Thermodynamics

JEE Main level, single correct, medium

106 of 190

Question 106 of 190

Match List I with List II : List I List II A Isothermal Process I Work done by the gas decreases internal energy B Adiabatic Process II No change in internal energy C Isochoric Process III The heat absorbed goes partly to increase internal energy and partly to do work D Isobaric Process IV No work is done on or by the gas Choose the correct answer from the options given below :

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

(A) Change in internal energy is expressed as∆U=nCv∆T∆U=n{C}_{v}∆T, here, ∆T∆T is change in temperature.
Since, in an isothermal process temperature remains constant, thus, ∆U=0∆U=0
A→IIA\rightarrow II
(B) In an adiabatic process, heat transfer, Q=0Q=0.
So, from first law of thermodynamics,
Q=∆U+WQ=∆U+W
∆U=−W∆U=-W
Since, work done by gas is positive, thus, ∆U∆U is negative
B→IB\rightarrow I
(C) In an isochoric process, volume remains constant, so work done by or on the gas W=P∆V=0W=P∆V=0
C→IVC\rightarrow IV
(D) In an isobaric process, pressure remains constant, so work doneW=P∆V≠0W=P∆V\neq 0 and change in internal energy ∆U=nCv∆T≠0∆U=n{C}_{v}∆T\neq 0
Thus, heat absorbed goes partly to increase internal energy and partly do work.
D→IIID\rightarrow III

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