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)`.

A carnot cycle is performed by `1 mol e of air (gamma=1.4)` initially

1.	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)`.
Answer» 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%`

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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)`.

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