Six moles of an ideal gas performs a cycle shown in figure. If the temperature are `T_(A) = 600 K`, `T_(B) = 800 K`, `T_(C) = 2200 K` and `T_(D) = 1200 K`, the work done per cycle is

Options

A. `20 kJ`
B. `30 kJ`
C. `40 kJ`
D. `60 kJ`

Answer

C

Explanation

Prosesses `A` to `B` and `C` to `D` are parts of straigt line graph of the form `y=mx`
Also `P =(muR)/(V)T(mu=6)`
`implies P prop T` so volume remains constant for the graph `AB` and `CD`

So no work is done during process for `A` to `B` and `C`to `D` ie., `W_(AB) = W_(CD) = 0` and `W_(BC) = P_(2) (V_(C)-V_(B))=muR`
`(T_(C)-T_(B))`
`=6R(2200-800)= 6Rxx1400J`
Also `W_(DA) = P_(1)(V_(A)-V_(D)) = muR(T_(A)-T_(B))`
`6R(600-1200)=-6Rxx600J`
Hence work done in complete cycle
`W = W_(AB) +W_(BC) +W_(CD)+W_(DA)`
`=0+6Rxx1400+0-6Rxx600`
`=6Rxx900=6xx8.3xx800~~40KJ`.