A mono-atomic ideal gas of two moles is taken through a cyclic process starting from `A` as shwon in the figure below. The volume ratios are `V_(B)//V_(A) = 2` and `V_(D)//V_(A) = 4`. If the temperature `T_(A)` at `A` is `27^(@)C`. Calculate a. The temperature of gas at `B`. b. Heat absorbed or evolved in each process. c. Total work done in cyclic process.

A mono-atomic ideal gas of two moles is taken through a cyclic process

1.	A mono-atomic ideal gas of two moles is taken through a cyclic process starting from `A` as shwon in the figure below. The volume ratios are `V_(B)//V_(A) = 2` and `V_(D)//V_(A) = 4`. If the temperature `T_(A)` at `A` is `27^(@)C`. Calculate a. The temperature of gas at `B`. b. Heat absorbed or evolved in each process. c. Total work done in cyclic process.
Answer» `AtoB` (It is isobaric process) `(V_(A))/(T_(A))=(V_(B))/(T_(B))` `thereforeT_(B)=(V_(B))/(V_(A))xxT_(A)=2xx300=600K` `q_(AB)=nC_(P)DeltaT=2xx(5)/(2)RDeltaT` `=2xx(5)/(2)xx2xx300=3000cal` `BtoC`, (Isothermal process) `DeltaU=0` `thereforeq_(BC)=W=2.303nRT" log "((V_(C))/(V_(B)))` `=2.303xx2xx2xx600log((4)/(2))` `=1.663xx10^(3)cal` C`to`D (Isobaric process) `q_(CD)=nC_(V)DeltaT=2xx(3)/(2)xx2(-300)=-1800cal` `DtoA` (Isothermal process) ltbr `q_(DA)=2.303nRT_(A)" log "(V_(A))/(V_D))` `=2.303xx2xx2xx300" log "(1)/(4)` `=-1 .663xx10^(3)cal` ,brgt Total heat change `=3000+1.663xx10^(3)-1800-1.663xx10^(3)` `=1200cal` work done `=-1200cal`

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