Home
Class 12
CHEMISTRY
{:("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

AI Generated Solution

The correct Answer is:
To solve the matching question, we need to analyze the statements in Column 1 and find the corresponding statements in Column 2 based on thermodynamic principles. ### Step-by-Step Solution: 1. **Analyze Statement A**: - **Statement**: ΔG(system) at constant temperature and pressure = 0. - **Interpretation**: This condition indicates that the system is in equilibrium. At equilibrium, no net change occurs, and the system is not able to do useful work. - **Match**: This corresponds to (p) "Process is in equilibrium". 2. **Analyze Statement B**: - **Statement**: ΔS(system) + ΔS(surrounding) > 0. - **Interpretation**: A positive change in the total entropy (system + surroundings) indicates that the process is spontaneous. A spontaneous process can do useful work. - **Match**: This corresponds to (r) "Process is spontaneous". 3. **Analyze Statement C**: - **Statement**: ΔS(system) + ΔS(surrounding) < 0. - **Interpretation**: A negative change in the total entropy indicates that the process is non-spontaneous. A non-spontaneous process cannot do useful work. - **Match**: This corresponds to (q) "Process is nonspontaneous". 4. **Analyze Statement D**: - **Statement**: ΔG(system) at constant temperature and pressure > 0. - **Interpretation**: A positive Gibbs free energy change indicates that the process is non-spontaneous. A non-spontaneous process cannot do useful work. - **Match**: This corresponds to (s) "System is unable to do useful work". ### Final Matches: - A → P ("Process is in equilibrium") - B → R ("Process is spontaneous") - C → Q ("Process is nonspontaneous") - D → S ("System is unable to do useful work") ### Summary of Matches: - A (ΔG = 0) → P - B (ΔS(system) + ΔS(surrounding) > 0) → R - C (ΔS(system) + ΔS(surrounding) < 0) → Q - D (ΔG > 0) → S

To solve the matching question, we need to analyze the statements in Column 1 and find the corresponding statements in Column 2 based on thermodynamic principles. ### Step-by-Step Solution: 1. **Analyze Statement A**: - **Statement**: ΔG(system) at constant temperature and pressure = 0. - **Interpretation**: This condition indicates that the system is in equilibrium. At equilibrium, no net change occurs, and the system is not able to do useful work. - **Match**: This corresponds to (p) "Process is in equilibrium". ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • THERMODYNAMICS

    RESONANCE ENGLISH|Exercise exercise-2 Part:(I) :Only one option correct|17 Videos
  • THERMODYNAMICS

    RESONANCE ENGLISH|Exercise exercise-2 Part:(II): Single and Double value integer type|11 Videos
  • THERMODYNAMICS

    RESONANCE ENGLISH|Exercise exercise-3 Part-3 :(Only one option correct type) section C|11 Videos
  • TEST SERIES

    RESONANCE ENGLISH|Exercise CHEMISTRY|50 Videos

Similar Questions

Explore conceptually related problems

A process is non-spontaneous when DeltaS_("total") lt 0 .

{:(,"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

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

    A
    `DeltaS_("total")=DeltaS_("system")+DeltaS_("surr")gt0`
    B
    `DeltaS_("total")=DeltaS_("system")+DeltaS_("surr")lt0`
    C
    `DeltaS_("system")=DeltaS_("total")+DeltaS_("surr") gt 0`
    D
    `DeltaS_("surr")=DeltaS_("total")+DeltaS_("system")lt0`
  • Similar Questions

    Explore conceptually related problems

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

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

    Comment on the spontaneity of a process when DeltaH gt 0, T DeltaS lt 0

    For the given process :- {:("Column-I","Column-II"),((A)Process " "JtoK,(P) Wgt0),((B)Process" "KtoL ,(q) Wlt0),((C)Process " "LtoM,(r)Qgt0),((D)Process" "MtoJ,(s)Qlt0):}

    Comment on the spontaneity of a process when Delta H gt 0, T Delta S lt 0

    {:("Column-I","Column-II"),((A)"Acetone +"CHCl_3, (p)DeltaS_("mix") gt0),((B)"Ethanol + Water ",(q)DeltaV_(mix)gt0),((C )C_2H_5Br+C_2H_5I,(r)DeltaH_(mix) lt 0),((D)"Aceton + Benzene",(s)"Maximum boiling azeotropes"),(,(t)"Minimum boiling azeotropes"):}

    Comment on the spontaneity of a process when DeltaH lt 0, T Delta S gt 0