In column-II, some situations are given, and in column-I, their results are given. Mathc the proper entries from column-2 to column-1 using the codes iven below the columns

Options

A. `{:(p,q,r,s),(3,1,4,2):}`
B. `{:(p,q,r,s),(2,1,3,4):}`
C. `{:(p,q,r,s),(3,2,1,4):}`
D. `{:(p,q,r,s),(2,4,3,1):}`

Explanation

(A) When `0^(@)C` water volume decreases slightly, so `W_("system")=-ve`. To melt the ice, some heat has to be given `(Q=mL_(f))` which is almost equal to increase in internal energy
(B)
Since, `P-v` cycle is clockwise, so `W_("net")=+ve`
and `(DeltaU)_("cycle")=0`
(C) By the fan, some work is done on the room air. Done to this, temperature of the gas increaes slightly so internal energy will increase slightly Matematically.
`Q=W+Deltau`
`Q=-ve+DeltaurArr DeltaU=+ve`.
(D) `P-V` diagram for the process is

From the diagram
`W_(ArarrBrarrC)=-ve`
`(PV)_(c)lt(PV)_(A)rArrT_(C)ltT_(A)`
So, internal energy decrease.
`(E) dQ=-2dU`
`dQ=2h(5)/(2)RdT`
`C=(dQ//dT)/(n)=5R` .....(i)
`C=C_(V)+(R)/(1-x)=(5)/(2)R+(R)/(1-x)` ....(ii)
From equ. (i) `&` (ii).
get `x=(3)/(5)`
So, process equ. is
`PV^(3//5)=` const.
if `PdownarrowrArrVuparrow rArrW=+ve`
To find relation between `T` and `V`, put `P=(nRT)/(V)`
`((nRT)/(V))(V^(3//5))=` constant
`VuparrowrArrTuparrow rArr` internal energy will increase.