A carnot cycle is performed by `1 mol e of air (gamma=1.4)` initially at `327^(@)C`. Each stage represents a compression or expansion in the ratio `1:6`. Calcultate (a) the lowest temperature (b) net work done during each cycle and (C ) efficiency of the engine.
Take `R=8.31J mol e^(-1)K^(-1)`.

Explanation

Here, `(V_(1))/(V_(2))= 1/6 , gamma= 1.4`,
`T_(1)= 327^(@)C= 327+273= 600K`
(a) For an adiabatic change,
`T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)`
`T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)= 600(1/6)^(1.4-1)`
`=600xx0.4884= 293K= 293-273`
`=20^(@)C`
This is lowest temperature.
(b) Net work done during each cycle by 1 mole of air.
`W=RT_(1) "log"_(e) (V_(2))/(V_(1))- RT_(2) "log"_(e)(V_(3))/(V_(4))`
`=R(T_(1)-T_(2)) "log"_(e) (V_(2))/(V_(1))( :. (V_(3))/(V_(4))=(V_(2))/(V_(1)))`
`= 2.303R (T_(1)-T_(2))"log"_(10) (V_(2))/(V_(1))`
`=2.303xx8.31(600-293)log_(10)6`
`W= 2.303xx8.31xx307xx0.7782`
`=4572.2J`
(C ) `eta= 1 -(T_(2))/(T_(1))= 1 - (293)/(600)= (307)/600`
`=0.512= 51.2%`