Thermodynamics

JEE Main level, single correct, medium

135 of 190

Question 135 of 190

Match the thermodynamics processes taking place in a system with the correct conditions. In the table : ∆Q∆Qis the heat supplied, ∆W∆W is the work done and ∆U∆U is change in internal energy of the system. Process Condition (I) Adiabatic (A) ∆W=0∆W=0 (II) Isothermal (B) ∆Q=0∆Q=0 (III) Isochoric (C) ∆U≠0,∆W≠0,∆Q≠0∆U\neq 0,∆W\neq 0,∆Q\neq 0 (IV) Isobaric (D) ∆U=0∆U=0

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Answer: (D) (I) - (B), (II) - (D), (III) - (A), (IV) - (C)

In Adiabatic ΔQ=0\Delta Q=0,i.e.,,i.e., Exchange of heat does not take place.
In Isothermal ΔU=0\Delta U=0,i.e.,,i.e., Temperature does not change therefore internal energy also remains constant.
In Isochoric ΔW=0\Delta W=0,i.e.,,i.e., Volume remains constant therefore work done is zero.
In Isobaric ∆U≠0,∆W≠0,∆Q≠0∆U\neq 0,∆W\neq 0,∆Q\neq 0
W=P.∆VW=P.∆V, Volume changes so work done is not equal to zero.
∆U=f2P∆V∆U=\dfrac{f}{2}P∆V, Volume changes therefore internal energy changes.
∆Q=nCp∆T∆Q=n{C}_{p}∆T, Temperature changes therefore heat also changes.

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