JEE Physics 2023 Question with Solution

Given: Match thermodynamic processes in List I with the correct statements in List II.

Find: The correct correspondence between A, B, C, D and I, II, III, IV.

For an ideal gas,

$$\Delta U = nC_v \Delta T$$

So internal energy depends only on temperature change.

For isothermal process, temperature remains constant, so

$$\Delta T = 0$$

Hence,

$$\Delta U = 0$$

Therefore, A (\to) II.

For adiabatic process,

$$\Delta Q = 0$$

Using the first law of thermodynamics,

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

So,

$$\Delta U = -\Delta W$$

If work is done by the gas, $\Delta W > 0$, hence $\Delta U < 0$. Therefore, work done by the gas decreases internal energy. So B (\to) I.

For isochoric process, volume remains constant, hence no work is done:

$$\Delta W = 0$$

Therefore, C (\to) IV.

For isobaric process,

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

Also,

$$\Delta U = nC_v \Delta T \neq 0$$

So the heat absorbed is used partly to increase internal energy and partly to do work. Therefore, D (\to) III.

Thus the correct matching is A-II, B-I, C-IV, D-III.

The correct option is B.

Given: Four thermodynamic processes and four statements.

Find: The correct one-to-one matching.

1. Isothermal process: temperature is constant, so internal energy of an ideal gas does not change.
2. Adiabatic process: no heat exchange occurs. If the gas does work, that energy comes from its internal energy, so internal energy decreases.
3. Isochoric process: volume is constant, therefore

$$\Delta W = 0$$

4. Isobaric process: pressure is constant, and generally both temperature and volume change. Hence supplied heat is split between increasing internal energy and doing work.

This gives:

• A-II
• B-I
• C-IV
• D-III

So the correct option is B.