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 :

Step-by-step solution

(A) Change in internal energy is expressed as ∆ U = n C v ∆ T , here, ∆ T is change in temperature. Since, in an isothermal process temperature remains constant, thus, ∆ U = 0 A → I I (B) In an adiabatic process, heat transfer, Q = 0 . So, from first law of thermodynamics, Q = ∆ U + W ∆ U = - W Since, work done by gas is positive, thus, ∆ U is negative B → I (C) In an isochoric process, volume remains constant, so work done by or on the gas W = P ∆ V = 0 C → I V (D) In an isobaric process, pressure remains constant