Some gas (C_(p)//C_(V)=gamma=1.25) follows the cycle ABCDA as shown in the figure. The ratio of the energy given out bythe gas to its surrounding durning the isochoric section of the cycle to the expansion work done during the isobaric section of the cycle is

Some gas (C_(p)//C_(V)=gamma=1.25) follows the cycle ABCDA as shown in

1.	Some gas (C_(p)//C_(V)=gamma=1.25) follows the cycle ABCDA as shown in the figure. The ratio of the energy given out bythe gas to its surrounding durning the isochoric section of the cycle to the expansion work done during the isobaric section of the cycle is
Answer» 2 0.04 6 0.08 Solution :Let n MOLES of gas follows the cycle ABCDA `Q_(isochoric)=nC_(V)DeltaT=(nR)/(gamma-1)(T_(B)-T_(C))` `W_(ISOBARIC)=nRDeltaT=nR(T_(A)-T_(D))=nR(T_(B-T_(C)))` `therefore` Required RATIO =`(Q_(isochoric))/(W_(isobaric))=(1)/(gamma-1)=4`