The efficiency of the cycle shown below in the figure (consisting of one isobar, one adiabat and one isotherm) is 50 % . The ratio, x , between the highest and lowest temperatures attained in this cycle obeys (the working substance is a ideal gas):-

Step-by-step solution

C P C P - R = γ C P = γ C P - γ R γ R ⇒ ( γ - 1 ) C p γ R γ - 1 = C P η =  Work done   Heat sup plied  η = n C P Δ T - n R T ℓ n P 1 P 2 n C p Δ T = 1 2 n C p Δ T = 2 n R T ℓ n P 1 P 2 P 1 P 2 1 - γ = T 1   T 2 γ γ R γ - 1   T 2 - T 1 = 2 R T 1 ℓ n T 1   T 2 γ 1 - γ T 2 - T 1 = 2   T 1 ln T 2   T 1 x - 1 = ln x 2 x ²= e ^ x -1