JIPMER2018PhysicsThermodynamics
The efficiency of an ideal gas with adiabatic exponent for the shown cyclic process would be
Options
- A(2 2-1) /( -1)
- B(1-1 2) /( -1)
- C(2 2+1) /( -1)
- D(2 2-1) /( +1)
Correct answer
A. (2 2-1) /( -1)
Step-by-step solution
As , W _ BC = p V = nR T =- nR (2 ~T ₀- T ₀ )= nRT ₀ aligned & and W _ CA =+2 nRT ₀ 2 & Work done = W _ CA + W _ BC & =2 nRT ₀ 2- nRT ₀ & = nRT ₀(2 2-1) & aligned Also, input heat, Q _ BC = nC _ p T = nR T ₀ -1 Efficiency = Work Input heat = (2 2-1) /( -1)