An ideal gas is taken through a cyclic thermodynamic process though 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 = 3635 J respectively. The corresponding works involved are w_1= 2200 J, W_2= -825 J, W_3= -1100 J, and W_4respectively (i) Find the value of W_4(ii) What is the efficiency of the cycle?

An ideal gas is taken through a cyclic thermodynamic process though fo

1.	An ideal gas is taken through a cyclic thermodynamic process though 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 = 3635 J respectively. The corresponding works involved are w_1= 2200 J, W_2= -825 J, W_3= -1100 J, and W_4respectively (i) Find the value of W_4(ii) What is the efficiency of the cycle?
Answer» Solution :In a cyclic PROCESS `Delta `U = 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_1= W_2 +W_3 +W_4= 2200 -825-1100 + W_4= 275+ W_4` From 1 st law of thermodynamics, `Q = Delta U + W (or) 1040 = 0 + 275+W_4` ` therefore W_4 = 1040 - 275 = 765 J` Efficiency of cycle `=("workdone(W)")/(" Heat absorved"(Q_1+Q_4)) =( 275+ 465)/( 5960 +3645 )=(1040 )/( 9605 )` % efficiency `=( 1040 )/( 9605 )xx 100 = 10.83%`