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.

Explanation

`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`