Q67 / 75 · JEE Main 2020 (04 Sep Shift 2) · Physics

Match the thermodynamics processes taking place in a system with the correct conditions. In the table : ∆Q is the heat supplied, ∆W is the work done and ∆U is change in internal energy of the system.

  Process   Condition
(I) Adiabatic (A) ∆W=0
(II) Isothermal (B) ∆Q=0
(III) Isochoric (C) ∆U≠0,∆W≠0,∆Q≠0
(IV)

Isobaric

 (D) ∆U=0

  • A
    (I) - (A), (II) - (B), (III) - (D), (IV) - (D)
  • B
    (I) - (B), (II) - (A), (III) - (D), (IV) - (C)
  • C
    (I) - (A), (II) - (A), (III) - (B), (IV) - (C)
  • Correct answer
    D
    (I) - (B), (II) - (D), (III) - (A), (IV) - (C)

Solution

In Adiabatic ΔQ=0,i.e., Exchange of heat does not take place.

In Isothermal ΔU=0,i.e., Temperature does not change therefore internal energy also remains constant.

In Isochoric ΔW=0,i.e., Volume remains constant therefore work done is zero.

In Isobaric ∆U≠0, ∆W≠0, ∆Q≠0
W=P.∆V, Volume changes so work done is not equal to zero.
∆U=f2P∆V, Volume changes therefore internal energy changes.
 ∆Q=nCp∆T, Temperature changes therefore heat also changes.

 

 

Hence, the correct answer is option: 'D'.

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