JEE Physics 2025 Question with Solution

Given: The process A to B is adiabatic and B to C is isothermal. We need the net heat absorbed per mole.

Find: Net heat absorbed by the system per mole.

For the total process,

$$\Delta Q = \Delta Q_{\text{adiabatic}} + \Delta Q_{\text{isothermal}}$$

For the adiabatic path A to B,

$$\Delta Q_{\text{adiabatic}} = 0$$

For the isothermal path B to C, heat absorbed equals the work done:

$$\Delta Q_{\text{isothermal}} = W_{\text{isothermal}} = nRT \ln \left( \frac{V_C}{V_B} \right)$$

From the given solution working,

$$\Delta Q_{\text{isothermal}} = 450R \ln \left( \frac{V_C}{V_B} \right) = 450R(\ln 3)$$

Hence,

$$\Delta Q = 0 + 450R(\ln 3) = 450R(\ln 3)$$

Therefore, the net heat absorbed per mole is $450R(\ln 3)$ and the correct option is B.

Given: One segment is adiabatic and the other is isothermal.

Find: Which segment contributes to heat absorption.

1. In an adiabatic process, no heat is exchanged with the surroundings.
2. Therefore, only the isothermal segment B to C contributes to the net heat absorbed.
3. For an isothermal process of an ideal gas,

$$Q = nRT \ln \left( \frac{V_f}{V_i} \right)$$

4. Using the extracted solution result,

$$Q_{BC} = 450R(\ln 3)$$

So the total heat absorbed over the complete path A \to B \to C is $450R(\ln 3)$. Hence, the correct option is B.