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 wrok done in cyclic process.

Explanation

`A rarrB:` (It is isobaric procee)
`(V_(A))/(T_(A)) - (V_(B))/(T_(B))`
`:. T_(B) = (V_(B))/(V_(A)) xx T_(A) = 2 xx 300 = 600K`
`q_(AB) = nC_(P)Delta_(T) = 2xx (5)/(2)R DeltaT`
`= 2 xx (5)/(2) xx2xx300 = 3000cal`
`B rarrC`: (Isothermal process)
`DeltaU = 0`
`:. q_(BC) = w = 2.303nRT log_(10) ((V_(C))/(V_(B)))`
`= 2.303 xx2xx2xx600 log.(4)/(2)`
`C rarrD:` (Isochric process)
`q_(CD) = nC_(v) DeltaT = 2xx (3)/(2) xx 2 (-300) =- 1800cal`
`D rarA`: (Isochoric procee)
`q_(DA) = 2.303 nRT_(A) log_(10). (V_(A))/(V_(D))`
`= 2.303 xx 2xx 2xx200 log.(1)/(4)`
Total heat change
`= 3000 + (1.663 xx 10^(3)) - 1800 - (1.663 xx 10^(3)) = 1200 cal`
Work done `=- 1200 cal`