Figure below shows different processes for given amount of ideal gas

`{:(,"Column I",,"Column II",),((A),"In figure (i)",(P),DeltaQ gt 0,),((B),"In figure (ii)",(Q),Delta W lt 0,),((C),"In figure (iii)",(R),Delta Q lt 0,),((D),"In figure (iv)",(S),Delta W gt 0,),(,,(T),dU = 0,):}`

Answer

(A) = PST, (B) = S, (C) = PS, (D) = QRT

Explanation

`Delta Q = Delta U + Delta W`
`Delta U = Delta C_(v) Delta T`
`Delta T = 0 rArr Delta U = 0` (option `- T)`.
Volume increase then work done will be positive (option s).
Volume increases then work done will be positive (options).

Expansion takes places then work done will be positive (ooptions s)
`Delta Q = n C_(P) Delta T = nC_(P) (T_(2) - T_(1))`
`Delta Q = nC_(P) ((PV_(2))/(nR) - (PV_(1))/(nR))`
`Delta Q = (nC_(P) P)/(nR) (V_(2) - V_(1))`
`rArr Delta Q = (PC_(P))/(R) (V_(2) - V_(1))`
Since `Delta V` is + ve hence `Delta Q` is positive For a cyclic process `phi U = 0` (option - T) If cycle is anticlock the work done will be negative i.e. `Delta W gt 0` (option - Q)
`Delta Q = Delta U + Delta W`
`Delta Q = Delta W`
Hence `Delta Q gt O` (option - R)