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

Step-by-step 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 = f 2 P ∆ V , Volume changes therefore internal energy changes. ∆ Q = n C p ∆ T , Temperature changes therefore heat also changes.