This question requires matching different thermodynamic processes with their corresponding expressions derived from the First Law of Thermodynamics.

First Law of Thermodynamics Basis

The First Law of Thermodynamics states that the change in internal energy ($\Delta U$) of a system is equal to the heat supplied to the system ($\Delta Q$) minus the work done by the system ($\Delta W$). Mathematically, it is expressed as:

$ \Delta Q = \Delta U + \Delta W $

Where:

• $ \Delta Q $ = Heat supplied
• $ \Delta U $ = Change in internal energy
• $ \Delta W $ = Work done by the system

For different thermodynamic processes, the conditions change, leading to specific forms of this law.

Isobaric Process Match (A)

An Isobaric process occurs at constant pressure ($P = \text{constant}$). The work done by the system is given by $ \Delta W = P\Delta V $, where $ \Delta V $ is the change in volume.

Substituting this into the First Law:

$ \Delta Q = \Delta U + P\Delta V $

This matches expression (IV) in List-II. So, (A)-(IV).

Isochoric Process Match (B)

An Isochoric process occurs at constant volume ($V = \text{constant}$). This means the change in volume is zero ($ \Delta V = 0 $).

Consequently, the work done by the system is zero:

$ \Delta W = P\Delta V = P \times 0 = 0 $

Substituting this into the First Law:

$ \Delta Q = \Delta U + 0 $

$ \Delta Q = \Delta U $

This matches expression (II) in List-II. So, (B)-(II).

Adiabatic Process Match (C)

An Adiabatic process is defined by the condition that no heat is exchanged between the system and its surroundings. Therefore, the heat supplied is zero.

$ \Delta Q = \text{zero} $

This matches expression (III) in List-II. So, (C)-(III).

Isothermal Process Match (D)

An Isothermal process occurs at constant temperature ($T = \text{constant}$). For an ideal gas, the internal energy ($U$) depends only on temperature. Thus, if the temperature is constant, the change in internal energy is zero.

$ \Delta U = 0 $

Substituting this into the First Law:

$ \Delta Q = 0 + \Delta W $

$ \Delta Q = \Delta W $

This matches expression (I) in List-II. So, (D)-(I).

Correct Matching Summary

Based on the analysis:

• (A) Isobaric matches with (IV) $ \Delta Q = \Delta U + P\Delta V $
• (B) Isochoric matches with (II) $ \Delta Q = \Delta U $
• (C) Adiabatic matches with (III) $ \Delta Q = \text{zero} $
• (D) Isothermal matches with (I) $ \Delta Q = \Delta W $

Therefore, the correct combination is (A)-(IV), (B)-(II), (C)-(III), (D)-(I).