The efficiency of the cycle shown below in the figure ( consisting of one isobar , one adiabatic and one isotherm ( is 50 % the ratio ,x, between the highest and lowest temperature attained in this cycle obeys ( the working substance is an ideal gas )

Options

A. `x= e^(x-1)`
B. `x^2 = e^(x-1)`
C. `x=e^(x^2-1)`
D. ` x^2 = e^(x^2-1)`

Answer

B

Explanation

` (C_P)/( C_p -R)= gamma`
` C_p= gamma - gamma R `
` gamma implies ( gamma -1) C_p`
` (gamma R)/( gamma-1)=C_P`
` ( gamma R)/( gamma -1) =C_P`
` eta = (" work done")/("Heat supplied") `
`eta =(n C_P Delta T - nRT ln ((P_1)/(P_2)))/( nC_P DeltaT) =1/2`
` nC_p Delta T = 2 n RT l n (P_1)/(P_2)`
`((P_1)/(P_2))^(1-gamma)= ((T_1)/(T_2))^(gamma)`
` ( gamma R )/(gamma =1) (T_2 - T_1) = 2RT _1 l n ((T_1)/(T_2))^(gamma/(1-gamma)`
` T_2 - T_1 = 2T_1 l ((T_2)/(T_1) )`
` x -1 = l n (x^2)`
` x^2 = e^(x-1)`
option (B)