JEE Physics 2025 Question with Solution

Given: List-I contains the thermodynamic processes (A) Isothermal, (B) Adiabatic, (C) Isobaric, and (D) Isochoric.

Find: The correct matching with the characteristics in List-II and hence the correct option.

From the given solution:

• Isothermal process occurs at constant temperature, so for an ideal gas

$$\Delta T = 0 \Rightarrow \Delta U = 0$$

Hence, (A) \rightarrow (IV).

• Adiabatic process has no heat exchange with surroundings

$$\Delta Q = 0$$

Hence, (B) \rightarrow (II).

• Isobaric process occurs at constant pressure. In general, internal energy changes, so

$$\Delta U \neq 0$$

Hence, (C) \rightarrow (III).

• Isochoric process occurs at constant volume, therefore work done is zero

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

Hence, (D) \rightarrow (I).

So the complete matching is:

• (A)-(IV)
• (B)-(II)
• (C)-(III)
• (D)-(I)

This corresponds to option (3). Therefore, the correct option is C.

Given: A match-the-following question on basic thermodynamic processes.

Find: Which option gives the correct correspondence between each process and its defining characteristic.

Step 1: Use the defining condition of each process.

For an isothermal process, temperature remains constant:

$$\Delta T = 0$$

For an ideal gas, internal energy depends only on temperature. Therefore,

$$\Delta U = 0$$

So, (A) \rightarrow (IV).

For an adiabatic process, no heat is transferred:

$$\Delta Q = 0$$

So, (B) \rightarrow (II).

For an isobaric process, pressure remains constant:

$$\Delta P = 0$$

In such a process, internal energy can change, so the listed characteristic is

$$\Delta U \neq 0$$

So, (C) \rightarrow (III).

For an isochoric process, volume remains constant:

$$\Delta V = 0$$

Hence, work done is

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

So, (D) \rightarrow (I).

Step 2: Compare with the options.

The matching obtained is (A)-(IV), (B)-(II), (C)-(III), (D)-(I), which is exactly option (3).

Conclusion: The correct option is C.