Figure shows a cyclic process ABCDBEA performed on an ideal cycle.If P_A =2 atm, P_B5 atm, and P_6 =6atm, V_E - V_A= 20 litre, find the work done by the gas in the complete process (latm, pressure = 1 xx 10^5Pa)

Figure shows a cyclic process ABCDBEA performed on an ideal cycle.If P

1.	Figure shows a cyclic process ABCDBEA performed on an ideal cycle.If P_A =2 atm, P_B5 atm, and P_6 =6atm, V_E - V_A= 20 litre, find the work done by the gas in the complete process (latm, pressure = 1 xx 10^5Pa)
Answer» Solution :Evidently, the entire cycle can be visualised as MADE up of two CYCLES, i.e., cycle ABEA and cycle BCDB. Work DONE during the cycle ABEA, by the gas ` W_1=` area of the loop in the P-V diagram. `=1/2 (" (BASE)")/("(altitude) ")=1/2(V_E- V_A ) (P_B-P_A) =1/2( 20 xx 10^(-3)m^3) (5-2 ) xx 10^5Nm^(-2) = 3KJ` Work done in the cycle BCDB, by the gas `W_2 =`-( area of loop BCDB) Now, evidently, triangles ABE and BCD are similiar, the corresponding angles being equal ` therefore("Areaof " DeltaBCD)/( "Areaof ABE ") =((" lengthof normalfromB ONCD")/(" lengthof normalfromB ofEA"))^2=((P_c -P_B )/(P_B -P_A))^2 = ((6 atm- 5 atm)/( 5 atm -2atm ))^2 =1/9` `therefore` Areaof ` DeltaABCD= 1/9`Areaof `DeltaABE= 1/9 xx 3 KJ= 0.33KJthereforeW_2= 0.33KJ` ` therefore` Totalworkdoneby thegas ` DeltaW= W_1 +W_2= 3KJ- 0.33 KJ= 2.67KJ .`