Helium gas goes through a cycle A B C D A (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

In case of a cyclic process, work done is equal to the area under the cycle and is taken to be positive if the cycle is clockwise. Work done by the gas W= Area of the rectangle A B C D=P₀ V₀ Helium gas is a monoatomic gas. C_V= 3 2 R and C_P= 5 2 R Along the path A B , heat supplied to the gas at constant volume, Q_ A B =n C_V T=n 3 2 R T= 3 2 V₀ P= 3 2 P₀ V₀ Along the path B C , heat supplied to the gas at constant pressure, Q_ B C =n C_P T=n 5 2 R T= 5 2 (2 P₀ ) V=5 P₀ V₀ Along the path C D and D A , heat is reje