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 ?

An ideal gas is taken through a cyclic thermodynamic process through f

1.	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 ?
Answer» Solution :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%`