +W_(4)`
`:.W_(4)=1040 -275 =765J`
Efficiency of cycle `=("Work done (W)")/("Heat absorved"(Q_(1)+Q_(4)))`
`=(275+765)/(5960+3645)=(1040)/(9605)`
% efficiency `=(1040)/(9605)xx100=10.83%`