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{:("column"-1,"Column"-2),((A)(DeltaG("s...

`{:("column"-1,"Column"-2),((A)(DeltaG_("system"))_(T.P) = 0,(p) "Process is in equilibrium"),((B)DeltaS_("system")+DeltaS_("surrounding") gt 0, (q)"Process is nonspontaneous"),((C)DeltaS_("system") +DeltaS_("surroumding") lt 0 , (r) "Process is spontaneous"),((D)(DeltaG_("system"))_(T.P) gt 0,(s)"System is unable to do useful work"):}`

Text Solution

Verified by Experts

The correct Answer is:
`(A) rarr (p,s); (B) rarr (r); (C) rarr(q,s); (D)rarr (q,s)`
`(A)DeltaG_("sys") =0 implies` equilibrium and free energy is zero. So, no useful work.
`(B) DeltaS_("universe") gt 0 implies` Spontaneous process and able to do usefull work.
`(C) DeltaS_("universe") lt 0 implies` Nonspontaneous process and unable to do usefull work.
`(D) DeltaG_("sys") gt 0 implies` Nonspontaneous process and unable to do useful work.
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{:(,"Column-I",,"Column-II"),((a),(DeltaG_(system))_(T.P^(=0)),(p),"Process in equilibrium"),((b),DeltaS_(system)+DeltaS_("sorrrounding")gt0,(q),"Process is non-spontaneous"),((c),DeltaS_(system)+DeltaS_("Surrounding")lt0,(r),"Process is spontaneous"),((d),(DeltaG_(system))+_(T,P)gt0,(s),"System is unable to do useful work"):}

{:(,"Column-I",,"Column-II"),((a),"Reversible adiabatic compression",(p),"Process in equilibrium"),((b),"Reversible vaporisation of liquid",(q),DeltaS_(system)lt0),((c),2N(g)rarrN_(2)(g),(r),DeltaS_("surrounding")lt0),((d),MgCO_(3)(s)oversetDeltararrMg(s)+CO_(2)(g),(s),DeltaS_("sublimation")=0):}

Knowledge Check

  • In a reversible process, DeltaS_("system") + DeltaS_("surrounds") is:

    A
    `gt 0`
    B
    `lt 0`
    C
    `ge 0`
    D
    `=0`

  • DeltaS_("surroundings")=+959.1 JK^(-1) mol^(-1) DeltaS_("system")=-163.1 JK^(-1) mol^(-1) Then the process is

    A
    Spontaneous
    B
    Non spontaneous
    C
    At equilibrium
    D
    Cannot be predicted from the information

  • Statement-1: For a process to be spontaneous, Delta G as well as Delta S has to be less than zero. Statement-2 : For spontaneous change, Delta S_("total") gt 0

    A
    Statement-1 is true, Statement-2, is true, Statement -2 is a correct explanation for statement-1
    B
    Statement-1 is true, Statement-2, is true, Statement -2 is not a correct explanation for statement-1
    C
    Statement-1 is true, Statement-2 is false
    D
    Statement-1 is false, Statement-2 is true

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    Match the column-1 {:(,"Column I",,"Column II"),((A),DeltaS="n"C_(v)l"n"(T_(2))/(T_(1)),(P),DeltaS_("sys")" for an irreversible isochoric process"),((B),DeltaS=-(DeltaU_("sys"))/(T),(Q),|DeltaS|_("surr")" for a reversible isochroric process"),((C),DeltaS=nC_(v)l"n"(T_(2))/(T_(1))+nRl"n"(V_(2))/(V_(1)),(R),DeltaS_("surr")" for a reversible isochoric process"),(,,(S),DeltaS_("sys")" for a reversible process"):}

    Match the column-I with Column - II Note that column- I may have more than one matching options in column-II. {:(,"Column-I" , "Column-II",),(("A"),"Reversible adiabatic compression ",("P")DeltaS_("system") gt 0,),(("B"),"Reversible vaposiation",("Q")DeltaS_("system") lt0,),(("C"),"A diabatic free expansion of ideal gas in vacuum at constant temperature ",("R")DeltaS_("surrounding") gt 0,),(("D"),"Dissocaiation of " CaCo_(3) to CaO(s) + CO_(2)(g) "",("S")DeltaS_("surrounding")= 0,) :}

    Match the following: {:("Column-1", "Column-2"),((A) K_(P) gt Q_(P),(p)"Non-spontaneous"),((B) DeltaG^(@) lt Rtlog_(e)Q_(p),(q)"Equilibrium"),(( C)K_(p)=Q_(p),( r) "Spontaneous and endothermic"),( (D) T gt (DeltaH)/(DeltaS), (s) "Spontaneous"),(,(t)"Spontanous and exothermic"):}

    If Delta G gt 0, then the nature of the process will be (spontaneous/non-spontaneous)?

    The total entropy change (DeltaS_("total")) for the system and surrounding of a spontaneous process is given by

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