The efficiency of an ideal gas with adiabatic exponent `'gamma'` for the shown cyclic process would be
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))`
Text Solution
Verified by Experts
The correct Answer is:
A
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)))`
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