Q.

Match the thermodynamic processes taking place in a system with the correct conditions. In the table: ΔQ is the heat supplied, ΔW is the work done and ΔU is change in internal energy of the system.

  Process   Condition
I) Adiabatic A) ΔW=0
II) Isothermal B) ΔQ=0
III) Isochoric C) ΔU≠0, ΔW≠0, ΔQ≠0
IV) Isobaric D) ΔU=0

 

1 (I)-(A), (II)-(B), (III)-(D), (IV)-(D)  
2 (I)-(B), (II)-(A), (III)-(D), (IV)-(C)  
3 (I)-(A), (II)-(A), (III)-(B), (IV)-(C)  
4 (I)-(B), (II)-(D), (III)-(A), (IV)-(C)  

Ans.

(4)

I) Adiabatic process : No exchange of heat takes place with surroundings. ⇒ΔQ=0

II) Isothermal process : Temperature remains constant

∴ ΔT=0⇒ΔU=f2nRΔT⇒ΔU=0

No change in internal energy [ΔU=0].

III) Isochoric process volume remains constant

ΔV=0⇒W=∫P dV=0

Hence work done is zero.

IV) In isobaric process pressure remains constant

ΔU=f2nRΔT=f2[PΔV]≠0

∴ ΔQ=nCpΔT≠0