A cyclic process consisting of two isobaric and two adiabatic processes is shown in the figure. If P_2 = nP_1 then prove that efficiency of thisprocess is eta=1-n^((1-1/gamma)) where gamma=C_P/C_V

A cyclic process consisting of two isobaric and two adiabatic processe

1.	A cyclic process consisting of two isobaric and two adiabatic processes is shown in the figure. If P_2 = nP_1 then prove that efficiency of thisprocess is eta=1-n^((1-1/gamma)) where gamma=C_P/C_V
Answer» Solution :For adiabatic process `TP^((1-gamma)/gamma)` = constant `therefore T_1P^((1-gamma)/gamma)=T_2P_2^((1-gamma)/gamma)` `therefore T_2/T_1=(P_1/P_2)^((1-gamma)/gamma)` But `P_2=nP_1 therefore P_1/P_2=1/n` `T_2/T_1=(1/n)^((1-gamma)/gamma)=(n)^((1-gamma)/gamma)=n^(1-1/gamma)` Now EFFICIENCY `eta=1-T_2/T_1` `therefore eta=1-(n)^(1-1/gamma)` (`because` PUTTING VALUE of `T_2/T_1`)