Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas)

Step-by-step solution

Work done in complete cycle = Area under P-V graph = P ₀ V ₀ from A to B , heat given to the gas = nC _ v T = n 3 2 R T = 3 2 ~V ₀ P = 3 2 P ₀ V ₀ from B to C , heat given to the system =n C_ p T = n ( 5 2 R ) T = 5 2 (2 P ₀ ) V =5 P ₀ V ₀ from C to D and D to A , heat is rejected. efficiency, = work done by gas heat given to the gas 100 = P₀ V₀ 3 2 P₀ V₀+5 P₀ V₀ =15.4 %