`3` moles of an ideal mono atomic gas performs a cycle as shown in fig. `If` gas temperature `T_(A)=400 K` `T_(B)=800K,T_(C)=2400K`, and `T_(D)=1200K`. Then total work done by gas is

Explanation

Processes A to B and C to D are parts os straightline graphs of forms y=mx
and `P=(nR)/(V)T (n=3) i.e. PaT`
So, volume remains constant for the graphs AB and CD.
So, no work is done during processes for A to B and C to D.
`W_(AB)=W_(CD)=0` and `W_(BC)=P_(2)(V_(C)-V_(B))`
`=nR(T_(C)-T_(B))=3R(2400-800)=4800R`
`W_(DA)=P_(1)(V_(A)-V_(D))=nR(T_(A)T_(D))`
`=3R(400-1200)=-2400R`
Work done in the complete cycle
`W=W_(AB)+W_(BC)+W_(CD)+W_(DA)`
`=0+4800R+0+(-2400)R`
`=2400R=19953.6 Japprox20KJ`