The efficiency of an ideal gas with adiabatic exponent `'gamma'` for the shown cyclic process would be

Options

A. `((gamma-1)("2 ln 2"-1))/(1+(gamma-1)"2 ln 2")`
B. `((gamma-1)(1-"2 ln 2"))/((gamma-1)"2 ln 2"-1)`
C. `(("2 ln 2"+1)(gamma-1))/((gamma-1)"2 ln 2"+1)`
D. `(("2 ln 2"-1))/(gamma//(gamma-1))`

Answer

A

Explanation

AB process
`0=q_(1)+w_(1)`
`=q_(1)-nR(2T_(0))ln2`
BC process `DeltaU_(2)=q_(2)+w_(2)`
`(nR)/((gamma-1))(T_(0)-2T_(0))=q_(2)-((nRT_(0))/(V_(0))).(V_(0)-2V_(0))`
CA process
`DeltaU_(3)=q_(3)+w_(3)`
`(nR)/((gamma-1))(2T_(0)-T_(0))=q_(3)+0`
`" Efficiency "= ("Total work done")/("Total heat absorbed")=(w_(1)+w_(2))/(q_(1)+q_(3))`
`=((-2RT_(0)ln2)+(nRT_(0)))/((2RT_(0)ln2)+((nRT_(0))/(gamma-1)))`